Chapter 11 Area Related To Circles

Given:

Area of the circle = $154\ \text{cm}^2$

---

To Find:

The perimeter (circumference) of the circle.

---

Solution:

The formula for the area of a circle is given by:

Area $= \pi r^2$, where $r$ is the radius of the circle.

We are given the area is $154\ \text{cm}^2$. Let's use $\pi = \frac{22}{7}$.

Now, we solve for $r^2$ from equation (i):

$r^2 = 154 \times \frac{7}{22}$

$r^2 = \frac{\cancel{154}^{7} \times 7}{\cancel{22}_{1}}$

$r^2 = 7 \times 7$

$r^2 = 49$

Taking the square root of both sides:

$r = \sqrt{49}$

$r = 7\ \text{cm}$

The radius of the circle is $7\ \text{cm}$.

---

The formula for the perimeter (circumference) of a circle is given by:

Perimeter $= 2\pi r$

Substitute the value of $r = 7\ \text{cm}$ and $\pi = \frac{22}{7}$ into the formula:

Perimeter $= 2 \times \frac{22}{7} \times 7$

Perimeter $= 2 \times \frac{22}{\cancel{7}} \times \cancel{7}$

Perimeter $= 2 \times 22$

Perimeter $= 44\ \text{cm}$

---

The perimeter of the circle is $44\ \text{cm}$.

Comparing this result with the given options, we find that it matches option (C).

Therefore, the correct answer is (C) 44 cm.

Given:

Angle of the sector = $\theta$ degrees

Radius of the circle = $r$ units

---

To Find:

Area of the sector.

---

Solution:

The area of a full circle with radius $r$ is given by the formula $\pi r^2$.

A sector of a circle is a portion of the circle enclosed by two radii and an arc.

The angle of a full circle at the centre is $360^\circ$.

The area of a sector is proportional to the angle subtended at the centre.

If the angle is $\theta$ degrees, the sector represents a fraction $\frac{\theta}{360^\circ}$ of the total circle's area.

Therefore, the area of the sector is given by:

Area of sector $= \left(\text{Fraction of the circle}\right) \times \left(\text{Area of the full circle}\right)$

Area of sector $= \frac{\theta}{360^\circ} \times \pi r^2$

Area of sector $= \frac{\pi r^2 \theta}{360^\circ}$

---

Comparing this derived formula with the given options:

Option (A) is $\frac{\pi r^{2}\theta}{360^\circ}$. This matches our formula.

Option (B) is $\frac{\pi r^{2}\theta}{180^\circ}$. This is incorrect.

Option (C) is $\frac{2\pi r\theta}{360^\circ}$. This relates to the length of the arc of the sector (which is $\frac{\theta}{360^\circ} \times 2\pi r$), not the area.

Option (D) is $\frac{2\pi r\theta}{180^\circ}$. This is also incorrect.

The correct formula for the area of the sector with angle $\theta$ (in degrees) and radius $r$ is $\frac{\pi r^{2}\theta}{360^\circ}$.

Therefore, the correct answer is (A) $\frac{\pi r^{2}\theta}{360^\circ}$.

Given:

Radius of the first circle = $R_1$

Radius of the second circle = $R_2$

Radius of the third circle = $R$

The sum of the areas of the two circles with radii $R_1$ and $R_2$ is equal to the area of the circle with radius $R$.

---

To Find:

The relationship between $R_1$, $R_2$, and $R$.

---

Solution:

The area of a circle with radius $r$ is given by the formula $\text{Area} = \pi r^2$.

According to the given information:

Area of circle with radius $R_1$ + Area of circle with radius $R_2$ = Area of circle with radius $R$

Using the area formula, we can write this as:

$\pi R_1^2 + \pi R_2^2 = \pi R^2$

We can factor out $\pi$ from the left side of the equation:

$\pi (R_1^2 + R_2^2) = \pi R^2$

Now, we can divide both sides of the equation by $\pi$ (since $\pi \neq 0$):

$\frac{\pi (R_1^2 + R_2^2)}{\pi} = \frac{\pi R^2}{\pi}$

$R_1^2 + R_2^2 = R^2$

This equation represents the relationship between the radii $R_1$, $R_2$, and $R$.

---

Comparing this result with the given options:

Option (A) is $R_1 + R_2 = R$. This is incorrect.

Option (B) is $R_1^2 + R_2^2 = R^2$. This matches our derived relationship.

Option (C) is $R_1 + R_2 < R$. This is incorrect.

Option (D) is $R_1^2 + R_2^2 < R^2$. This is incorrect.

The relationship between the radii is $R_1^2 + R_2^2 = R^2$.

Therefore, the correct answer is (B) $R_1^2 + R_2^2 = R^2$.

Given:

Radius of the first circle = $R_1$

Radius of the second circle = $R_2$

Radius of the third circle = $R$

The sum of the circumferences of the two circles with radii $R_1$ and $R_2$ is equal to the circumference of the circle with radius $R$.

---

To Find:

The relationship between $R_1$, $R_2$, and $R$.

---

Solution:

The circumference of a circle with radius $r$ is given by the formula $\text{Circumference} = 2\pi r$.

According to the given information:

Circumference of circle with radius $R_1$ + Circumference of circle with radius $R_2$ = Circumference of circle with radius $R$

Using the circumference formula, we can write this as:

$2\pi R_1 + 2\pi R_2 = 2\pi R$

We can factor out $2\pi$ from the left side of the equation:

$2\pi (R_1 + R_2) = 2\pi R$

Now, we can divide both sides of the equation by $2\pi$ (since $2\pi \neq 0$):

$\frac{2\pi (R_1 + R_2)}{2\pi} = \frac{2\pi R}{2\pi}$

$R_1 + R_2 = R$

This equation represents the relationship between the radii $R_1$, $R_2$, and $R$.

---

Comparing this result with the given options:

Option (A) is $R_1 + R_2 = R$. This matches our derived relationship.

Option (B) is $R_1 + R_2 > R$. This is incorrect.

Option (C) is $R_1 + R_2 < R$. This is incorrect.

Option (D) is Nothing definite can be said about the relation among $R_1$, $R_2$ and $R$. This is incorrect as we found a definite relationship.

The relationship between the radii is $R_1 + R_2 = R$.

Therefore, the correct answer is (A) $R_1 + R_2 = R$.

Given:

The circumference of a circle is equal to the perimeter of a square.

---

To Find:

The relationship between the area of the circle and the area of the square.

---

Solution:

Let $r$ be the radius of the circle and $s$ be the side length of the square.

The circumference of the circle is given by the formula $C = 2\pi r$.

The perimeter of the square is given by the formula $P = 4s$.

According to the problem statement, the circumference and perimeter are equal:

$2\pi r = 4s$

We can express the side length of the square $s$ in terms of the radius of the circle $r$:

$s = \frac{2\pi r}{4} = \frac{\pi r}{2}$

---

Now, let's find the area of the circle and the area of the square in terms of the same variable (e.g., $r$).

The area of the circle is given by the formula $A_{\text{circle}} = \pi r^2$.

The area of the square is given by the formula $A_{\text{square}} = s^2$.

Substitute the expression for $s$ from the equality of circumference and perimeter into the area of the square formula:

$A_{\text{square}} = \left(\frac{\pi r}{2}\right)^2 = \frac{\pi^2 r^2}{4}$

---

Now we compare the area of the circle and the area of the square:

$A_{\text{circle}} = \pi r^2$

$A_{\text{square}} = \frac{\pi^2 r^2}{4}$

To compare these two values, we can compare the coefficients of $r^2$ (assuming $r > 0$, which must be true for a circle with a defined circumference).

We need to compare $\pi$ and $\frac{\pi^2}{4}$.

We know that $\pi \approx 3.14159$.

Let's compare $\pi$ with $\frac{\pi^2}{4}$. This is equivalent to comparing $1$ with $\frac{\pi}{4}$ after dividing both sides by $\pi$ (since $\pi > 0$).

Comparing $1$ and $\frac{\pi}{4}$:

$\frac{\pi}{4} \approx \frac{3.14159}{4} \approx 0.7854$

Since $1 > 0.7854$, we have $1 > \frac{\pi}{4}$.

Multiplying both sides by $4\pi$ (which is positive), we get:

$4\pi > \pi^2$

Now, divide both sides by 4:

$\pi > \frac{\pi^2}{4}$

Since $\pi > \frac{\pi^2}{4}$ and $r^2 > 0$, multiplying both sides by $r^2$ preserves the inequality:

$\pi r^2 > \frac{\pi^2 r^2}{4}$

Thus, $A_{\text{circle}} > A_{\text{square}}$.

---

Conclusion:

If the circumference of a circle and the perimeter of a square are equal, the area of the circle is greater than the area of the square.

Comparing this result with the given options, it matches option (B).

Therefore, the correct answer is (B) Area of the circle > Area of the square.

Given:

A semi-circle of radius $r$ units.

---

To Find:

The area of the largest triangle that can be inscribed in the semi-circle.

---

Solution:

Consider a semi-circle with center O and radius $r$. Let the diameter be AB.

The largest triangle inscribed in a semi-circle will have its base along the diameter of the semi-circle.

Let the base of the triangle be the diameter AB.

The length of the diameter AB is $2r$.

Let C be the third vertex of the triangle, which lies on the arc of the semi-circle.

The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.

In this case, the base is AB ($2r$). The height of the triangle is the perpendicular distance from the vertex C to the base AB.

To maximize the area of the triangle with a fixed base AB, the height must be as large as possible.

The maximum perpendicular distance from a point on the arc to the diameter occurs at the midpoint of the arc.

Let C be the midpoint of the arc. The distance from C to the diameter AB is equal to the radius of the semi-circle, $r$.

So, the maximum height of the triangle is $h = r$ when the base is the diameter AB ($2r$).

The base of the largest triangle is $b = 2r$ and the corresponding height is $h = r$.

Area of the triangle $= \frac{1}{2} \times \text{base} \times \text{height}$

Area $= \frac{1}{2} \times (2r) \times r$

Area $= \frac{1}{\cancel{2}} \times \cancel{2}r \times r$

Area $= r^2$ sq. units

---

The area of the largest triangle that can be inscribed in a semi-circle of radius $r$ units is $r^2$ sq. units.

Comparing this result with the given options:

Option (A) is $r^2$ sq. units. This matches our calculated area.

Option (B) is $\frac{1}{2} r^2$ sq. units. This would be the area if the height was $r/2$ or the base was $r$.

Option (C) is $2 r^2$ sq. units. This is twice the correct area.

Option (D) is $\sqrt{2} r^2$ sq. units. This is incorrect.

Therefore, the correct answer is (A) $r^2$ sq. units.

Given:

The perimeter of a circle is equal to the perimeter of a square.

---

To Find:

The ratio of the area of the circle to the area of the square.

---

Solution:

Let $r$ be the radius of the circle and $s$ be the side length of the square.

The perimeter of the circle (circumference) is given by $C = 2\pi r$.

The perimeter of the square is given by $P = 4s$.

According to the problem, the perimeters are equal:

$2\pi r = 4s$

We can express the side $s$ in terms of the radius $r$:

$s = \frac{2\pi r}{4} = \frac{\pi r}{2}$

---

The area of the circle is given by $A_{\text{circle}} = \pi r^2$.

The area of the square is given by $A_{\text{square}} = s^2$.

Substitute the expression for $s$ into the formula for the area of the square:

$A_{\text{square}} = \left(\frac{\pi r}{2}\right)^2 = \frac{\pi^2 r^2}{4}$

---

Now, we find the ratio of the area of the circle to the area of the square:

$\frac{A_{\text{circle}}}{A_{\text{square}}} = \frac{\pi r^2}{\frac{\pi^2 r^2}{4}}$

$\frac{A_{\text{circle}}}{A_{\text{square}}} = \pi r^2 \times \frac{4}{\pi^2 r^2}$

Assuming $r \neq 0$, we can cancel $r^2$ from the numerator and denominator:

$\frac{A_{\text{circle}}}{A_{\text{square}}} = \frac{4\pi}{\pi^2}$

Assuming $\pi \neq 0$, we can cancel $\pi$ from the numerator and denominator:

$\frac{A_{\text{circle}}}{A_{\text{square}}} = \frac{4}{\pi}$

---

Using the value $\pi = \frac{22}{7}$:

$\frac{A_{\text{circle}}}{A_{\text{square}}} = \frac{4}{\frac{22}{7}}$

$\frac{A_{\text{circle}}}{A_{\text{square}}} = 4 \times \frac{7}{22}$

$\frac{A_{\text{circle}}}{A_{\text{square}}} = \frac{\cancel{4}^{2} \times 7}{\cancel{22}_{11}}$

$\frac{A_{\text{circle}}}{A_{\text{square}}} = \frac{14}{11}$

---

The ratio of their areas is $14:11$.

Comparing this result with the given options, it matches option (B).

Therefore, the correct answer is (B) 14 : 11.

Given:

Diameter of the first circular park ($d_1$) = $16\ \text{m}$

Diameter of the second circular park ($d_2$) = $12\ \text{m}$

The area of a new single circular park is equal to the sum of the areas of the two given circular parks.

---

To Find:

The radius of the new circular park.

---

Solution:

Let the radius of the first circular park be $r_1$.

$r_1 = \frac{d_1}{2} = \frac{16}{2} = 8\ \text{m}$

Let the radius of the second circular park be $r_2$.

$r_2 = \frac{d_2}{2} = \frac{12}{2} = 6\ \text{m}$

---

The area of a circle with radius $r$ is given by $A = \pi r^2$.

Area of the first park ($A_1$) $= \pi r_1^2 = \pi (8)^2 = 64\pi\ \text{m}^2$

Area of the second park ($A_2$) $= \pi r_2^2 = \pi (6)^2 = 36\pi\ \text{m}^2$

---

Let the radius of the new circular park be $R$.

Area of the new park ($A_{\text{new}}$) $= \pi R^2$

According to the problem, the area of the new park is the sum of the areas of the two smaller parks:

$A_{\text{new}} = A_1 + A_2$

$\pi R^2 = 64\pi + 36\pi$

$\pi R^2 = (64 + 36)\pi$

Divide both sides of equation (i) by $\pi$ (since $\pi \neq 0$):

$\frac{\pi R^2}{\pi} = \frac{100\pi}{\pi}$

$R^2 = 100$

Taking the positive square root (since radius must be positive):

$R = \sqrt{100}$

$R = 10\ \text{m}$

---

The radius of the new park is $10\ \text{m}$.

Comparing this result with the given options, it matches option (A).

Therefore, the correct answer is (A) 10 m.

Given:

Side of the square = $6\ \text{cm}$

A circle is inscribed in the square.

---

To Find:

The area of the inscribed circle.

---

Solution:

When a circle is inscribed in a square, the diameter of the circle is equal to the side length of the square.

Let $s$ be the side length of the square and $d$ be the diameter of the inscribed circle.

Given $s = 6\ \text{cm}$.

So, the diameter of the circle is $d = 6\ \text{cm}$.

The radius of the circle $(r)$ is half of the diameter.

From equation (i):

$r = 3\ \text{cm}$

---

The area of a circle with radius $r$ is given by the formula $A = \pi r^2$.

Substitute the value of the radius $r = 3\ \text{cm}$ into the area formula:

From equation (ii):

Area $= \pi \times 9\ \text{cm}^2$

Area $= 9\pi\ \text{cm}^2$

---

The area of the circle that can be inscribed in a square of side 6 cm is $9\pi\ \text{cm}^2$.

Comparing this result with the given options, it matches option (D).

Therefore, the correct answer is (D) $9 \pi\ \text{cm}^2$.

Given:

Radius of the circle ($r$) = $8\ \text{cm}$

A square is inscribed in the circle.

---

To Find:

The area of the inscribed square.

---

Solution:

Let the circle have center O and radius $r$. Let the inscribed square be ABCD, with its vertices on the circle.

The diagonal of the inscribed square is equal to the diameter of the circle.

Diameter of the circle ($d$) $= 2 \times \text{radius}$

Let $s$ be the side length of the square.

In a square, the diagonal ($d_{\text{square}}$) is related to the side length by the formula $d_{\text{square}} = s\sqrt{2}$.

So, $s\sqrt{2} = d$

Using equation (i):

From equation (ii), we solve for $s$:

$s = \frac{16}{\sqrt{2}}$

To simplify, we rationalize the denominator:

$s = \frac{16}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{16\sqrt{2}}{2} = 8\sqrt{2}\ \text{cm}$

The side length of the inscribed square is $8\sqrt{2}\ \text{cm}$.

---

The area of a square with side length $s$ is given by $A_{\text{square}} = s^2$.

Substitute the value of $s = 8\sqrt{2}\ \text{cm}$ into the area formula:

From equation (iii):

Area $= 8^2 \times (\sqrt{2})^2\ \text{cm}^2$

Area $= 64 \times 2\ \text{cm}^2$

Area $= 128\ \text{cm}^2$

---

The area of the square that can be inscribed in a circle of radius 8 cm is $128\ \text{cm}^2$.

Comparing this result with the given options, it matches option (B).

Therefore, the correct answer is (B) $128\ \text{cm}^2$.

Given:

Diameter of the first circle ($d_1$) = $36\ \text{cm}$

Diameter of the second circle ($d_2$) = $20\ \text{cm}$

The circumference of a new circle is equal to the sum of the circumferences of the two given circles.

---

To Find:

The radius of the new circle ($R$).

---

Solution:

Let the radius of the first circle be $r_1$.

Let the radius of the second circle be $r_2$.

Let the radius of the new circle be $R$.

---

The circumference of a circle with radius $r$ is given by the formula $C = 2\pi r$.

Circumference of the first circle ($C_1$) $= 2\pi r_1 = 2\pi (18)$

Circumference of the second circle ($C_2$) $= 2\pi r_2 = 2\pi (10)$

Circumference of the new circle ($C_{\text{new}}$) $= 2\pi R$

---

According to the problem statement:

Substitute the circumference formulas into equation (iii):

$2\pi R = 2\pi (18) + 2\pi (10)$

Factor out $2\pi$ from the right side:

$2\pi R = 2\pi (18 + 10)$

$2\pi R = 2\pi (28)$

Divide both sides by $2\pi$ (since $2\pi \neq 0$):

From equation (iv):

$R = 28\ \text{cm}$

---

The radius of the new circle is $28\ \text{cm}$.

Comparing this result with the given options, it matches option (C).

Therefore, the correct answer is (C) 28 cm.

Given:

Radius of the first circle ($r_1$) = $24\ \text{cm}$

Radius of the second circle ($r_2$) = $7\ \text{cm}$

The area of a new circle is equal to the sum of the areas of the two given circles.

---

To Find:

The diameter of the new circle ($D$).

---

Solution:

The area of a circle with radius $r$ is given by the formula $A = \pi r^2$.

Area of the first circle ($A_1$) $= \pi r_1^2 = \pi (24\ \text{cm})^2 = 576\pi\ \text{cm}^2$

Area of the second circle ($A_2$) $= \pi r_2^2 = \pi (7\ \text{cm})^2 = 49\pi\ \text{cm}^2$

Let the radius of the new circle be $R$. Its area is $A_{\text{new}} = \pi R^2$.

According to the problem statement, the area of the new circle is the sum of the areas of the two smaller circles:

Substitute the areas into equation (i):

$\pi R^2 = 576\pi + 49\pi$

Combine the terms on the right side:

Divide both sides of equation (ii) by $\pi$ (since $\pi \neq 0$):

$R^2 = 625$

Taking the positive square root to find the radius (since radius must be a positive value):

From equation (iii):

$R = 25\ \text{cm}$

The radius of the new circle is $25\ \text{cm}$.

---

The diameter of a circle ($D$) is twice its radius ($R$).

Substitute the value of $R$ from equation (iii) into equation (iv):

$D = 2 \times 25\ \text{cm}$

$D = 50\ \text{cm}$

---

The diameter of the new circle is $50\ \text{cm}$.

Comparing this result with the given options, it matches option (D).

Therefore, the correct answer is (D) 50 cm.

Given:

The statement: Area of a segment of a circle = area of the corresponding sector – area of the corresponding triangle.

---

To Verify:

Whether the given statement is true and provide a reason.

---

Reasoning:

Consider a circle with center O and radius $r$. Let a chord AB divide the circle into a minor segment and a major segment.

The chord AB, along with the arc AB, forms a segment.

The radii OA and OB, along with the arc AB, form a sector (sector OAB).

The radii OA and OB, along with the chord AB, form a triangle (triangle OAB).

By observing the diagram (which is not provided, but can be mentally visualised), the area of the sector OAB is composed of two parts:

1. The area of the triangle OAB.

2. The area of the segment formed by the chord AB and the arc AB.

Thus, we can write the relationship between the areas as:

To find the area of the segment, we can rearrange equation (i):

This derived relationship matches the statement given in the question.

Therefore, the statement is consistent with the geometric definition of a segment and a sector.

---

Conclusion:

Yes, the statement is true.

The reason is that a sector is formed by the union of the corresponding triangle and the corresponding segment. Hence, the area of the segment is obtained by subtracting the area of the triangle from the area of the sector.

Given:

Side of the square = $5\ \text{cm}$.

An inner circle is inscribed in the square.

An outer circle is circumscribing the square.

Statement to verify: Area of the outer circle is two times the area of the inner circle.

---

To Verify:

Whether the statement "Area of the outer circle = 2 $\times$ Area of the inner circle" is true.

---

Reasoning:

Let the side of the square be $s = 5\ \text{cm}$.

For the inner circle (inscribed in the square):

The diameter of the inscribed circle is equal to the side of the square.

Radius of inner circle ($r_{\text{inner}}$) $= \frac{\text{Diameter}}{2} = \frac{5}{2}\ \text{cm}$.

Area of inner circle ($A_{\text{inner}}$) $= \pi (r_{\text{inner}})^2 = \pi \left(\frac{5}{2}\right)^2 = \pi \frac{25}{4}\ \text{cm}^2$.

---

For the outer circle (circumscribing the square):

The diameter of the circumscribing circle is equal to the diagonal of the square.

The diagonal of a square with side $s$ is $s\sqrt{2}$.

Radius of outer circle ($r_{\text{outer}}$) $= \frac{\text{Diameter}}{2} = \frac{5\sqrt{2}}{2}\ \text{cm}$.

Area of outer circle ($A_{\text{outer}}$) $= \pi (r_{\text{outer}})^2 = \pi \left(\frac{5\sqrt{2}}{2}\right)^2 = \pi \frac{(5\sqrt{2})^2}{2^2} = \pi \frac{25 \times 2}{4} = \pi \frac{50}{4} = \pi \frac{25}{2}\ \text{cm}^2$.

---

Now, let's compare the area of the outer circle with two times the area of the inner circle:

$2 \times A_{\text{inner}} = 2 \times \left(\pi \frac{25}{4}\right) = \pi \frac{50}{4} = \pi \frac{25}{2}\ \text{cm}^2$.

We found that $A_{\text{outer}} = \pi \frac{25}{2}\ \text{cm}^2$ and $2 \times A_{\text{inner}} = \pi \frac{25}{2}\ \text{cm}^2$.

Thus, $A_{\text{outer}} = 2 \times A_{\text{inner}}$.

---

Conclusion:

Yes, the statement is true.

The reason is that the area of the outer circle is $\pi \frac{25}{2}\ \text{cm}^2$ and the area of the inner circle is $\pi \frac{25}{4}\ \text{cm}^2$, and $\pi \frac{25}{2} = 2 \times \pi \frac{25}{4}$.

Given:

Side of the square = $a\ \text{cm}$.

A circle is inscribed in the square.

Statement to verify: The area of the inscribed circle is $\pi a^2\ \text{cm}^2$.

---

To Verify:

Whether the statement "Area of inscribed circle = $\pi a^2\ \text{cm}^2$" is true.

---

Reasoning:

When a circle is inscribed in a square, the diameter of the circle is equal to the side length of the square.

Let $s$ be the side length of the square and $d$ be the diameter of the inscribed circle.

Given $s = a\ \text{cm}$.

So, the diameter of the inscribed circle is $d = a\ \text{cm}$.

The radius of the inscribed circle $(r)$ is half of its diameter.

---

The area of a circle with radius $r$ is given by the formula $A = \pi r^2$.

Substitute the value of the radius $r = \frac{a}{2}\ \text{cm}$ into the area formula:

From equation (ii):

Area $= \pi \left(\frac{a^2}{2^2}\right)\ \text{cm}^2$

Area $= \pi \frac{a^2}{4}\ \text{cm}^2$

Area $= \frac{\pi a^2}{4}\ \text{cm}^2$

---

The calculated area of the inscribed circle is $\frac{\pi a^2}{4}\ \text{cm}^2$.

The statement claims the area is $\pi a^2\ \text{cm}^2$.

Comparing the two areas: $\frac{\pi a^2}{4}$ and $\pi a^2$.

Since $\frac{1}{4} \neq 1$, the two expressions are not equal (assuming $a \neq 0$).

---

Conclusion:

No, the statement is false.

The reason is that the radius of the circle inscribed in a square of side $a$ is $\frac{a}{2}$, and the area of the circle is $\pi (\frac{a}{2})^2 = \frac{\pi a^2}{4}$, not $\pi a^2$.

Given:

Radius of the circle = $a\ \text{cm}$.

A square is circumscribing the circle.

Statement to verify: The perimeter of the circumscribing square is $8a\ \text{cm}$.

---

To Verify:

Whether the statement "Perimeter of circumscribing square = $8a\ \text{cm}$" is true.

---

Reasoning:

When a square circumscribes a circle, each side of the square is tangent to the circle.

The distance between two opposite sides of the square is equal to the diameter of the circle.

Let the radius of the circle be $r = a\ \text{cm}$.

The diameter of the circle is $d = 2r = 2a\ \text{cm}$.

Let the side length of the circumscribing square be $s$.

So, $s = 2a\ \text{cm}$.

---

The perimeter of a square with side length $s$ is given by the formula $P = 4s$.

Substitute the value of $s = 2a\ \text{cm}$ into the perimeter formula:

From equation (i):

Perimeter $= 8a\ \text{cm}$

---

The calculated perimeter of the circumscribing square is $8a\ \text{cm}$.

The statement claims the perimeter is $8a\ \text{cm}$.

The calculated perimeter matches the statement.

---

Conclusion:

Yes, the statement is true.

The reason is that the side of a square circumscribing a circle of radius $a$ is equal to the diameter of the circle, which is $2a$. The perimeter of the square is 4 times its side length, so it is $4 \times (2a) = 8a\ \text{cm}$.

Given:

Diameter of the circle = $d$.

An inner square is inscribed in the circle.

An outer square is circumscribing the circle.

Statement to verify: The area of the outer square is four times the area of the inner square.

---

To Verify:

Whether the statement "Area of outer square = 4 $\times$ Area of inner square" is true.

---

Reasoning:

For the inner square (inscribed in the circle):

The diagonal of the inscribed square is equal to the diameter of the circle.

Let the side length of the inner square be $s_{\text{inner}}$.

The diagonal of a square is $s \sqrt{2}$. So, $s_{\text{inner}}\sqrt{2} = d$.

The area of the inner square is $A_{\text{inner}} = s_{\text{inner}}^2$.

Substitute the value of $s_{\text{inner}}$ from equation (i):

$A_{\text{inner}} = \left(\frac{d}{\sqrt{2}}\right)^2 = \frac{d^2}{(\sqrt{2})^2} = \frac{d^2}{2}$.

---

For the outer square (circumscribing the circle):

The side length of the circumscribing square is equal to the diameter of the circle.

Let the side length of the outer square be $s_{\text{outer}}$.

The area of the outer square is $A_{\text{outer}} = s_{\text{outer}}^2$.

Substitute the value of $s_{\text{outer}}$ from equation (ii):

$A_{\text{outer}} = d^2$.

---

Now, let's compare the area of the outer square ($A_{\text{outer}}$) with the area of the inner square ($A_{\text{inner}}$):

$A_{\text{outer}} = d^2$

$A_{\text{inner}} = \frac{d^2}{2}$

We can see that $d^2 = 2 \times \frac{d^2}{2}$.

So, $A_{\text{outer}} = 2 \times A_{\text{inner}}$.

The area of the outer square is two times the area of the inner square, not four times.

---

Conclusion:

No, the statement is false.

The reason is that the area of the inner square is $\frac{d^2}{2}$ and the area of the outer square is $d^2$, where $d$ is the diameter of the circle. The ratio of the areas is $\frac{A_{\text{outer}}}{A_{\text{inner}}} = \frac{d^2}{d^2/2} = 2$. Thus, the area of the outer square is two times the area of the inner square.

Given:

The statement: Area of a segment of a circle is less than the area of its corresponding sector.

---

To Verify:

Whether the given statement is true and provide a reason.

---

Reasoning:

Consider a circle with center O and radius $r$. Let the corresponding sector be formed by radii OA and OB and arc AB. The corresponding segment is formed by the chord AB and arc AB.

The area of the sector is the sum of the area of the triangle formed by the radii and the chord (Triangle OAB) and the area of the corresponding segment.

Rearranging equation (i) to find the area of the segment:

The area of a triangle is always non-negative ($\ge 0$).

Area of Triangle OAB $\ge 0$.

From equation (ii), if Area of Triangle OAB $> 0$, then Area of Segment $<$ Area of Sector.

If Area of Triangle OAB $= 0$, then Area of Segment $=$ Area of Sector.

The area of Triangle OAB is 0 if and only if the vertices O, A, and B are collinear. This happens when the chord AB is a diameter of the circle. In this case, the segment is a semicircle, and the corresponding sector is also a semicircle.

For any segment other than a semicircle or a full circle (where the chord is not a diameter), the triangle OAB has a positive area, so the area of the segment is strictly less than the area of the sector.

However, for a semicircle, the segment and the corresponding sector are the same region (a semicircle), and their areas are equal. The triangle formed by the radii along the diameter has an area of 0.

The statement says the area of a segment is "less than" the area of its corresponding sector. This is not true for the case of a semicircle, where the areas are equal.

---

Conclusion:

No, the statement is false.

The reason is that the area of a segment is given by the area of the corresponding sector minus the area of the triangle formed by the chord and the radii to the endpoints of the chord. While for most segments the area of the triangle is positive, leading to the segment's area being less than the sector's area, in the case of a semicircle, the chord is a diameter, the triangle is degenerate (area is 0), and the area of the segment is equal to the area of the sector.

A more accurate statement would be: Area of a segment of a circle is less than or equal to the area of its corresponding sector.

Given:

Diameter of the circular wheel = $d\ \text{cm}$.

Statement to verify: The distance travelled by the wheel in one revolution is $2\pi d\ \text{cm}$.

---

To Verify:

Whether the statement "Distance travelled in one revolution = $2\pi d\ \text{cm}$" is true and provide a reason.

---

Reasoning:

When a circular wheel completes one revolution, every point on its circumference touches the surface exactly once.

Therefore, the distance travelled by the wheel in one revolution is equal to the length of its circumference.

The formula for the circumference of a circle with diameter $d$ is given by:

The statement claims the distance travelled in one revolution is $2\pi d\ \text{cm}$.

The actual distance travelled in one revolution is the circumference, which is $\pi d\ \text{cm}$ (from equation (i)).

Comparing the claimed distance ($2\pi d$) with the actual distance ($\pi d$), we see that they are not equal (unless $d=0$, which is not a valid wheel).

Note that $2\pi d$ is actually twice the circumference, or the circumference of a circle with diameter $2d$ (or radius $d$).

---

Conclusion:

No, the statement is false.

The reason is that the distance travelled by a circular wheel in one revolution is equal to its circumference, which is given by the formula $\pi \times \text{diameter} = \pi d$, not $2\pi d$. The formula $2\pi d$ would correspond to the circumference of a circle with diameter $2d$, or radius $d$.

Given:

Distance covered by the wheel = $s$ metres.

Radius of the circular wheel = $r$ metres.

Statement to verify: Number of revolutions made by the wheel = $\frac{s}{2\pi r}$.

---

To Verify:

Whether the given statement about the number of revolutions is true and provide a reason.

---

Reasoning:

The distance covered by a circular wheel in one complete revolution is equal to its circumference.

The circumference of a circle with radius $r$ is given by the formula $C = 2\pi r$.

The total distance covered by the wheel is $s$ metres.

The total distance covered is the product of the number of revolutions and the distance covered in one revolution.

Substitute the given total distance ($s$) and the distance covered in one revolution ($2\pi r$) into equation (ii):

$s = \text{Number of revolutions} \times (2\pi r)$

To find the number of revolutions, divide the total distance by the distance covered in one revolution:

The derived formula for the number of revolutions (equation iii) is $\frac{s}{2\pi r}$, which is exactly the same as the statement given in the question.

---

Conclusion:

Yes, the statement is true.

The reason is that the distance covered by a circular wheel in one revolution is equal to its circumference, which is $2\pi r$. To cover a total distance $s$, the wheel must make a number of revolutions such that the sum of the distances covered in each revolution equals $s$. Therefore, the number of revolutions is the total distance divided by the circumference, i.e., $\frac{s}{2\pi r}$.

Given:

The statement: The numerical value of the area of a circle is greater than the numerical value of its circumference.

---

To Verify:

Whether the given statement is true and provide a reason.

---

Reasoning:

Let the radius of the circle be $r$.

The numerical value of the area of the circle is $A = \pi r^2$.

The numerical value of the circumference of the circle is $C = 2\pi r$.

We need to check if the statement $A > C$ is always true for any valid radius $r$ ($r \ge 0$).

We are comparing $\pi r^2$ and $2\pi r$.

Let's set up the inequality $\pi r^2 > 2\pi r$.

Since $\pi > 0$, we can divide both sides by $\pi$ without changing the direction of the inequality:

$r^2 > 2r$

Now, rearrange the inequality:

$r^2 - 2r > 0$

Factor out $r$:

$r(r - 2) > 0$

For this inequality to be true, the product $r(r-2)$ must be positive. This occurs in two cases:

Case 1: Both factors are positive. $r > 0$ and $r - 2 > 0 \implies r > 2$.

Case 2: Both factors are negative. $r < 0$ and $r - 2 < 0 \implies r < 0$. However, the radius of a circle cannot be negative, so this case is not relevant for a real circle.

The inequality $r(r-2) > 0$ (which means $A > C$) is true only when $r > 2$.

What happens for other values of $r$ ($r \ge 0$)?

If $r = 0$ (a point circle), $A = \pi (0)^2 = 0$ and $C = 2\pi (0) = 0$. In this case, $A = C$, so $A \ngtr C$.

If $0 < r < 2$, then $r > 0$ and $r - 2 < 0$. The product $r(r-2)$ is negative ($r(r-2) < 0$). This means $r^2 < 2r$, and thus $\pi r^2 < 2\pi r$, so $A < C$.

If $r = 2$, then $r - 2 = 0$. The product $r(r-2) = 0$ ($2(2-2) = 0$). This means $r^2 = 2r$, and thus $\pi r^2 = 2\pi r$, so $A = C$. For example, if $r=2$, Area = $\pi(2^2) = 4\pi$ and Circumference = $2\pi(2) = 4\pi$. Numerically they are equal.

The statement "the numerical value of the area is greater than the numerical value of its circumference" is only true when the radius $r$ is greater than 2 units. It is not true for all radii (e.g., $r=1$, $r=2$).

---

Conclusion:

No, the statement is false.

The reason is that the numerical value of the area of a circle ($\pi r^2$) is greater than the numerical value of its circumference ($2\pi r$) only when the radius $r > 2$. When $0 < r < 2$, the area is less than the circumference, and when $r=0$ or $r=2$, the area is equal to the circumference.

Given:

Radius of the first circle = $r$.

Radius of the second circle = $2r$.

The length of an arc of the first circle is equal to the length of an arc of the second circle.

The statement to evaluate is: The angle of the corresponding sector of the first circle is double the angle of the corresponding sector of the other circle.

---

To Verify:

Whether the given statement (that the angle of the first sector is double the angle of the second sector) is false.

---

Reasoning:

Let the angle of the sector of the first circle (with radius $r$) be $\theta_1$ (in degrees).

Let the angle of the sector of the second circle (with radius $2r$) be $\theta_2$ (in degrees).

The formula for the length of an arc of a sector with angle $\theta$ (in degrees) and radius $R$ is given by: Arc Length $= \frac{\theta}{360^\circ} \times 2\pi R$.

Length of the arc of the first circle ($L_1$) $= \frac{\theta_1}{360^\circ} \times 2\pi r$.

Length of the arc of the second circle ($L_2$) $= \frac{\theta_2}{360^\circ} \times 2\pi (2r) = \frac{\theta_2}{360^\circ} \times 4\pi r$.

According to the problem, the lengths of the arcs are equal:

Substitute the arc length formulas into equation (i):

$\frac{\theta_1}{360^\circ} \times 2\pi r = \frac{\theta_2}{360^\circ} \times 4\pi r$

Assuming $r > 0$, $\pi \neq 0$, and $360^\circ \neq 0$, we can cancel the common terms $\frac{2\pi r}{360^\circ}$ from both sides:

From equation (ii), we get the relationship:

$\theta_1 = 2\theta_2$

---

The statement given in the question is that "the angle of the corresponding sector of the first circle ($\theta_1$) is double the angle of the corresponding sector of the other circle ($\theta_2$)", which is exactly $\theta_1 = 2\theta_2$.

Our derivation shows that $\theta_1 = 2\theta_2$ is true if the arc lengths are equal.

The question asks if the statement "$\theta_1 = 2\theta_2$" is false.

Since we found $\theta_1 = 2\theta_2$ to be true under the given conditions, the statement "$\theta_1 = 2\theta_2$" is not false.

---

Conclusion:

No, the statement is not false; it is true.

The reason is that if the arc lengths are equal, $\frac{\theta_1}{360^\circ} \times 2\pi r = \frac{\theta_2}{360^\circ} \times 2\pi (2r)$, which simplifies to $\theta_1 r = \theta_2 (2r)$, and since $r \neq 0$, we get $\theta_1 = 2\theta_2$. Thus, the angle of the sector with radius $r$ is indeed double the angle of the sector with radius $2r$ when their arc lengths are equal.

The statement is false.

---

Let the two different circles have radii $r_1$ and $r_2$, where $r_1 \neq r_2$. Let the corresponding arc lengths of the two sectors be $l_1$ and $l_2$. We are given that $l_1 = l_2 = l$ (say).

---

The formula for the arc length ($l$) of a sector with radius $r$ and angle $\theta$ (in radians) is $l = r\theta$.

The formula for the area ($A$) of a sector in terms of arc length is given by $A = \frac{1}{2}lr$.

---

For the first sector, the area is $A_1 = \frac{1}{2} l_1 r_1$.

For the second sector, the area is $A_2 = \frac{1}{2} l_2 r_2$.

---

Since $l_1 = l_2 = l$, we have:

$A_1 = \frac{1}{2} l r_1$

$A_2 = \frac{1}{2} l r_2$

---

For the areas to be equal, we would need $A_1 = A_2$, which means:

$\frac{1}{2} l r_1 = \frac{1}{2} l r_2$

Assuming $l > 0$ (for non-zero sectors with positive arc length), we can divide both sides by $\frac{1}{2}l$:

$r_1 = r_2$

---

However, the problem states that the two circles are different, which implies that their radii are not equal, i.e., $r_1 \neq r_2$.

---

Therefore, if $r_1 \neq r_2$ and $l > 0$, then $\frac{1}{2} l r_1 \neq \frac{1}{2} l r_2$, which means $A_1 \neq A_2$.

---

Hence, the areas of the two sectors are not equal.

The statement is false.

---

Let the two different circles have radii $r_1$ and $r_2$, where $r_1 \neq r_2$. Let the corresponding arc lengths of the two sectors be $l_1$ and $l_2$. We are given that their areas are equal, i.e., $A_1 = A_2 = A$ (say).

---

The formula for the area ($A$) of a sector with radius $r$ and corresponding arc length $l$ is given by $A = \frac{1}{2}lr$.

---

For the first sector, the area is $A_1 = \frac{1}{2} l_1 r_1$.

For the second sector, the area is $A_2 = \frac{1}{2} l_2 r_2$.

---

Since $A_1 = A_2$, we have:

$\frac{1}{2} l_1 r_1 = \frac{1}{2} l_2 r_2$

Multiplying both sides by 2, we get:

$l_1 r_1 = l_2 r_2$

---

This equation can be rewritten as:

$\frac{l_1}{l_2} = \frac{r_2}{r_1}$

---

Since the two circles are different, their radii are not equal. So, $r_1 \neq r_2$.

This implies that the ratio $\frac{r_2}{r_1}$ is not equal to 1 (assuming $r_1, r_2 \neq 0$).

---

Therefore, from the equation $\frac{l_1}{l_2} = \frac{r_2}{r_1}$, we must have $\frac{l_1}{l_2} \neq 1$.

This means $l_1 \neq l_2$.

---

Thus, if the areas of sectors from two different circles are equal, it is not necessary that their corresponding arc lengths are equal; in fact, if the radii are different, the arc lengths must be different.

The statement is false.

---

To draw the largest possible circle inside a rectangle, the diameter of the circle must be equal to the shorter side of the rectangle.

---

Given the rectangle has length $a$ cm and breadth $b$ cm, with $a > b$.

The shorter side is the breadth, which is $b$ cm.

---

The diameter of the largest circle that can be drawn inside the rectangle is equal to the breadth, $b$ cm.

Therefore, the radius of this circle is $r = \frac{b}{2}$ cm.

---

The area of a circle with radius $r$ is given by the formula $A = \pi r^2$.

Substituting the radius $r = \frac{b}{2}$, the area of the largest inscribed circle is:

$A = \pi \left(\frac{b}{2}\right)^2$

$A = \pi \left(\frac{b^2}{4}\right)$

$A = \frac{\pi b^2}{4}$ cm$^2$

---

The area given in the statement is $\pi b^2$ cm$^2$.

Since $\frac{\pi b^2}{4} \neq \pi b^2$ (unless $b=0$, which is not the case for a rectangle), the statement is false.

The statement is true.

---

Let the radii of the two circles be $r_1$ and $r_2$.

Let their circumferences be $C_1$ and $C_2$, respectively.

The formula for the circumference of a circle with radius $r$ is $C = 2\pi r$.

---

For the first circle, $C_1 = 2\pi r_1$.

For the second circle, $C_2 = 2\pi r_2$.

---

We are given that the circumferences of the two circles are equal:

---

Substituting the formulas for circumference:

$2\pi r_1 = 2\pi r_2$

---

Since $2\pi$ is a non-zero constant, we can divide both sides by $2\pi$:

$\frac{2\pi r_1}{2\pi} = \frac{2\pi r_2}{2\pi}$

$r_1 = r_2$

---

So, the radii of the two circles must be equal.

---

Now, let the areas of the two circles be $A_1$ and $A_2$, respectively.

The formula for the area of a circle with radius $r$ is $A = \pi r^2$.

---

For the first circle, $A_1 = \pi r_1^2$.

For the second circle, $A_2 = \pi r_2^2$.

---

Since $r_1 = r_2$, it follows that $r_1^2 = r_2^2$.

Multiplying both sides by $\pi$, we get:

$\pi r_1^2 = \pi r_2^2$

This means $A_1 = A_2$.

---

Therefore, if the circumferences of two circles are equal, their areas must also be equal.

The statement is true.

---

Let the radii of the two circles be $r_1$ and $r_2$.

Let their areas be $A_1$ and $A_2$, respectively.

Let their circumferences be $C_1$ and $C_2$, respectively.

---

The formula for the area of a circle with radius $r$ is $A = \pi r^2$.

The formula for the circumference of a circle with radius $r$ is $C = 2\pi r$.

---

For the first circle, $A_1 = \pi r_1^2$ and $C_1 = 2\pi r_1$.

For the second circle, $A_2 = \pi r_2^2$ and $C_2 = 2\pi r_2$.

---

We are given that the areas of the two circles are equal:

---

Substituting the formulas for area:

$\pi r_1^2 = \pi r_2^2$

---

Since $\pi$ is a non-zero constant, we can divide both sides by $\pi$:

$\frac{\pi r_1^2}{\pi} = \frac{\pi r_2^2}{\pi}$

$r_1^2 = r_2^2$

---

Since radius is a non-negative value, taking the square root of both sides gives:

$\sqrt{r_1^2} = \sqrt{r_2^2}$

$r_1 = r_2$

---

So, the radii of the two circles must be equal.

---

Now, we compare their circumferences. Since $r_1 = r_2$, we can write:

$2\pi r_1 = 2\pi r_2$

Substituting the formulas for circumference:

$C_1 = C_2$

---

Therefore, if the areas of two circles are equal, their circumferences must also be equal.

The statement is false.

---

When a square is inscribed in a circle, the diagonal of the square is equal to the diameter of the circle.

---

Given that the diameter of the circle is $p$ cm.

So, the diagonal of the inscribed square is $d = p$ cm.

---

Let the side length of the square be $s$ cm.

In a square, the relationship between the diagonal ($d$) and the side length ($s$) is given by $d^2 = s^2 + s^2$, which simplifies to $d^2 = 2s^2$. Thus, $d = s\sqrt{2}$.

---

Using the given diagonal $d=p$, we have:

$p = s\sqrt{2}$

---

Solving for the side length $s$:

$s = \frac{p}{\sqrt{2}}$ cm.

---

The area of the square is given by $A = s^2$.

Substituting the value of $s$:

$A = \left(\frac{p}{\sqrt{2}}\right)^2$

$A = \frac{p^2}{(\sqrt{2})^2}$

$A = \frac{p^2}{2}$ cm$^2$

---

The area of the square is $\frac{p^2}{2}$ cm$^2$, not $p^2$ cm$^2$.

Since $\frac{p^2}{2} \neq p^2$ (for $p \neq 0$), the statement is false.

Given:

Diameter of the first circle, $d_1 = 20$ cm.

Diameter of the second circle, $d_2 = 48$ cm.

The area of a third circle ($A$) is equal to the sum of the areas of the first two circles ($A_1$ and $A_2$).

To Find:

The diameter of the third circle.

---

Solution:

The radius of the first circle is $r_1 = \frac{d_1}{2} = \frac{20}{2} = 10$ cm.

The area of the first circle is $A_1 = \pi r_1^2 = \pi (10)^2 = 100\pi$ cm$^2$.

---

The radius of the second circle is $r_2 = \frac{d_2}{2} = \frac{48}{2} = 24$ cm.

The area of the second circle is $A_2 = \pi r_2^2 = \pi (24)^2 = 576\pi$ cm$^2$.

---

Let the radius of the third circle be $R$. Its area is $A = \pi R^2$.

According to the problem, the area of the third circle is the sum of the areas of the first two circles.

---

Substituting the area formulas:

$\pi R^2 = 100\pi + 576\pi$

$\pi R^2 = (100 + 576)\pi$

$\pi R^2 = 676\pi$

---

Divide both sides by $\pi$ (since $\pi \neq 0$):

$R^2 = 676$

---

Taking the square root of both sides (and considering the positive radius):

$R = \sqrt{676}$

$R = 26$ cm

---

The diameter of the third circle is $D = 2R$.

$D = 2 \times 26$

$D = 52$ cm

---

The diameter of the circle whose area is equal to the sum of the areas of the two given circles is 52 cm.

Given:

Radius of the circle, $r = 21$ cm.

Central angle of the sector, $\theta = 120^\circ$.

---

To Find:

Area of the sector.

---

Solution:

The formula for the area of a sector of a circle with radius $r$ and central angle $\theta$ (in degrees) is given by:

$A = \frac{\theta}{360^\circ} \times \pi r^2$

---

Substitute the given values $r = 21$ cm and $\theta = 120^\circ$ into the formula. We will use $\pi = \frac{22}{7}$.

$A = \frac{120^\circ}{360^\circ} \times \frac{22}{7} \times (21)^2$

---

Simplify the fraction $\frac{120}{360}$:

$\frac{\cancel{120}^{1}}{\cancel{360}_{3}} = \frac{1}{3}$

---

So, the area becomes:

$A = \frac{1}{3} \times \frac{22}{7} \times 21 \times 21$

---

Cancel out common factors:

$A = \frac{1}{3} \times \frac{22}{\cancel{7}^{1}} \times \cancel{21}^{3} \times 21$

$A = \frac{1}{3} \times 22 \times 3 \times 21$

$A = \frac{1}{\cancel{3}^{1}} \times 22 \times \cancel{3}^{1} \times 21$

$A = 22 \times 21$

---

Calculate the product:

$22 \times 21 = 462$

---

The area of the sector is 462 cm$^2$.

Given:

Radius of the inscribed circle, $r = 7.5$ cm.

The circle is inscribed in a square.

Use $\pi = 3.14$.

---

To Find:

The area of the shaded region (Area of square - Area of circle).

---

Solution:

When a circle is inscribed in a square, the diameter of the circle is equal to the side length of the square.

---

Diameter of the circle = $2 \times \text{radius}$

Diameter = $2 \times 7.5 = 15$ cm.

---

Side length of the square, $s =$ Diameter of the circle

$s = 15$ cm.

---

Area of the square = $s^2$

Area of square = $(15)^2 = 15 \times 15 = 225$ cm$^2$.

---

Area of the circle = $\pi r^2$

Area of circle = $3.14 \times (7.5)^2$

Area of circle = $3.14 \times 56.25$

Area of circle = $176.625$ cm$^2$.

---

The shaded region is the area of the square minus the area of the circle.

Area of shaded region = Area of square - Area of circle

Area of shaded region = $225 - 176.625$

Area of shaded region = $48.375$ cm$^2$.

---

The area of the shaded region is 48.375 cm$^2$.

Given:

Radius of the circle, $r = 36$ cm.

Area of the sector, $A = 54\pi$ cm$^2$.

---

To Find:

The length of the corresponding arc of the sector, $l$.

---

Solution:

The area of a sector of a circle can be expressed in terms of its radius and corresponding arc length using the formula:

$A = \frac{1}{2} \times \text{arc length} \times \text{radius}$

or

$A = \frac{1}{2}lr$

---

Substitute the given values for the Area ($A$) and Radius ($r$) into this formula:

$54\pi = \frac{1}{2} \times l \times 36$

---

Simplify the right side of the equation:

$54\pi = \frac{36}{2} \times l$

$54\pi = 18l$

---

Now, solve for $l$ by dividing both sides by 18:

$l = \frac{54\pi}{18}$

---

Simplify the fraction:

$l = \frac{\cancel{54}^{3}\pi}{\cancel{18}^{1}}$

$l = 3\pi$

---

The length of the corresponding arc is $3\pi$ cm.

Given:

Radius of the first circle, $r_1 = 15$ cm.

Radius of the second circle, $r_2 = 18$ cm.

The circumference of a new circle is equal to the sum of the circumferences of the two given circles.

---

To Find:

The radius of the new circle.

---

Solution:

The formula for the circumference of a circle with radius $r$ is given by $C = 2\pi r$.

---

Let $C_1$ be the circumference of the first circle and $r_1$ be its radius.

$C_1 = 2\pi r_1 = 2\pi (15) = 30\pi$ cm.

---

Let $C_2$ be the circumference of the second circle and $r_2$ be its radius.

$C_2 = 2\pi r_2 = 2\pi (18) = 36\pi$ cm.

---

Let the radius of the new circle be $R$ and its circumference be $C_{new}$.

$C_{new} = 2\pi R$

---

According to the problem statement, the circumference of the new circle is the sum of the circumferences of the two given circles:

$C_{new} = C_1 + C_2$

---

Substitute the expressions for the circumferences:

$2\pi R = 30\pi + 36\pi$

$2\pi R = (30 + 36)\pi$

$2\pi R = 66\pi$

---

To find $R$, divide both sides of the equation by $2\pi$ (since $2\pi$ is non-zero):

$\frac{2\pi R}{2\pi} = \frac{66\pi}{2\pi}$

$R = \frac{66}{2}$

$R = 33$ cm.

---

The radius of the circle whose circumference is equal to the sum of the circumferences of the two given circles is 33 cm.

Given:

A square of diagonal 8 cm is inscribed in a circle.

Diagonal of the square, $d = 8$ cm.

---

To Find:

The area of the shaded region (Area of Circle - Area of Square).

---

Solution:

When a square is inscribed in a circle, the diagonal of the square is equal to the diameter of the circle.

---

Diameter of the circle = Diagonal of the square

Diameter, $D = 8$ cm.

---

The radius of the circle, $r = \frac{D}{2} = \frac{8}{2} = 4$ cm.

---

The area of the circle is given by $A_{\text{circle}} = \pi r^2$.

$A_{\text{circle}} = \pi (4)^2 = 16\pi$ cm$^2$.

---

The area of a square can be calculated using its diagonal $d$ with the formula $A_{\text{square}} = \frac{1}{2}d^2$.

$A_{\text{square}} = \frac{1}{2} (8)^2 = \frac{1}{2} (64) = 32$ cm$^2$.

---

The shaded region is the area of the circle minus the area of the square.

Area of shaded region = $A_{\text{circle}} - A_{\text{square}}$

Area of shaded region = $16\pi - 32$ cm$^2$.

---

The area of the shaded region is $(16\pi - 32)$ cm$^2$.

Alternatively, factoring out 16:

Area of shaded region = $16(\pi - 2)$ cm$^2$.

Given:

Radius of the circle, $r = 28$ cm.

Central angle of the sector, $\theta = 45^\circ$.

---

To Find:

Area of the sector.

---

Solution:

The formula for the area of a sector of a circle with radius $r$ and central angle $\theta$ (in degrees) is given by:

$A = \frac{\theta}{360^\circ} \times \pi r^2$

---

Substitute the given values $r = 28$ cm and $\theta = 45^\circ$ into the formula. We will use $\pi = \frac{22}{7}$.

$A = \frac{45^\circ}{360^\circ} \times \frac{22}{7} \times (28)^2$

---

Simplify the fraction $\frac{45}{360}$:

$\frac{\cancel{45}^{1}}{\cancel{360}_{8}} = \frac{1}{8}$

---

So, the area becomes:

$A = \frac{1}{8} \times \frac{22}{7} \times 28 \times 28$

---

Cancel out common factors:

$A = \frac{1}{8} \times \frac{22}{\cancel{7}^{1}} \times \cancel{28}^{4} \times 28$

$A = \frac{1}{8} \times 22 \times 4 \times 28$

$A = \frac{1}{\cancel{8}^{2}} \times 22 \times \cancel{4}^{1} \times 28$

$A = \frac{1}{2} \times 22 \times 28$

$A = \cancel{22}^{11} \times \frac{28}{\cancel{2}^{1}}$

$A = 11 \times 28$

---

Calculate the product:

$11 \times 28 = 308$

---

The area of the sector is 308 cm$^2$.

Given:

Radius of the motor cycle wheel, $r = 35$ cm.

Speed of the motor cycle, $v = 66$ km/h.

---

To Find:

Number of revolutions the wheel makes per minute.

---

Solution:

First, we need to convert the speed from km/h to cm/min to match the unit of the radius.

Speed $v = 66$ km/h

Convert km to cm: $1 \text{ km} = 1000 \text{ m} = 1000 \times 100 \text{ cm} = 100000 \text{ cm}$.

Convert hours to minutes: $1 \text{ hour} = 60 \text{ minutes}$.

---

So, speed in cm/min is:

$v = \frac{66 \text{ km}}{1 \text{ h}} = \frac{66 \times 100000 \text{ cm}}{60 \text{ min}}$

$v = \frac{6600000}{60}$ cm/min

$v = \frac{660000}{6}$ cm/min

$v = 110000$ cm/min.

---

Next, calculate the circumference of the wheel. The distance covered in one revolution is equal to the circumference of the wheel.

Circumference, $C = 2\pi r$. Using $\pi = \frac{22}{7}$.

$C = 2 \times \frac{22}{7} \times 35$

$C = 2 \times 22 \times \frac{\cancel{35}^{5}}{\cancel{7}^{1}}$

$C = 2 \times 22 \times 5$

$C = 44 \times 5$

$C = 220$ cm.

---

The number of revolutions per minute is the total distance covered in one minute divided by the distance covered in one revolution (circumference).

Number of revolutions per minute $= \frac{\text{Distance covered per minute}}{\text{Circumference}}$

Number of revolutions per minute $= \frac{110000 \text{ cm/min}}{220 \text{ cm/revolution}}$

---

Calculate the value:

Number of revolutions per minute $= \frac{110000}{220}$

$= \frac{11000}{22}$

$= \frac{\cancel{11000}^{500}}{\cancel{22}^{1}}$

$= 500$ revolutions per minute.

---

The wheel must make 500 revolutions per minute to maintain a speed of 66 km/h.

Given:

Length of the rope = 14 m.

Dimensions of the rectangular field = 20 m $\times$ 16 m.

The cow is tied at one corner of the rectangular field.

---

To Find:

The area of the field in which the cow can graze.

---

Solution:

The cow is tied at a corner of the rectangular field. The rope allows the cow to graze in a circular area with the corner as the center and the length of the rope as the radius. Since the cow is restricted by the rectangular field boundaries, the grazing area will be a sector of a circle.

---

The angle at each corner of a rectangle is $90^\circ$.

The rope length is the radius of the circle that the cow can potentially cover, $r = 14$ m.

---

Since the cow is in a corner of the rectangle, the area it can graze within the field boundaries is a sector of a circle with a central angle of $90^\circ$.

The area of a sector of a circle with radius $r$ and central angle $\theta$ (in degrees) is given by:

$A = \frac{\theta}{360^\circ} \times \pi r^2$

---

Here, $r = 14$ m and $\theta = 90^\circ$. We will use $\pi = \frac{22}{7}$.

$A = \frac{90^\circ}{360^\circ} \times \frac{22}{7} \times (14)^2$

---

Simplify the fraction $\frac{90}{360}$:

$\frac{\cancel{90}^{1}}{\cancel{360}_{4}} = \frac{1}{4}$

---

So, the area becomes:

$A = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14$

---

Cancel out common factors:

$A = \frac{1}{4} \times \frac{22}{\cancel{7}^{1}} \times \cancel{14}^{2} \times 14$

$A = \frac{1}{4} \times 22 \times 2 \times 14$

$A = \frac{1}{\cancel{4}^{2}} \times 22 \times \cancel{2}^{1} \times 14$

$A = \frac{1}{2} \times 22 \times 14$

$A = \cancel{22}^{11} \times \frac{14}{\cancel{2}^{1}}$

$A = 11 \times 14$

---

Calculate the product:

$11 \times 14 = 154$

---

The area of the field in which the cow can graze is 154 m$^2$.

---

Note: The rope length (14 m) is less than both the length (20 m) and the breadth (16 m) of the rectangle. This ensures that the entire $90^\circ$ sector is within the rectangular field boundaries, and no part of the grazing area is outside the field.

Given:

The flower bed consists of a rectangular middle section and two semi-circular ends.

Dimensions of the rectangular part: Length = 38 cm, Width = 10 cm.

The semi-circular ends are attached to the sides of length 10 cm.

---

To Find:

The total area of the flower bed.

---

Solution:

The area of the flower bed is the sum of the area of the rectangular part and the areas of the two semi-circular ends.

---

Area of the rectangular part = Length $\times$ Width

Area of rectangle = $38 \times 10 = 380$ cm$^2$.

---

The two semi-circular ends are attached to the sides of length 10 cm. This means the diameter of each semi-circle is 10 cm.

Diameter of semi-circle = 10 cm.

Radius of semi-circle, $r = \frac{\text{Diameter}}{2} = \frac{10}{2} = 5$ cm.

---

The area of a semi-circle is $\frac{1}{2}\pi r^2$.

Area of one semi-circle = $\frac{1}{2} \pi (5)^2 = \frac{1}{2} \pi (25) = \frac{25}{2}\pi$ cm$^2$.

---

The flower bed has two such semi-circular ends. The combined area of the two semi-circles is equal to the area of a full circle with radius 5 cm.

Combined area of two semi-circles = $2 \times \left(\frac{1}{2}\pi r^2\right) = \pi r^2$

Combined area of two semi-circles = $\pi (5)^2 = 25\pi$ cm$^2$.

---

Total area of the flower bed = Area of rectangle + Combined area of two semi-circles

Total area = $380 + 25\pi$ cm$^2$.

---

If a numerical value is required, we can use $\pi \approx 3.14$ or $\pi \approx \frac{22}{7}$. The image does not specify which value to use, so we will leave it in terms of $\pi$. If a specific value was intended, it should have been mentioned in the question text as in Sample Question 3.

Using $\pi \approx 3.14$ (if needed):

Total area $\approx 380 + 25 \times 3.14$

Total area $\approx 380 + 78.5$

Total area $\approx 458.5$ cm$^2$.

---

The area of the flower bed is $(380 + 25\pi)$ cm$^2$.

Given:

AB is the diameter of the circle.

AC = 6 cm, BC = 8 cm.

Use $\pi = 3.14$.

The shaded region is the area of the circle minus the area of triangle ABC.

---

To Find:

The area of the shaded region.

---

Solution:

Since AB is the diameter of the circle and C is a point on the circle, the angle subtended by the diameter at any point on the circumference is $90^\circ$. Therefore, $\triangle$ABC is a right-angled triangle with the right angle at C.

---

We can find the length of the diameter AB using the Pythagorean theorem in $\triangle$ABC:

$AB^2 = AC^2 + BC^2$

$AB^2 = (6)^2 + (8)^2$

$AB^2 = 36 + 64$

$AB^2 = 100$

$AB = \sqrt{100}$

$AB = 10$ cm.

---

The diameter of the circle is 10 cm.

The radius of the circle is $r = \frac{\text{Diameter}}{2} = \frac{10}{2} = 5$ cm.

---

The area of the circle is $A_{\text{circle}} = \pi r^2$.

$A_{\text{circle}} = 3.14 \times (5)^2$

$A_{\text{circle}} = 3.14 \times 25$

$A_{\text{circle}} = 78.5$ cm$^2$.

---

The area of the right-angled triangle ABC is $\frac{1}{2} \times \text{base} \times \text{height}$. We can take AC as the base and BC as the height (or vice versa).

$A_{\text{triangle}} = \frac{1}{2} \times AC \times BC$

$A_{\text{triangle}} = \frac{1}{2} \times 6 \times 8$

$A_{\text{triangle}} = \frac{1}{2} \times 48$

$A_{\text{triangle}} = 24$ cm$^2$.

---

The area of the shaded region is the area of the circle minus the area of triangle ABC.

Area of shaded region = $A_{\text{circle}} - A_{\text{triangle}}$

Area of shaded region = $78.5 - 24$

Area of shaded region = $54.5$ cm$^2$.

---

The area of the shaded region is 54.5 cm$^2$.

Given:

A square with four semi-circles drawn on each side as diameter, pointing inwards. The shaded region consists of the four corner areas within the square not covered by the semi-circles.

Note: The side length of the square is not provided in the problem statement.

---

To Find:

The area of the shaded region.

---

Solution:

Let the side length of the square be $s$.

The area of the square is $A_{\text{square}} = s^2$.

---

A semi-circle is drawn on each side of the square as its diameter. The radius of each semi-circle is half the side length of the square, so $r = \frac{s}{2}$.

The area of one semi-circle is $A_{\text{semi-circle}} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi \left(\frac{s}{2}\right)^2 = \frac{1}{2} \pi \frac{s^2}{4} = \frac{\pi s^2}{8}$.

---

Consider the areas covered by the semi-circles drawn on two opposite sides, for example, on side AB and side CD. The union of these two areas covers the entire area of the square except for the two shaded corner regions adjacent to the other two sides (AD and BC).

---

Let $C$ be the area of one shaded corner region. There are four identical shaded corner regions.

Area(Semi-circle on AB) $\cup$ Area(Semi-circle on CD) = Area(Semi-circle on AB) + Area(Semi-circle on CD) - Area(Intersection of the two semi-circles)

In this figure, the semi-circles on opposite sides (AB and CD) intersect in the central unshaded region. The union of the two opposite semi-circles covers the square entirely except for the two shaded corners adjacent to the other two sides.

So, Area(Union of semi-circles on AB and CD) = Area(Square) - Area(Shaded corner near AD) - Area(Shaded corner near BC).

Area(Union of semi-circles on AB and CD) = Area(Semi-circle on AB) + Area(Semi-circle on CD).

Therefore, Area(Semi-circle on AB) + Area(Semi-circle on CD) = Area(Square) - $2 \times C$.

---

Substituting the area formula for the semi-circles:

$\frac{\pi s^2}{8} + \frac{\pi s^2}{8} = s^2 - 2C$

$\frac{2\pi s^2}{8} = s^2 - 2C$

$\frac{\pi s^2}{4} = s^2 - 2C$

---

Now, we solve for $2C$:

$2C = s^2 - \frac{\pi s^2}{4}$

$2C = s^2 \left(1 - \frac{\pi}{4}\right)$

---

The total shaded area is $4C$. Multiply the equation by 2:

$4C = 2 \times s^2 \left(1 - \frac{\pi}{4}\right)$

$4C = 2s^2 - \frac{2\pi s^2}{4}$

$4C = 2s^2 - \frac{\pi s^2}{2}$

---

The total area of the shaded region is $s^2 \left(2 - \frac{\pi}{2}\right)$.

---

Since the side length $s$ is not given in the problem statement, the area of the shaded region is expressed in terms of $s$ and $\pi$. If a numerical value for the side length $s$ and the value of $\pi$ were provided, the area could be calculated numerically.

---

The area of the shaded region is $s^2 \left(2 - \frac{\pi}{2}\right)$ square units, where $s$ is the side length of the square.

Given:

From Fig. 11.9, a square is shown with four identical circles inscribed within it.

The side length of the square is given as 14 cm in the figure.

The shaded region is the area of the square not covered by the four circles.

---

To Find:

The area of the shaded region.

---

Solution:

The side length of the square is $s = 14$ cm.

The four identical circles are arranged in a 2x2 pattern, touching the sides of the square and each other. This means that along any side of the square, the side length is equal to the sum of the diameters of two circles.

Let $d$ be the diameter of each circle. Then $2d = s$.

So, the diameter of each circle is $d = \frac{s}{2} = \frac{14}{2} = 7$ cm.

The radius of each circle is $r = \frac{d}{2} = \frac{7}{2} = 3.5$ cm.

---

The area of one circle is given by the formula $A_{\text{circle}} = \pi r^2$.

We will use $\pi = \frac{22}{7}$ since the radius is a multiple of 3.5 or 7/2.

$A_{\text{circle}} = \frac{22}{7} \times (3.5)^2$

$A_{\text{circle}} = \frac{22}{7} \times \left(\frac{7}{2}\right)^2$

$A_{\text{circle}} = \frac{22}{7} \times \frac{49}{4}$

Cancel the common factor 7:

$A_{\text{circle}} = \frac{22}{\cancel{7}^{1}} \times \frac{\cancel{49}^{7}}{4}$

$A_{\text{circle}} = \frac{22 \times 7}{4} = \frac{154}{4}$

$A_{\text{circle}} = 38.5$ cm$^2$.

---

There are four identical circles. The total area occupied by the four circles is $4 \times A_{\text{circle}}$.

Total area of circles $= 4 \times 38.5$

Total area of circles $= 154$ cm$^2$.

---

The area of the square is given by $A_{\text{square}} = s^2$.

$A_{\text{square}} = (14)^2 = 14 \times 14 = 196$ cm$^2$.

---

The shaded region is the area of the square minus the total area of the four circles.

Area of shaded region $= A_{\text{square}} - \text{Total area of circles}$

Area of shaded region $= 196 - 154$

Area of shaded region $= 42$ cm$^2$.

---

The area of the shaded region is 42 cm$^2$.

Given:

Radius of the circle, $r = 14$ cm.

Central angle of the corresponding sector, $\theta = 60^\circ$.

---

To Find:

The area of the minor segment.

---

Solution:

The area of the minor segment is the area of the corresponding sector minus the area of the triangle formed by the radii and the chord.

Area of segment = Area of sector - Area of triangle

---

First, calculate the area of the sector with radius $r = 14$ cm and central angle $\theta = 60^\circ$.

Area of sector $= \frac{\theta}{360^\circ} \times \pi r^2$

Using $\pi = \frac{22}{7}$:

Area of sector $= \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times (14)^2$

Area of sector $= \frac{1}{6} \times \frac{22}{7} \times 14 \times 14$

Area of sector $= \frac{1}{6} \times 22 \times \frac{\cancel{14}^{2}}{\cancel{7}^{1}} \times 14$

Area of sector $= \frac{1}{6} \times 22 \times 2 \times 14$

Area of sector $= \frac{1}{\cancel{6}^{3}} \times \cancel{44}^{22} \times 14$

Area of sector $= \frac{22 \times 14}{3} = \frac{308}{3}$ cm$^2$.

---

Next, calculate the area of the triangle formed by the radii and the chord. Let the center of the circle be O and the endpoints of the chord be A and B. $\triangle$OAB has two sides OA and OB which are radii, so OA = OB = $r = 14$ cm. The angle between these radii is $\angle$AOB = $60^\circ$.

Since OA = OB and $\angle$AOB = $60^\circ$, $\triangle$OAB is an isosceles triangle with the angle between the equal sides being $60^\circ$. This implies that $\triangle$OAB is an equilateral triangle.

In an equilateral triangle with side length $a$, the area is $\frac{\sqrt{3}}{4} a^2$. Here, the side length $a = r = 14$ cm.

Area of $\triangle$OAB $= \frac{\sqrt{3}}{4} (14)^2$

Area of $\triangle$OAB $= \frac{\sqrt{3}}{4} \times 196$

Area of $\triangle$OAB $= \sqrt{3} \times \frac{196}{4}$

Area of $\triangle$OAB $= 49\sqrt{3}$ cm$^2$.

---

Alternatively, the area of $\triangle$OAB can be found using the formula $\frac{1}{2} r^2 \sin\theta$ for an isosceles triangle with two sides of length $r$ and the angle between them $\theta$.

Area of $\triangle$OAB $= \frac{1}{2} (14)^2 \sin(60^\circ)$

Area of $\triangle$OAB $= \frac{1}{2} \times 196 \times \frac{\sqrt{3}}{2}$

Area of $\triangle$OAB $= \frac{196\sqrt{3}}{4} = 49\sqrt{3}$ cm$^2$. (This confirms the previous calculation)

---

Area of the minor segment = Area of sector - Area of $\triangle$OAB

Area of minor segment $= \frac{308}{3} - 49\sqrt{3}$ cm$^2$.

---

The area of the minor segment is $\left(\frac{308}{3} - 49\sqrt{3}\right)$ cm$^2$.

Given:

ABCD is a square.

P, Q, R, S are the mid-points of AB, BC, CD, and DA respectively.

Arcs are drawn with centres A, B, C, and D such that they intersect at P, Q, R, S.

From the figure, the side length of the square is $s = 14$ cm.

Use $\pi = 3.14$.

---

To Find:

The area of the shaded region.

---

Solution:

The arcs are drawn from the vertices (A, B, C, D) and pass through the midpoints of the adjacent sides (P, Q, R, S). This means the radius of each arc is equal to half the side length of the square.

Radius of each arc, $r = \frac{\text{Side length}}{2} = \frac{14}{2} = 7$ cm.

---

The shaded region is the central area enclosed by the four arcs. This is a standard geometric figure.

The area of the shaded region in this configuration (a square of side $s=2r$ with four arcs of radius $r$ drawn from the vertices to the midpoints of adjacent sides) is given by the formula:

$A_{\text{shaded}} = r^2(4 - \pi)$

---

Substitute the values $r = 7$ cm and $\pi = 3.14$ into the formula:

$A_{\text{shaded}} = (7)^2 (4 - 3.14)$

$A_{\text{shaded}} = 49 (0.86)$

---

Calculate the product:

$49 \times 0.86 = 42.14$

---

The area of the shaded region is 42.14 cm$^2$.

---

Note: This formula $r^2(4-\pi)$ represents the area of the square $4r^2$ minus the area of four sectors of radius $r$ and angle $90^\circ$ (summing to $\pi r^2$), which results in the shaded area plus four unshaded lens-shaped regions. However, for this specific configuration where arcs meet at midpoints, the formula $r^2(4-\pi)$ directly gives the shaded area. A rigorous derivation involves calculating the area of the lens shapes or using integration, which is beyond the scope expected at this level, suggesting the formula is either known or derived through a simpler method not immediately obvious from segment/sector formulas alone.

Given:

ABC is an equilateral triangle with side length $a = 10$ cm.

D, E, F are the mid-points of BC, CA, AB respectively.

Arcs are drawn with centres A, B, C, passing through the midpoints of the opposite sides.

Use $\pi = 3.14$.

---

To Find:

The area of the shaded region.

---

Solution:

The arcs are drawn from the vertices and pass through the midpoints of the opposite sides. For the arc with center A, it passes through D and E, which are midpoints of BC and CA respectively. This means the radius of the arc is equal to the length of the median from A to BC, or the length of the line segment AD or AE. However, the description says the arcs intersect the sides at their respective mid-points. This implies the radius of the arc centered at A is AF or AE, which are halves of the side length AB and AC.

Let's assume the interpretation from the image is correct: Arcs are drawn from vertices A, B, C with radius equal to half the side length, i.e., $r = \frac{10}{2} = 5$ cm.

Since ABC is an equilateral triangle, each interior angle is $60^\circ$.

---

The shaded region is the area of the equilateral triangle minus the areas of the three sectors at the vertices that are not shaded.

---

First, calculate the area of the equilateral triangle with side length $a = 10$ cm.

Area of equilateral triangle $= \frac{\sqrt{3}}{4} a^2$

Area of equilateral triangle $= \frac{\sqrt{3}}{4} (10)^2 = \frac{\sqrt{3}}{4} \times 100 = 25\sqrt{3}$ cm$^2$.

Using $\sqrt{3} \approx 1.732$ (a common approximation if needed, but not explicitly given).

Area of triangle $\approx 25 \times 1.732 = 43.3$ cm$^2$.

---

Now, calculate the area of the three sectors. Each sector has a radius $r = 5$ cm and a central angle equal to the angle of the equilateral triangle, $\theta = 60^\circ$.

Area of one sector $= \frac{\theta}{360^\circ} \times \pi r^2$

Area of one sector $= \frac{60^\circ}{360^\circ} \times \pi (5)^2$

Area of one sector $= \frac{1}{6} \times \pi \times 25 = \frac{25\pi}{6}$ cm$^2$.

---

There are three such sectors, one at each vertex. The total area of the three sectors is $3 \times \text{Area of one sector}$.

Total area of three sectors $= 3 \times \frac{25\pi}{6} = \frac{75\pi}{6} = \frac{25\pi}{2}$ cm$^2$.

---

Using $\pi = 3.14$:

Total area of three sectors $= \frac{25 \times 3.14}{2}$

Total area of three sectors $= \frac{78.5}{2} = 39.25$ cm$^2$.

---

The shaded region is the area of the equilateral triangle minus the total area of the three sectors.

Area of shaded region = Area of triangle - Total area of three sectors

Area of shaded region = $25\sqrt{3} - \frac{25\pi}{2}$ cm$^2$.

---

Using the numerical values:

Area of shaded region $\approx 43.3 - 39.25$

Area of shaded region $\approx 4.05$ cm$^2$.

---

The question asks for the area of the shaded region using $\pi = 3.14$. It is likely intended that we calculate the area using numerical values for both the triangle and the sectors.

---

Area of equilateral triangle $= \frac{\sqrt{3}}{4} (10)^2 = 25\sqrt{3}$. Using $\sqrt{3} \approx 1.732$, Area $\approx 43.3$ cm$^2$.

Area of the three sectors $= 39.25$ cm$^2$.

Area of shaded region = Area of triangle - Sum of areas of three sectors

Area of shaded region $= 25\sqrt{3} - 3 \times \left(\frac{60}{360} \times \pi (5)^2\right)$

Area of shaded region $= 25\sqrt{3} - 3 \times \left(\frac{1}{6} \times \pi \times 25\right)$

Area of shaded region $= 25\sqrt{3} - \frac{25\pi}{2}$

Area of shaded region $\approx 25 \times 1.732 - \frac{25 \times 3.14}{2}$

Area of shaded region $\approx 43.3 - 39.25$

Area of shaded region $\approx 4.05$ cm$^2$.

---

The area of the shaded region is approximately 4.05 cm$^2$.

Given:

Radius of arcs, $r = 14$ cm.

Centres of arcs are P, Q, R.

Use $\pi = 3.14$.

---

To Find:

Area of shaded region.

---

Solution:

From the figure, we assume PQR is an equilateral triangle where the side length $a$ is twice the radius of the arcs, i.e., $a = 2r = 2 \times 14 = 28$ cm.

---

The shaded region is the area of the equilateral triangle PQR minus the area of the three sectors of radius $r=14$ cm at the vertices P, Q, R.

---

The angle of each sector at the vertex of an equilateral triangle is $60^\circ$. The sum of the angles of the three sectors is $3 \times 60^\circ = 180^\circ$.

---

Area of the three sectors combined $= \frac{\text{Sum of angles}}{360^\circ} \times \pi r^2 = \frac{180^\circ}{360^\circ} \times \pi (14)^2 = \frac{1}{2} \times \pi \times 196 = 98\pi$ cm$^2$.

---

Area of equilateral triangle with side $a = 28$ cm is $\frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (28)^2 = \frac{\sqrt{3}}{4} \times 784 = 196\sqrt{3}$ cm$^2$.

---

Area of shaded region = Area of $\triangle$PQR - Area of three sectors

Area of shaded region $= 196\sqrt{3} - 98\pi$ cm$^2$.

---

Using $\pi = 3.14$:

Area of shaded region $= 196\sqrt{3} - 98(3.14) = 196\sqrt{3} - 307.72$ cm$^2$.

---

Using $\sqrt{3} \approx 1.732$ for a numerical answer:

Area of shaded region $\approx 196(1.732) - 307.72$

Area of shaded region $\approx 339.472 - 307.72$

Area of shaded region $\approx 31.752$ cm$^2$.

---

The area of the shaded region is $(196\sqrt{3} - 98\pi)$ cm$^2$ or approximately 31.75 cm$^2$ (using $\pi = 3.14$ and $\sqrt{3} \approx 1.732$).

Given:

Radius of the circular park (inner radius), $r_1 = 105$ m.

Width of the road, $w = 21$ m.

---

To Find:

The area of the road.

---

Solution:

The road surrounds the park, forming an annular region (area between two concentric circles).

The radius of the outer circle (park including the road) is the inner radius plus the width of the road.

Outer radius, $r_2 = r_1 + w = 105 + 21 = 126$ m.

---

The area of the road is the area of the outer circle minus the area of the inner circle.

Area of road $= \text{Area of outer circle} - \text{Area of inner circle}$

Area of road $= \pi r_2^2 - \pi r_1^2 = \pi (r_2^2 - r_1^2)$.

---

Substitute the values $r_1 = 105$ and $r_2 = 126$. We will use $\pi = \frac{22}{7}$.

Area of road $= \pi (126^2 - 105^2)$

Using the difference of squares formula $a^2 - b^2 = (a-b)(a+b)$:

$126^2 - 105^2 = (126 - 105)(126 + 105)$

$126 - 105 = 21$

$126 + 105 = 231$

$126^2 - 105^2 = 21 \times 231 = 4851$

---

Area of road $= \pi \times 4851$

Using $\pi = \frac{22}{7}$:

Area of road $= \frac{22}{7} \times 4851$

Area of road $= 22 \times \frac{\cancel{4851}^{693}}{\cancel{7}^{1}}$

Area of road $= 22 \times 693$

Area of road $= 15246$

---

The area of the road is 15246 m$^2$.

Given:

ABCD is a quadrilateral.

Arcs are drawn with radius $r = 21$ cm from each vertex A, B, C, D as centres.

The shaded region is the sum of the areas of the four sectors formed at the vertices within the quadrilateral.

---

To Find:

The area of the shaded region.

---

Solution:

The shaded region consists of four sectors, one at each vertex of the quadrilateral ABCD. The radius of each sector is given as $r = 21$ cm.

---

Let the angles at the vertices of the quadrilateral be $\angle$A, $\angle$B, $\angle$C, and $\angle$D.

The sum of the interior angles of a quadrilateral is $360^\circ$.

$\angle$A + $\angle$B + $\angle$C + $\angle$D = $360^\circ$

---

The area of the sector at vertex A is $A_A = \frac{\angle\text{A}}{360^\circ} \times \pi r^2$.

The area of the sector at vertex B is $A_B = \frac{\angle\text{B}}{360^\circ} \times \pi r^2$.

The area of the sector at vertex C is $A_C = \frac{\angle\text{C}}{360^\circ} \times \pi r^2$.

The area of the sector at vertex D is $A_D = \frac{\angle\text{D}}{360^\circ} \times \pi r^2$.

---

The area of the shaded region is the sum of the areas of these four sectors.

Area of shaded region $= A_A + A_B + A_C + A_D$

Area of shaded region $= \frac{\angle\text{A}}{360^\circ} \pi r^2 + \frac{\angle\text{B}}{360^\circ} \pi r^2 + \frac{\angle\text{C}}{360^\circ} \pi r^2 + \frac{\angle\text{D}}{360^\circ} \pi r^2$

Area of shaded region $= \frac{\pi r^2}{360^\circ} (\angle\text{A} + \angle\text{B} + \angle\text{C} + \angle\text{D})$

---

Substitute the sum of the angles $\angle$A + $\angle$B + $\angle$C + $\angle$D = $360^\circ$ and the radius $r = 21$ cm.

Area of shaded region $= \frac{\pi (21)^2}{360^\circ} (360^\circ)$

Area of shaded region $= \pi (21)^2$

Area of shaded region $= \pi \times 441$ cm$^2$.

---

Using $\pi = \frac{22}{7}$:

Area of shaded region $= \frac{22}{7} \times 441$

Area of shaded region $= 22 \times \frac{\cancel{441}^{63}}{\cancel{7}^{1}}$

Area of shaded region $= 22 \times 63$

Area of shaded region $= 1386$

---

The area of the shaded region is 1386 cm$^2$.

---

Note: The shape of the quadrilateral does not affect the sum of the areas of the four sectors at the vertices with the same radius, as long as the sectors do not overlap outside the quadrilateral or overlap excessively within the shaded region in a way that alters the sum of the areas defined by the angles at the vertices.

Given:

Length of the arc, $l = 20$ cm.

Central angle subtended by the arc, $\theta = 60^\circ$.

---

To Find:

The radius of the circle, $r$.

---

Solution:

The formula for the length of an arc ($l$) of a sector with radius $r$ and central angle $\theta$ (in degrees) is given by:

$l = \frac{\theta}{360^\circ} \times 2\pi r$

---

Substitute the given values into the formula:

$20 = \frac{60^\circ}{360^\circ} \times 2\pi r$

---

Simplify the fraction $\frac{60}{360}$:

$\frac{\cancel{60}^{1}}{\cancel{360}_{6}} = \frac{1}{6}$

---

The equation becomes:

$20 = \frac{1}{6} \times 2\pi r$

$20 = \frac{2\pi r}{6}$

$20 = \frac{\pi r}{3}$

---

To find the radius $r$, multiply both sides of the equation by 3:

$20 \times 3 = \pi r$

$60 = \pi r$

---

Now, divide both sides by $\pi$:

$r = \frac{60}{\pi}$

---

The radius of the circle is $\frac{60}{\pi}$ cm.

(Note: If a numerical value for $\pi$ were provided, the radius could be calculated as a decimal value).

Given:

Radius of the circle, $r = 20$ cm.

Central angle subtended by the chord, $\theta = 90^\circ$.

Use $\pi = 3.14$.

---

To Find:

The area of the corresponding major segment.

---

Solution:

The area of the major segment can be found by subtracting the area of the minor segment from the area of the entire circle.

Area of major segment = Area of circle - Area of minor segment

---

First, find the area of the circle:

Area of circle $= \pi r^2 = 3.14 \times (20)^2 = 3.14 \times 400 = 1256$ cm$^2$.

---

Next, find the area of the minor segment. The area of the minor segment is the area of the sector corresponding to the $90^\circ$ angle minus the area of the triangle formed by the radii and the chord.

Area of minor segment = Area of minor sector - Area of triangle

---

Area of minor sector $= \frac{\theta}{360^\circ} \times \pi r^2 = \frac{90^\circ}{360^\circ} \times 3.14 \times (20)^2$

Area of minor sector $= \frac{1}{4} \times 3.14 \times 400 = 3.14 \times 100 = 314$ cm$^2$.

---

The triangle formed by the radii and the chord for a central angle of $90^\circ$ is a right-angled triangle with legs equal to the radius (20 cm). The area of this triangle is:

Area of triangle $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 20 = \frac{1}{2} \times 400 = 200$ cm$^2$.

---

Area of minor segment = Area of minor sector - Area of triangle

Area of minor segment $= 314 - 200 = 114$ cm$^2$.

---

Finally, calculate the area of the major segment:

Area of major segment = Area of circle - Area of minor segment

Area of major segment $= 1256 - 114 = 1142$ cm$^2$.

---

The area of the corresponding major segment of the circle is 1142 cm$^2$.

Given:

Triangle ABC with sides AB = 14 cm, BC = 48 cm, CA = 50 cm.

Arcs drawn with centres A, B, C and radius $r = 5$ cm each.

Use $\pi = 3.14$.

The shaded region is the area of the triangle minus the areas of the three sectors at the vertices.

---

To Find:

The area of the shaded region.

---

Solution:

First, let's determine the type of triangle ABC. Check if it is a right-angled triangle using the Pythagorean theorem. The longest side is CA = 50 cm.

Consider $AB^2 + BC^2$:

$14^2 + 48^2 = 196 + 2304 = 2500$

Consider $CA^2$:

$50^2 = 2500$

---

Since $AB^2 + BC^2 = CA^2$, the triangle ABC satisfies the Pythagorean theorem. Therefore, $\triangle$ABC is a right-angled triangle with the right angle at vertex B.

---

The area of the right-angled triangle ABC is given by:

Area($\triangle$ABC) $= \frac{1}{2} \times \text{base} \times \text{height}$

Area($\triangle$ABC) $= \frac{1}{2} \times AB \times BC$

Area($\triangle$ABC) $= \frac{1}{2} \times 14 \times 48$

Area($\triangle$ABC) $= 7 \times 48 = 336$ cm$^2$.

---

The shaded region is the area of the triangle minus the areas of the three sectors at vertices A, B, and C. Each sector has a radius of 5 cm.

The central angles of these sectors are the angles of the triangle, i.e., $\angle$A, $\angle$B, and $\angle$C.

The sum of the angles in any triangle is $180^\circ$. So, $\angle$A + $\angle$B + $\angle$C = $180^\circ$.

---

The area of a sector with radius $r$ and angle $\theta$ is $\frac{\theta}{360^\circ} \times \pi r^2$.

The sum of the areas of the three sectors is:

Area of sectors $= \frac{\angle\text{A}}{360^\circ} \pi r^2 + \frac{\angle\text{B}}{360^\circ} \pi r^2 + \frac{\angle\text{C}}{360^\circ} \pi r^2$

Area of sectors $= \frac{\pi r^2}{360^\circ} (\angle\text{A} + \angle\text{B} + \angle\text{C})$

Area of sectors $= \frac{\pi (5)^2}{360^\circ} (180^\circ)$

Area of sectors $= \frac{25\pi}{360} \times 180$

Area of sectors $= \frac{25\pi}{\cancel{360}^{2}} \times \cancel{180}^{1}$

Area of sectors $= \frac{25\pi}{2}$ cm$^2$.

---

Using $\pi = 3.14$:

Area of sectors $= \frac{25 \times 3.14}{2} = \frac{78.5}{2} = 39.25$ cm$^2$.

---

The area of the shaded region is the area of the triangle minus the sum of the areas of the three sectors.

Area of shaded region = Area($\triangle$ABC) - Area of sectors

Area of shaded region $= 336 - 39.25$

Area of shaded region $= 296.75$ cm$^2$.

---

The area of the shaded region is 296.75 cm$^2$.

Given:

Side of the square grassy lawn = 20 m.

Initial length of the rope, $r_1 = 6$ m.

The rope length is increased by 5.5 m.

---

To Find:

The increase in the area of the grassy lawn in which the calf can graze.

---

Solution:

The calf is tied at a corner of the square lawn. The area the calf can graze is a sector of a circle with the corner as the center and the rope length as the radius. Since it is a square corner, the central angle of the sector is $90^\circ$. The side length of the square (20 m) is greater than the initial and final rope lengths, so the entire grazing area will be within the lawn.

---

Case 1: Initial rope length

Initial radius, $r_1 = 6$ m.

Central angle, $\theta = 90^\circ$.

Area grazed initially, $A_1 = \frac{\theta}{360^\circ} \times \pi r_1^2$

$A_1 = \frac{90^\circ}{360^\circ} \times \pi (6)^2 = \frac{1}{4} \times \pi \times 36 = 9\pi$ m$^2$.

---

Case 2: Increased rope length

The length of the rope is increased by 5.5 m.

New radius, $r_2 = r_1 + 5.5 = 6 + 5.5 = 11.5$ m.

Central angle, $\theta = 90^\circ$.

Area grazed after increasing the rope, $A_2 = \frac{\theta}{360^\circ} \times \pi r_2^2$

$A_2 = \frac{90^\circ}{360^\circ} \times \pi (11.5)^2 = \frac{1}{4} \times \pi \times (11.5)^2$

$A_2 = \frac{1}{4} \times \pi \times 132.25 = 33.0625\pi$ m$^2$.

---

Increase in grazing area:

Increase in area $= A_2 - A_1$

Increase in area $= 33.0625\pi - 9\pi$

Increase in area $= (33.0625 - 9)\pi = 24.0625\pi$ m$^2$.

---

The problem does not specify the value of $\pi$. If we use $\pi = 3.14$ (as in previous examples), the numerical value is:

Increase in area $\approx 24.0625 \times 3.14$

Increase in area $\approx 75.568125$ m$^2$.

If we use $\pi = \frac{22}{7}$:

Increase in area $= 24.0625 \times \frac{22}{7} = \frac{240625}{10000} \times \frac{22}{7} = \frac{9625}{400} \times \frac{22}{7} = \frac{1375}{400} \times 22 = \frac{55}{16} \times 22 = \frac{55}{8} \times 11 = \frac{605}{8} = 75.625$ m$^2$.

---

Let's use the more common value $\pi \approx 3.14$ as in Sample Question 2 and 11.

Increase in area $\approx 75.57$ m$^2$ (rounded to two decimal places).

---

The increase in area of the grassy lawn in which the calf can graze is $24.0625\pi$ m$^2$, or approximately 75.57 m$^2$ using $\pi \approx 3.14$.

Given:

Area of the circular playground, $A = 22176$ m$^2$.

Rate of fencing = $\textsf{₹} 50$ per metre.

---

To Find:

The cost of fencing the ground.

---

Solution:

To find the cost of fencing, we first need to find the circumference of the circular playground. The circumference depends on the radius, which can be found from the given area.

---

The formula for the area of a circle with radius $r$ is $A = \pi r^2$.

Substitute the given area:

$22176 = \pi r^2$

We will use $\pi = \frac{22}{7}$ for the calculation.

$22176 = \frac{22}{7} r^2$

---

Solve for $r^2$:

$r^2 = \frac{22176 \times 7}{22}$

$r^2 = \frac{\cancel{22176}^{1008} \times 7}{\cancel{22}^{1}}$

$r^2 = 1008 \times 7$

$r^2 = 7056$

---

Now, find the radius $r$ by taking the square root of $r^2$:

$r = \sqrt{7056}$

$r = 84$ m

(Since radius must be positive)

---

The formula for the circumference of a circle with radius $r$ is $C = 2\pi r$.

Substitute the radius $r = 84$ m and $\pi = \frac{22}{7}$:

$C = 2 \times \frac{22}{7} \times 84$

$C = 2 \times 22 \times \frac{\cancel{84}^{12}}{\cancel{7}^{1}}$

$C = 2 \times 22 \times 12$

$C = 44 \times 12$

$C = 528$ m.

---

The cost of fencing the ground is the circumference multiplied by the rate per metre.

Cost = Circumference $\times$ Rate

Cost = $528 \text{ m} \times \textsf{₹} 50 \text{/m}$

Cost = $528 \times 50$

Cost = $26400$

---

The cost of fencing the ground is $\textsf{₹} 26400$.

Given:

Diameter of the front wheel, $d_f = 80$ cm.

Diameter of the rear wheel, $d_r = 2$ m.

Number of revolutions made by the front wheel = 1400.

---

To Find:

Number of revolutions the rear wheel will make to cover the same distance.

---

Solution:

First, ensure all measurements are in the same unit. Let's convert everything to centimeters.

Diameter of the front wheel, $d_f = 80$ cm.

Diameter of the rear wheel, $d_r = 2 \text{ m} = 2 \times 100 \text{ cm} = 200$ cm.

---

The distance covered by a wheel in one revolution is equal to its circumference.

Circumference of the front wheel, $C_f = \pi d_f = \pi \times 80 = 80\pi$ cm.

Circumference of the rear wheel, $C_r = \pi d_r = \pi \times 200 = 200\pi$ cm.

---

The total distance covered by the front wheel in 1400 revolutions is:

Distance $= \text{Number of revolutions} \times \text{Circumference per revolution}$

Distance covered by front wheel $= 1400 \times C_f = 1400 \times 80\pi = 112000\pi$ cm.

---

The rear wheel covers the same distance. Let $N$ be the number of revolutions made by the rear wheel.

Distance covered by rear wheel $= N \times C_r = N \times 200\pi$ cm.

---

Since the distance covered is the same:

Distance covered by rear wheel = Distance covered by front wheel

$N \times 200\pi = 112000\pi$

---

To find $N$, divide both sides by $200\pi$ (since $200\pi \neq 0$):

$N = \frac{112000\pi}{200\pi}$

$N = \frac{112000}{200}$

$N = \frac{1120}{2}$

$N = 560$

---

The rear wheel will make 560 revolutions in covering the same distance.

Given:

Sides of the triangular field are $a = 15$ m, $b = 16$ m, and $c = 17$ m.

Length of the rope for each animal (radius of the grazing sector), $r = 7$ m.

The animals are tied at the three corners (vertices) of the field.

---

To Find:

The area of the field which cannot be grazed by the three animals.

---

Solution:

The area of the field that cannot be grazed is the total area of the triangular field minus the total area grazed by the three animals inside the field.

Area not grazed = Area of triangle - Total grazed area inside the triangle.

---

The animals graze in sectors of circles at each vertex, with radius $r=7$ m. Since the rope length (7 m) is less than the shortest side of the triangle (15 m), the entire area grazed by each animal will be within the triangular field.

---

Let the angles at the three corners (vertices) of the triangle be $\angle$A, $\angle$B, and $\angle$C.

The sum of the angles in a triangle is $180^\circ$. So, $\angle$A + $\angle$B + $\angle$C = $180^\circ$.

---

The area grazed by the cow (at vertex A) is a sector with radius $r=7$ m and angle $\angle$A:

Area grazed at A $= \frac{\angle\text{A}}{360^\circ} \times \pi r^2$

Similarly, for the buffalo (at vertex B) and the horse (at vertex C):

Area grazed at B $= \frac{\angle\text{B}}{360^\circ} \times \pi r^2$

Area grazed at C $= \frac{\angle\text{C}}{360^\circ} \times \pi r^2$

---

The total area grazed by the three animals inside the field is the sum of the areas of these three sectors:

Total grazed area $= \frac{\angle\text{A}}{360^\circ} \pi r^2 + \frac{\angle\text{B}}{360^\circ} \pi r^2 + \frac{\angle\text{C}}{360^\circ} \pi r^2$

Total grazed area $= \frac{\pi r^2}{360^\circ} (\angle\text{A} + \angle\text{B} + \angle\text{C})$

Total grazed area $= \frac{\pi (7)^2}{360^\circ} (180^\circ)$

Total grazed area $= \frac{49\pi}{360} \times 180 = \frac{49\pi}{\cancel{360}^{2}} \times \cancel{180}^{1}$

Total grazed area $= \frac{49\pi}{2}$ m$^2$.

---

Now, we need to find the area of the triangular field with sides 15 m, 16 m, and 17 m. Since it's a general triangle, we can use Heron's formula.

The semi-perimeter $s = \frac{a+b+c}{2} = \frac{15+16+17}{2} = \frac{48}{2} = 24$ m.

Area of triangle $= \sqrt{s(s-a)(s-b)(s-c)}$

Area of triangle $= \sqrt{24(24-15)(24-16)(24-17)}$

Area of triangle $= \sqrt{24 \times 9 \times 8 \times 7}$

Area of triangle $= \sqrt{(8 \times 3) \times 9 \times 8 \times 7}$

Area of triangle $= \sqrt{8^2 \times 3 \times 9 \times 7}$

Area of triangle $= \sqrt{8^2 \times 3 \times 3^2 \times 7}$

Area of triangle $= \sqrt{8^2 \times 3^2 \times (3 \times 7)}$

Area of triangle $= 8 \times 3 \sqrt{21} = 24\sqrt{21}$ m$^2$.

---

Area of the field which cannot be grazed = Area of triangle - Total grazed area

Area not grazed $= 24\sqrt{21} - \frac{49\pi}{2}$ m$^2$.

---

Using $\pi = \frac{22}{7}$ and assuming $\sqrt{21}$ might need a numerical value if a numerical answer is expected (not explicitly stated):

Total grazed area $= \frac{49}{2} \times \frac{22}{7} = \frac{\cancel{49}^{7}}{2} \times \frac{\cancel{22}^{11}}{\cancel{7}^{1}} = 7 \times 11 = 77$ m$^2$.

Area not grazed $= 24\sqrt{21} - 77$ m$^2$.

---

Since no value for $\sqrt{21}$ is provided, the answer should be left in terms of $\sqrt{21}$ and calculated value for the sectors.

Area not grazed $= 24\sqrt{21} - 77$ m$^2$.

---

The area of the field which cannot be grazed by the three animals is $(24\sqrt{21} - 77)$ m$^2$.

Given:

Radius of the circle, $r = 12$ cm.

Central angle of the corresponding sector, $\theta = 60^\circ$.

Use $\pi = 3.14$.

---

To Find:

The area of the segment.

---

Solution:

The area of a segment of a circle is the area of the corresponding sector minus the area of the triangle formed by the radii and the chord.

Area of segment = Area of sector - Area of triangle

---

First, calculate the area of the sector with radius $r = 12$ cm and central angle $\theta = 60^\circ$.

Area of sector $= \frac{\theta}{360^\circ} \times \pi r^2$

Area of sector $= \frac{60^\circ}{360^\circ} \times 3.14 \times (12)^2$

Area of sector $= \frac{1}{6} \times 3.14 \times 144$

Area of sector $= \frac{3.14 \times \cancel{144}^{24}}{\cancel{6}^{1}}$

Area of sector $= 3.14 \times 24$

Area of sector $= 75.36$ cm$^2$.

---

Next, calculate the area of the triangle formed by the radii and the chord. Let the center of the circle be O and the endpoints of the chord be A and B. $\triangle$OAB has sides OA = OB = $r = 12$ cm, and the angle between them is $\angle$AOB = $60^\circ$.

Since OA = OB and $\angle$AOB = $60^\circ$, $\triangle$OAB is an isosceles triangle with the angle between the equal sides being $60^\circ$. This means $\triangle$OAB is an equilateral triangle with side length 12 cm.

The area of an equilateral triangle with side length $a$ is $\frac{\sqrt{3}}{4} a^2$.

Area of $\triangle$OAB $= \frac{\sqrt{3}}{4} (12)^2 = \frac{\sqrt{3}}{4} \times 144 = 36\sqrt{3}$ cm$^2$.

---

If a numerical answer is required, we need to use a value for $\sqrt{3}$. A common value is $\sqrt{3} \approx 1.732$.

Area of $\triangle$OAB $\approx 36 \times 1.732 = 62.352$ cm$^2$.

---

Area of segment = Area of sector - Area of triangle

Area of segment $= 75.36 - 36\sqrt{3}$ cm$^2$.

---

Using the numerical value for $\sqrt{3}$:

Area of segment $\approx 75.36 - 62.352 = 13.008$ cm$^2$.

---

The area of the segment is $(75.36 - 36\sqrt{3})$ cm$^2$, or approximately 13.01 cm$^2$ (rounded to two decimal places, using $\sqrt{3} \approx 1.732$). If the question expects an answer only using $\pi=3.14$ and not $\sqrt{3}$, there might be an issue with the problem statement or expected calculation method. However, for a $60^\circ$ angle, the equilateral triangle is the standard approach.

Given:

Diameter of the circular pond, $d_1 = 17.5$ m.

Width of the path, $w = 2$ m.

Rate of constructing the path = $\textsf{₹} 25$ per m$^2$.

---

To Find:

The cost of constructing the path.

---

Solution:

First, find the radius of the circular pond (inner radius).

Radius of the pond, $r_1 = \frac{d_1}{2} = \frac{17.5}{2} = 8.75$ m.

---

The path surrounds the pond, forming an annular region. The radius of the outer circle (pond plus path) is the inner radius plus the width of the path.

Outer radius, $r_2 = r_1 + w = 8.75 + 2 = 10.75$ m.

---

The area of the path is the area of the outer circle minus the area of the inner circle.

Area of path $= \text{Area of outer circle} - \text{Area of inner circle}$

Area of path $= \pi r_2^2 - \pi r_1^2 = \pi (r_2^2 - r_1^2)$.

---

Substitute the values $r_1 = 8.75$ and $r_2 = 10.75$. We will use $\pi = \frac{22}{7}$.

Area of path $= \pi ((10.75)^2 - (8.75)^2)$

Using the difference of squares formula $a^2 - b^2 = (a-b)(a+b)$:

$(10.75)^2 - (8.75)^2 = (10.75 - 8.75)(10.75 + 8.75)$

$10.75 - 8.75 = 2.00$

$10.75 + 8.75 = 19.50$

$(10.75)^2 - (8.75)^2 = 2 \times 19.5 = 39$

---

Area of path $= \pi \times 39$

Using $\pi = \frac{22}{7}$:

Area of path $= \frac{22}{7} \times 39$

Area of path $= \frac{858}{7}$ m$^2$.

---

The cost of constructing the path is the area of the path multiplied by the rate per m$^2$.

Cost = Area of path $\times$ Rate

Cost = $\frac{858}{7} \text{ m}^2 \times \textsf{₹} 25 \text{/m}^2$

Cost = $\frac{858 \times 25}{7}$

Cost = $\frac{21450}{7}$

---

Performing the division:

$\begin{array}{r} 3064.28\dots \\ 7{\overline{\smash{\big)}\,21450.00\dots}} \\ \underline{-~\phantom{()}(21)\phantom{...}} \\ 04\phantom{....} \\ \underline{-~\phantom{()}0\phantom{....}} \\ 45\phantom{...} \\ \underline{-~\phantom{()}(42)\phantom{...}} \\ 30\phantom{..} \\ \underline{-~\phantom{()}(28)\phantom{..}} \\ 20\phantom{.} \\ \underline{-~\phantom{()}(14)\phantom{.}} \\ 60\phantom{} \\ \underline{-~\phantom{()}(56)\phantom{}} \\ 4\phantom{} \end{array}$

The result is approximately $\textsf{₹} 3064.2857...$

---

The cost of constructing the path is $\textsf{₹} \frac{21450}{7}$ or approximately $\textsf{₹} 3064.29$ (rounded to two decimal places).

Given:

ABCD is a trapezium with AB || DC.

Parallel sides: AB = 18 cm, DC = 32 cm.

Height (distance between parallel sides) = 14 cm.

Arcs are drawn with centres A, B, C, D and equal radii $r = 7$ cm.

The shaded region is the area of the trapezium minus the areas of the four sectors at the vertices.

---

To Find:

The area of the shaded region.

---

Solution:

First, find the area of the trapezium ABCD.

Area of trapezium $= \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$

Area of trapezium $= \frac{1}{2} \times (AB + DC) \times \text{distance between AB and DC}$

Area of trapezium $= \frac{1}{2} \times (18 + 32) \times 14$

Area of trapezium $= \frac{1}{2} \times 50 \times 14$

Area of trapezium $= 25 \times 14 = 350$ cm$^2$.

---

The shaded region is the area of the trapezium minus the areas of the four sectors at the vertices A, B, C, and D. Each sector has a radius of 7 cm.

Let the angles at the vertices of the trapezium be $\angle$A, $\angle$B, $\angle$C, and $\angle$D.

The sum of the interior angles of any quadrilateral is $360^\circ$.

$\angle$A + $\angle$B + $\angle$C + $\angle$D = $360^\circ$

---

The area of a sector with radius $r$ and angle $\theta$ is $\frac{\theta}{360^\circ} \times \pi r^2$.

The sum of the areas of the four sectors is:

Area of sectors $= \frac{\angle\text{A}}{360^\circ} \pi r^2 + \frac{\angle\text{B}}{360^\circ} \pi r^2 + \frac{\angle\text{C}}{360^\circ} \pi r^2 + \frac{\angle\text{D}}{360^\circ} \pi r^2$

Area of sectors $= \frac{\pi r^2}{360^\circ} (\angle\text{A} + \angle\text{B} + \angle\text{C} + \angle\text{D})$

---

Substitute the sum of the angles $\angle$A + $\angle$B + $\angle$C + $\angle$D = $360^\circ$ and the radius $r = 7$ cm.

Area of sectors $= \frac{\pi (7)^2}{360^\circ} (360^\circ)$

Area of sectors $= \pi (7)^2 = 49\pi$ cm$^2$.

---

Using $\pi = \frac{22}{7}$:

Area of sectors $= 49 \times \frac{22}{7} = \cancel{49}^{7} \times \frac{22}{\cancel{7}^{1}} = 7 \times 22 = 154$ cm$^2$.

---

The area of the shaded region is the area of the trapezium minus the sum of the areas of the four sectors.

Area of shaded region = Area of trapezium - Area of sectors

Area of shaded region $= 350 - 154$

Area of shaded region $= 196$ cm$^2$.

---

The area of the shaded region of the figure is 196 cm$^2$.

Given:

Three circles, each with radius $r = 3.5$ cm.

Each circle touches the other two.

---

To Find:

The area enclosed between these three circles.

---

Solution:

When three circles of the same radius touch each other externally, the centers of the circles form an equilateral triangle. The distance between the centers of any two touching circles is the sum of their radii.

---

Let the centers of the three circles be P, Q, and R. Since each circle touches the other two and they have the same radius $r$, the distance between any two centers is $r + r = 2r$.

So, PQ = QR = RP = $2r = 2 \times 3.5 = 7$ cm.

Thus, $\triangle$PQR is an equilateral triangle with side length $a = 7$ cm.

---

The area enclosed between the three circles is the area of the equilateral triangle formed by their centers minus the areas of the three sectors (each with radius $r$ and central angle $60^\circ$) cut out from the triangle.

Area enclosed = Area($\triangle$PQR) - Area of three sectors.

---

Area of the equilateral triangle with side $a = 7$ cm:

Area($\triangle$PQR) $= \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (7)^2 = \frac{49\sqrt{3}}{4}$ cm$^2$.

---

Each sector at the vertices P, Q, R has a radius $r = 3.5$ cm and a central angle of $60^\circ$ (angle of equilateral triangle).

Area of one sector $= \frac{60^\circ}{360^\circ} \times \pi r^2 = \frac{1}{6} \times \pi (3.5)^2 = \frac{1}{6} \times \pi \times (7/2)^2 = \frac{1}{6} \times \pi \times \frac{49}{4} = \frac{49\pi}{24}$ cm$^2$.

---

Area of three sectors $= 3 \times \frac{49\pi}{24} = \frac{147\pi}{24} = \frac{49\pi}{8}$ cm$^2$.

---

Using $\pi = \frac{22}{7}$:

Area of three sectors $= \frac{49}{8} \times \frac{22}{7} = \frac{\cancel{49}^{7}}{8} \times \frac{\cancel{22}^{11}}{\cancel{7}^{1}} = \frac{7 \times 11}{8} = \frac{77}{8} = 9.625$ cm$^2$.

---

Area enclosed between the circles = Area($\triangle$PQR) - Area of three sectors.

Area enclosed $= \frac{49\sqrt{3}}{4} - \frac{49\pi}{8}$ cm$^2$.

---

Using $\pi = \frac{22}{7}$ and a value for $\sqrt{3}$ (if required for a numerical answer):

Area enclosed $= \frac{49\sqrt{3}}{4} - 9.625$ cm$^2$.

---

If a numerical answer is needed and assuming $\sqrt{3} \approx 1.732$:

Area($\triangle$PQR) $\approx \frac{49 \times 1.732}{4} = \frac{84.868}{4} = 21.217$ cm$^2$.

Area enclosed $\approx 21.217 - 9.625 = 11.592$ cm$^2$.

---

Using the values directly from the problem statement and common $\pi$ value:

Area of three sectors = 9.625 cm$^2$.

Area of equilateral triangle side 7 $= \frac{\sqrt{3}}{4} (7)^2 = \frac{49\sqrt{3}}{4}$.

Area enclosed $= \frac{49\sqrt{3}}{4} - 9.625$ cm$^2$.

---

The area enclosed between the circles is $\left(\frac{49\sqrt{3}}{4} - \frac{49\pi}{8}\right)$ cm$^2$ or $\left(\frac{49\sqrt{3}}{4} - 9.625\right)$ cm$^2$ using $\pi = \frac{22}{7}$. If a numerical answer is needed, a value for $\sqrt{3}$ is required.

Given:

Radius of the circle, $r = 5$ cm.

Corresponding arc length, $l = 3.5$ cm.

---

To Find:

The area of the sector.

---

Solution:

The area of a sector of a circle can be calculated directly if the radius and the corresponding arc length are known. The formula is:

Area of sector $= \frac{1}{2} \times \text{arc length} \times \text{radius}$

Area of sector $= \frac{1}{2} l r$

---

Substitute the given values into the formula:

$r = 5$ cm

$l = 3.5$ cm

Area of sector $= \frac{1}{2} \times 3.5 \times 5$

Area of sector $= \frac{1}{2} \times 17.5$

Area of sector $= 8.75$ cm$^2$.

---

The area of the sector is 8.75 cm$^2$.

Given:

Four circular cardboard pieces, each with radius $r = 7$ cm.

Each piece touches the other two pieces.

---

To Find:

The area of the portion enclosed between these pieces.

---

Solution:

When four identical circles are arranged such that each touches the other two, their centres form a square.

---

Let the centres of the four circles be P, Q, R, and S. The distance between the centres of two touching circles is the sum of their radii.

Since the radii are equal ($r=7$ cm), the distance between any two touching centres is $r + r = 2r$.

The centres form a square with side length $s = 2r = 2 \times 7 = 14$ cm.

---

The area of the square formed by the centres is:

Area of square $= s^2 = (14)^2 = 196$ cm$^2$.

---

The portion enclosed between the four circles is the area of the square formed by their centres minus the areas of the four sectors, one from each circle, that lie within this square.

---

At each corner of the square of centres, there is a sector of a circle with radius $r = 7$ cm. The angle of each corner of a square is $90^\circ$, so the central angle of each sector is $\theta = 90^\circ$.

---

Area of one sector $= \frac{\theta}{360^\circ} \times \pi r^2 = \frac{90^\circ}{360^\circ} \times \pi (7)^2 = \frac{1}{4} \times \pi \times 49 = \frac{49\pi}{4}$ cm$^2$.

---

There are four such sectors, one at each vertex of the square of centres. The total area of the four sectors within the square is:

Total area of four sectors $= 4 \times \frac{49\pi}{4} = 49\pi$ cm$^2$.

---

Using $\pi = \frac{22}{7}$:

Total area of four sectors $= 49 \times \frac{22}{7} = \cancel{49}^{7} \times \frac{22}{\cancel{7}^{1}} = 7 \times 22 = 154$ cm$^2$.

---

The area of the portion enclosed between the pieces is the area of the square minus the total area of the four sectors.

Area enclosed $= \text{Area of square} - \text{Total area of four sectors}$

Area enclosed $= 196 - 154 = 42$ cm$^2$.

---

The area of the portion enclosed between these pieces is 42 cm$^2$.

Given:

Area of square sheet = 784 cm$^2$.

Four congruent circular plates arranged as described.

---

To Find:

Area of square not covered by plates.

---

Solution:

Side length of the square, $s = \sqrt{784} = 28$ cm.

---

For four congruent circles placed as described, the side length of the square is equal to twice the diameter of a circle.

$s = 2 \times (2r) = 4r$

$28 = 4r \implies r = 7$ cm.

---

Area of one circular plate $= \pi r^2 = \pi (7)^2 = 49\pi$ cm$^2$.

---

Total area of four plates $= 4 \times 49\pi = 196\pi$ cm$^2$.

Using $\pi = \frac{22}{7}$:

Total area of four plates $= 196 \times \frac{22}{7} = 28 \times 22 = 616$ cm$^2$.

---

Area not covered = Area of square - Total area of plates

Area not covered $= 784 - 616 = 168$ cm$^2$.

---

The area of the square sheet not covered by the circular plates is 168 cm$^2$.

Given:

Dimensions of the rectangular floor = 5 m $\times$ 4 m.

Diameter of each circular tile = 50 cm.

Use $\pi = 3.14$.

The floor is covered with circular tiles arranged in rows and columns.

---

To Find:

The area of the floor that remains uncovered with tiles.

---

Solution:

First, convert the dimensions of the room to centimeters to match the tile diameter unit.

Length of the room = 5 m = $5 \times 100 = 500$ cm.

Width of the room = 4 m = $4 \times 100 = 400$ cm.

---

Diameter of each circular tile, $d = 50$ cm.

Radius of each circular tile, $r = \frac{d}{2} = \frac{50}{2} = 25$ cm.

---

The tiles are arranged such that their diameters align with the dimensions of the room. Along the length of the room (500 cm), the number of tiles that can fit is $\frac{\text{Length}}{\text{Diameter}} = \frac{500}{50} = 10$ tiles.

Along the width of the room (400 cm), the number of tiles that can fit is $\frac{\text{Width}}{\text{Diameter}} = \frac{400}{50} = 8$ tiles.

---

The total number of circular tiles covering the floor is the product of the number of tiles along the length and the width.

Total number of tiles $= 10 \times 8 = 80$ tiles.

---

Area of one circular tile $= \pi r^2 = 3.14 \times (25)^2 = 3.14 \times 625$ cm$^2$.

$3.14 \times 625 = 1962.5$ cm$^2$.

---

The total area covered by the 80 circular tiles is:

Total area of tiles $= 80 \times \text{Area of one tile}$

Total area of tiles $= 80 \times 1962.5$

$80 \times 1962.5 = 157000$ cm$^2$.

---

The area of the rectangular floor is:

Area of floor $= \text{Length} \times \text{Width} = 500 \times 400 = 200000$ cm$^2$.

---

The area of the floor that remains uncovered with tiles is the area of the floor minus the total area covered by the tiles.

Area uncovered $= \text{Area of floor} - \text{Total area of tiles}$

Area uncovered $= 200000 - 157000 = 43000$ cm$^2$.

---

We can convert the uncovered area back to square meters if needed.

$1 \text{ m}^2 = (100 \text{ cm})^2 = 10000 \text{ cm}^2$.

Area uncovered in m$^2 = \frac{43000}{10000} = 4.3$ m$^2$.

---

The area of the floor that remains uncovered with tiles is 43000 cm$^2$ or 4.3 m$^2$.

Given:

All vertices of a rhombus lie on a circle.

Area of the circle = 1256 cm$^2$.

Use $\pi = 3.14$.

---

To Find:

The area of the rhombus.

---

Solution:

If all the vertices of a rhombus lie on a circle, then the rhombus must be a square. A square is a rhombus with all angles equal to $90^\circ$. Since a square is also a cyclic quadrilateral, its vertices can lie on a circle.

---

Let the radius of the circle be $r$. The area of the circle is $\pi r^2$.

Given Area of circle = 1256 cm$^2$ and $\pi = 3.14$:

$1256 = 3.14 \times r^2$

$r^2 = \frac{1256}{3.14}$

$r^2 = \frac{125600}{314}$

$r^2 = 400$

Radius, $r = \sqrt{400} = 20$ cm.

---

Since the rhombus is a square and its vertices lie on the circle, the circle is the circumcircle of the square. The diagonal of the square is equal to the diameter of the circumcircle.

Diameter of the circle, $d = 2r = 2 \times 20 = 40$ cm.

Diagonal of the square = Diameter of the circle $= 40$ cm.

---

Let the side length of the square (rhombus) be $s$. In a square, the diagonal $d = s\sqrt{2}$.

$40 = s\sqrt{2}$

$s = \frac{40}{\sqrt{2}} = \frac{40\sqrt{2}}{2} = 20\sqrt{2}$ cm.

---

The area of a square can be found using the formula Area $= s^2$ or Area $= \frac{1}{2}d^2$.

Using the diagonal $d = 40$ cm:

Area of rhombus (square) $= \frac{1}{2} d^2 = \frac{1}{2} (40)^2 = \frac{1}{2} \times 1600 = 800$ cm$^2$.

---

Alternatively, using the side length $s = 20\sqrt{2}$ cm:

Area of rhombus (square) $= s^2 = (20\sqrt{2})^2 = 20^2 \times (\sqrt{2})^2 = 400 \times 2 = 800$ cm$^2$.

---

The area of the rhombus is 800 cm$^2$.

Given:

An archery target with three regions formed by three concentric circles.

The ratio of the diameters of the concentric circles is 1 : 2 : 3.

---

To Find:

The ratio of the areas of the three regions.

---

Solution:

Let the diameters of the three concentric circles be $d_1, d_2, d_3$.

Given $d_1 : d_2 : d_3 = 1 : 2 : 3$.

Let the radii of the three concentric circles be $R_1, R_2, R_3$. The ratio of radii is the same as the ratio of diameters.

$R_1 : R_2 : R_3 = 1 : 2 : 3$.

Let $R_1 = r$ for some unit radius $r$. Then $R_2 = 2r$ and $R_3 = 3r$.

---

The three regions are defined as follows:

Region 1: The innermost circle with radius $R_1$.

Region 2: The area between the first circle and the second circle (an annulus) with radii $R_1$ and $R_2$.

Region 3: The area between the second circle and the third circle (an annulus) with radii $R_2$ and $R_3$.

---

The area of the first circle is $A_{\text{circle } 1} = \pi R_1^2 = \pi (r)^2 = \pi r^2$.

This is the area of Region 1.

Area(Region 1) $= \pi r^2$.

---

The area of the second circle is $A_{\text{circle } 2} = \pi R_2^2 = \pi (2r)^2 = 4\pi r^2$.

The area of the third circle is $A_{\text{circle } 3} = \pi R_3^2 = \pi (3r)^2 = 9\pi r^2$.

---

The area of Region 2 is the area of the second circle minus the area of the first circle.

Area(Region 2) $= A_{\text{circle } 2} - A_{\text{circle } 1} = 4\pi r^2 - \pi r^2 = 3\pi r^2$.

---

The area of Region 3 is the area of the third circle minus the area of the second circle.

Area(Region 3) $= A_{\text{circle } 3} - A_{\text{circle } 2} = 9\pi r^2 - 4\pi r^2 = 5\pi r^2$.

---

The ratio of the areas of the three regions is:

Area(Region 1) : Area(Region 2) : Area(Region 3)

$\pi r^2 : 3\pi r^2 : 5\pi r^2$

Dividing by $\pi r^2$ (since $r > 0$ for a circle):

$1 : 3 : 5$

---

The ratio of the areas of the three regions is 1 : 3 : 5.

Given:

Length of the minute hand (radius of the circle), $r = 5$ cm.

Time period = from 6 : 05 a m to 6 : 40 a m.

---

To Find:

The area swept by the minute hand during the given time period.

---

Solution:

The time duration between 6 : 05 a m and 6 : 40 a m is 35 minutes (6:40 - 6:05).

---

The minute hand of a clock completes a full circle ($360^\circ$) in 60 minutes.

The angle swept by the minute hand in 1 minute is $\frac{360^\circ}{60} = 6^\circ$.

---

The angle swept by the minute hand in 35 minutes is:

Central angle, $\theta = 35 \times 6^\circ = 210^\circ$.

---

The area swept by the minute hand is the area of a sector of the circle with radius $r = 5$ cm and central angle $\theta = 210^\circ$.

The formula for the area of a sector is:

$A = \frac{\theta}{360^\circ} \times \pi r^2$

---

Substitute the values $r = 5$ cm and $\theta = 210^\circ$ into the formula:

$A = \frac{210^\circ}{360^\circ} \times \pi (5)^2$

---

Simplify the fraction $\frac{210}{360}$:

$\frac{\cancel{210}^{7}}{\cancel{360}_{12}} = \frac{7}{12}$

---

So, the area becomes:

$A = \frac{7}{12} \times \pi \times 25$

$A = \frac{175\pi}{12}$ cm$^2$.

---

The area swept by the minute hand is $\frac{175\pi}{12}$ cm$^2$.

(Note: If a numerical value for $\pi$ were provided, the answer could be given as a decimal value).

Given:

Area of the sector, $A = 770$ cm$^2$.

Central angle of the sector, $\theta = 200^\circ$.

---

To Find:

The length of the corresponding arc, $l$.

---

Solution:

We can relate the area of a sector, its central angle, and the radius using the formula:

$A = \frac{\theta}{360^\circ} \times \pi r^2$

We can also relate the arc length, central angle, and radius using the formula:

$l = \frac{\theta}{360^\circ} \times 2\pi r$

---

From the area formula, we can first find the radius $r$. Substitute the given values $A = 770$ and $\theta = 200^\circ$. We will use $\pi = \frac{22}{7}$.

$770 = \frac{200^\circ}{360^\circ} \times \frac{22}{7} \times r^2$

---

Simplify the fraction $\frac{200}{360}$:

$\frac{\cancel{200}^{10}}{\cancel{360}_{18}} = \frac{10}{18} = \frac{5}{9}$

---

The equation becomes:

$770 = \frac{5}{9} \times \frac{22}{7} \times r^2$

$770 = \frac{110}{63} r^2$

---

Solve for $r^2$:

$r^2 = \frac{770 \times 63}{110}$

$r^2 = \frac{\cancel{770}^{7} \times 63}{\cancel{110}^{1}}$

$r^2 = 7 \times 63 = 441$

Radius, $r = \sqrt{441} = 21$ cm.

---

Now, use the arc length formula with $r = 21$ cm and $\theta = 200^\circ$ and $\pi = \frac{22}{7}$.

$l = \frac{200^\circ}{360^\circ} \times 2\pi r$

$l = \frac{5}{9} \times 2 \times \frac{22}{7} \times 21$

---

Simplify the expression:

$l = \frac{5}{9} \times 2 \times \frac{22}{\cancel{7}^{1}} \times \cancel{21}^{3}$

$l = \frac{5}{9} \times 2 \times 22 \times 3$

$l = \frac{5}{\cancel{9}^{3}} \times 2 \times 22 \times \cancel{3}^{1}$

$l = \frac{5}{3} \times 2 \times 22$

$l = \frac{10 \times 22}{3} = \frac{220}{3}$

---

The length of the corresponding arc is $\frac{220}{3}$ cm.

---

Alternate Method:

The area of a sector can also be related to the arc length and radius by the formula:

$A = \frac{1}{2} l r$

We know $A = 770$ and we found $r = 21$. We can use this formula to find $l$ directly.

$770 = \frac{1}{2} \times l \times 21$

$770 = \frac{21}{2} l$

---

Solve for $l$:

$l = \frac{770 \times 2}{21}$

$l = \frac{\cancel{770}^{110} \times 2}{\cancel{21}^{3}}$

$l = \frac{110 \times 2}{3} = \frac{220}{3}$

---

This confirms the length of the corresponding arc is $\frac{220}{3}$ cm.

Given:

For Sector 1: Radius $r_1 = 7$ cm, Central angle $\theta_1 = 120^\circ$.

For Sector 2: Radius $r_2 = 21$ cm, Central angle $\theta_2 = 40^\circ$.

---

To Find:

Areas of the two sectors and the lengths of the corresponding arcs. Observation.

---

Solution:

We will use the formula for the area of a sector $A = \frac{\theta}{360^\circ} \times \pi r^2$ and the formula for the length of an arc $l = \frac{\theta}{360^\circ} \times 2\pi r$. We will use $\pi = \frac{22}{7}$.

---

For Sector 1:

Radius, $r_1 = 7$ cm

Central angle, $\theta_1 = 120^\circ$ (in degrees)

---

Area of Sector 1, $A_1 = \frac{\theta_1}{360^\circ} \times \pi r_1^2$

$A_1 = \frac{120^\circ}{360^\circ} \times \frac{22}{7} \times (7)^2$

$A_1 = \frac{1}{3} \times \frac{22}{7} \times 49$

$A_1 = \frac{1}{3} \times 22 \times \frac{\cancel{49}^{7}}{\cancel{7}^{1}}$

$A_1 = \frac{1}{3} \times 22 \times 7 = \frac{154}{3}$ cm$^2$.

---

Length of Arc 1, $l_1 = \frac{\theta_1}{360^\circ} \times 2\pi r_1$

$l_1 = \frac{120^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 7$

$l_1 = \frac{1}{3} \times 2 \times \frac{22}{\cancel{7}^{1}} \times \cancel{7}^{1}$

$l_1 = \frac{1}{3} \times 2 \times 22 = \frac{44}{3}$ cm.

---

For Sector 2:

Radius, $r_2 = 21$ cm

Central angle, $\theta_2 = 40^\circ$ (in degrees)

---

Area of Sector 2, $A_2 = \frac{\theta_2}{360^\circ} \times \pi r_2^2$

$A_2 = \frac{40^\circ}{360^\circ} \times \frac{22}{7} \times (21)^2$

$A_2 = \frac{1}{9} \times \frac{22}{7} \times 441$

$A_2 = \frac{1}{9} \times 22 \times \frac{\cancel{441}^{63}}{\cancel{7}^{1}}$

$A_2 = \frac{1}{\cancel{9}^{1}} \times 22 \times \cancel{63}^{7} = 22 \times 7 = 154$ cm$^2$.

---

Length of Arc 2, $l_2 = \frac{\theta_2}{360^\circ} \times 2\pi r_2$

$l_2 = \frac{40^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21$

$l_2 = \frac{1}{9} \times 2 \times \frac{22}{\cancel{7}^{1}} \times \cancel{21}^{3}$

$l_2 = \frac{1}{\cancel{9}^{3}} \times 2 \times 22 \times \cancel{3}^{1}$

$l_2 = \frac{1}{3} \times 2 \times 22 = \frac{44}{3}$ cm.

---

Summary of Results:

Area of Sector 1 = $\frac{154}{3}$ cm$^2$

Length of Arc 1 = $\frac{44}{3}$ cm

Area of Sector 2 = $154$ cm$^2$

Length of Arc 2 = $\frac{44}{3}$ cm

---

Observation:

We observe that the lengths of the corresponding arcs of the two sectors are equal ($l_1 = l_2 = \frac{44}{3}$ cm), but the areas of the two sectors are not equal ($A_1 = \frac{154}{3}$ cm$^2$ and $A_2 = 154$ cm$^2$).

Given:

From the figure, a square ABCD is shown with side length 14 cm.

Four semi-circles are drawn on each side of the square as diameter, pointing outwards.

The shaded region consists of the areas of these four semi-circles.

---

To Find:

The area of the shaded region.

---

Solution:

The shaded region is composed of four semi-circles, each drawn on a side of the square ABCD as its diameter.

---

The side length of the square is $s = 14$ cm.

The diameter of each semi-circle is equal to the side length of the square, $d = 14$ cm.

The radius of each semi-circle is $r = \frac{d}{2} = \frac{14}{2} = 7$ cm.

---

The area of one semi-circle with radius $r$ is given by $\frac{1}{2}\pi r^2$.

Area of one semi-circle $= \frac{1}{2} \pi (7)^2 = \frac{1}{2} \pi \times 49 = \frac{49\pi}{2}$ cm$^2$.

---

There are four such semi-circles. The total area of the shaded region is the sum of the areas of the four semi-circles.

Total shaded area $= 4 \times \text{Area of one semi-circle}$

Total shaded area $= 4 \times \frac{49\pi}{2} = 2 \times 49\pi = 98\pi$ cm$^2$.

---

Using $\pi = \frac{22}{7}$:

Total shaded area $= 98 \times \frac{22}{7} = \cancel{98}^{14} \times \frac{22}{\cancel{7}^{1}} = 14 \times 22 = 308$ cm$^2$.

---

The area of the shaded region is 308 cm$^2$.

Given:

Area of the circular wheel = 1.54 m$^2$.

Total distance rolled = 176 m.

---

To Find:

The number of revolutions made by the wheel.

---

Solution:

The distance covered by a circular wheel in one revolution is equal to its circumference. To find the number of revolutions, we need to divide the total distance rolled by the circumference of the wheel.

Number of revolutions $= \frac{\text{Total distance rolled}}{\text{Circumference}}$

---

First, find the radius of the wheel from its area.

Area of wheel $= \pi r^2$

$1.54 = \pi r^2$

Using $\pi = \frac{22}{7}$:

$1.54 = \frac{22}{7} r^2$

Convert 1.54 to a fraction: $1.54 = \frac{154}{100}$.

$\frac{154}{100} = \frac{22}{7} r^2$

---

Solve for $r^2$:

$r^2 = \frac{154}{100} \times \frac{7}{22}$

$r^2 = \frac{\cancel{154}^{7}}{100} \times \frac{7}{\cancel{22}^{1}}$

$r^2 = \frac{7 \times 7}{100} = \frac{49}{100}$

Radius, $r = \sqrt{\frac{49}{100}} = \frac{\sqrt{49}}{\sqrt{100}} = \frac{7}{10} = 0.7$ m.

---

Now, find the circumference of the wheel.

Circumference, $C = 2\pi r$

Using $r = 0.7$ m and $\pi = \frac{22}{7}$:

$C = 2 \times \frac{22}{7} \times 0.7$

$C = 2 \times \frac{22}{7} \times \frac{7}{10}$

$C = 2 \times \frac{\cancel{22}^{11}}{\cancel{7}^{1}} \times \frac{\cancel{7}^{1}}{\cancel{10}^{5}}$

$C = 2 \times \frac{11}{5} = \frac{22}{5} = 4.4$ m.

---

Now, calculate the number of revolutions.

Total distance rolled = 176 m.

Circumference = 4.4 m.

Number of revolutions $= \frac{176}{4.4} = \frac{1760}{44}$.

Number of revolutions $= \frac{\cancel{1760}^{40}}{\cancel{44}^{1}} = 40$.

---

The number of revolutions made by the circular wheel is 40.

Given:

Length of the chord = 5 cm.

Central angle subtended by the chord, $\theta = 90^\circ$.

---

To Find:

The difference between the areas of the major segment and the minor segment.

---

Solution:

Let the radius of the circle be $r$. Let the center of the circle be O, and the chord be AB. In the triangle OAB, OA = OB = $r$, and $\angle$AOB = $90^\circ$. $\triangle$OAB is a right-angled isosceles triangle with the right angle at O.

---

Using the Pythagorean theorem in $\triangle$OAB:

$AB^2 = OA^2 + OB^2$

$5^2 = r^2 + r^2$

$25 = 2r^2$

$r^2 = \frac{25}{2}$

Radius, $r = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2}$ cm.

---

The area of the triangle OAB is:

Area($\triangle$OAB) $= \frac{1}{2} \times OA \times OB \times \sin(\angle\text{AOB})$

Area($\triangle$OAB) $= \frac{1}{2} \times r \times r \times \sin(90^\circ)$

Area($\triangle$OAB) $= \frac{1}{2} r^2 \times 1 = \frac{1}{2} r^2$

Substitute $r^2 = \frac{25}{2}$:

Area($\triangle$OAB) $= \frac{1}{2} \times \frac{25}{2} = \frac{25}{4}$ cm$^2$.

---

The area of the sector corresponding to the $90^\circ$ angle is:

Area of sector $= \frac{\theta}{360^\circ} \times \pi r^2 = \frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \pi r^2$

Substitute $r^2 = \frac{25}{2}$:

Area of sector $= \frac{1}{4} \pi \left(\frac{25}{2}\right) = \frac{25\pi}{8}$ cm$^2$.

---

The area of the minor segment is the area of the sector minus the area of the triangle:

Area of minor segment $= \text{Area of sector} - \text{Area}(\triangle\text{OAB})$

Area of minor segment $= \frac{25\pi}{8} - \frac{25}{4}$ cm$^2$.

---

The area of the entire circle is:

Area of circle $= \pi r^2 = \pi \left(\frac{25}{2}\right) = \frac{25\pi}{2}$ cm$^2$.

---

The area of the major segment is the area of the circle minus the area of the minor segment:

Area of major segment $= \text{Area of circle} - \text{Area of minor segment}$

Area of major segment $= \frac{25\pi}{2} - \left(\frac{25\pi}{8} - \frac{25}{4}\right)$

Area of major segment $= \frac{25\pi}{2} - \frac{25\pi}{8} + \frac{25}{4}$

Area of major segment $= \frac{100\pi - 25\pi}{8} + \frac{25}{4} = \frac{75\pi}{8} + \frac{25}{4}$ cm$^2$.

---

The difference between the areas of the two segments is the area of the major segment minus the area of the minor segment.

Difference = Area of major segment - Area of minor segment

Difference $= \left(\frac{75\pi}{8} + \frac{25}{4}\right) - \left(\frac{25\pi}{8} - \frac{25}{4}\right)$

Difference $= \frac{75\pi}{8} + \frac{25}{4} - \frac{25\pi}{8} + \frac{25}{4}$

Difference $= \left(\frac{75\pi}{8} - \frac{25\pi}{8}\right) + \left(\frac{25}{4} + \frac{25}{4}\right)$

Difference $= \frac{50\pi}{8} + \frac{50}{4}$

Difference $= \frac{25\pi}{4} + \frac{25}{2}$ cm$^2$.

---

Alternatively, the difference between the major and minor segment areas is also given by: Area(Circle) - 2 $\times$ Area(Minor Segment).

Difference $= \frac{25\pi}{2} - 2 \times \left(\frac{25\pi}{8} - \frac{25}{4}\right)$

Difference $= \frac{25\pi}{2} - \frac{50\pi}{8} + \frac{50}{4}$

Difference $= \frac{25\pi}{2} - \frac{25\pi}{4} + \frac{25}{2}$

Difference $= \frac{50\pi - 25\pi}{4} + \frac{25}{2} = \frac{25\pi}{4} + \frac{25}{2}$ cm$^2$.

---

The difference of the areas of the two segments is $\left(\frac{25\pi}{4} + \frac{25}{2}\right)$ cm$^2$.

(Note: The problem does not specify which value of $\pi$ to use, so the answer is left in terms of $\pi$.)

Given:

Radius of the circle, $r = 21$ cm.

Central angle of the minor sector, $\theta_{\text{minor}} = 120^\circ$.

---

To Find:

The difference between the area of the major sector and the minor sector.

---

Solution:

The area of a sector with radius $r$ and central angle $\theta$ (in degrees) is given by $A = \frac{\theta}{360^\circ} \times \pi r^2$.

---

The central angle of the corresponding major sector is:

$\theta_{\text{major}} = 360^\circ - \theta_{\text{minor}} = 360^\circ - 120^\circ = 240^\circ$.

---

Area of the minor sector:

$A_{\text{minor}} = \frac{120^\circ}{360^\circ} \times \pi (21)^2 = \frac{1}{3} \times \pi \times 441 = 147\pi$ cm$^2$.

---

Area of the major sector:

$A_{\text{major}} = \frac{240^\circ}{360^\circ} \times \pi (21)^2 = \frac{2}{3} \times \pi \times 441 = 2 \times 147\pi = 294\pi$ cm$^2$.

---

The difference between the areas of the major sector and the minor sector is:

Difference $= A_{\text{major}} - A_{\text{minor}}$

Difference $= 294\pi - 147\pi = 147\pi$ cm$^2$.

---

Using $\pi = \frac{22}{7}$:

Difference $= 147 \times \frac{22}{7} = \cancel{147}^{21} \times \frac{22}{\cancel{7}^{1}} = 21 \times 22$

$21 \times 22 = 462$

---

The difference of the areas of the two sectors is 462 cm$^2$.