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Chapter 10 Practical Geometry
Welcome to the solutions guide for Chapter 10: Practical Geometry. This chapter elevates the foundational construction skills acquired earlier, focusing specifically on the precise construction of various types of triangles based on given measurements. Practical Geometry is a hands-on discipline that bridges theoretical knowledge with tangible creation, demanding accuracy and a methodical approach. It moves beyond simple sketching, requiring the disciplined use of fundamental geometric instruments – primarily the straightedge (ruler) for drawing lines and measuring lengths, and the compasses for drawing arcs and transferring lengths. Mastering these constructions not only enhances drawing skills but also provides a concrete understanding of the geometric properties and congruence conditions that define triangles.
The solutions provided offer meticulous, step-by-step instructions, essential for achieving accuracy in construction. The core focus is on building triangles when specific sets of their elements (sides and angles) are known, directly corresponding to the triangle congruence criteria learned theoretically:
- Constructing a triangle given SSS (Side-Side-Side) Criterion:
When the lengths of all three sides are provided, the solutions guide you to first draw one side as the base. Then, using compasses, set the radius to the length of the second side and draw an arc from one endpoint of the base. Similarly, set the compass radius to the length of the third side and draw an arc from the other endpoint of the base. The precise point where these two arcs intersect forms the third vertex of the triangle. Joining this point to the endpoints of the base completes the required triangle.
- Constructing a triangle given SAS (Side-Angle-Side) Criterion:
Here, two sides and the angle included between them are given. The process involves drawing one of the given sides. At one of its endpoints, construct the specified angle using either compasses (for standard angles like $60^\circ$, $90^\circ$, etc.) or a protractor. Measure and mark the length of the second given side along the newly drawn ray of the angle. Finally, join the marked point to the other endpoint of the initial side to form the triangle.
- Constructing a triangle given ASA (Angle-Side-Angle) Criterion:
In this case, two angles and the side included between them are known. The construction starts by drawing the included side. At each endpoint of this side, construct the given angles using compasses or a protractor. Extend the rays forming these angles until they intersect. This intersection point is the third vertex of the triangle.
- Constructing a right-angled triangle given RHS (Right angle-Hypotenuse-Side) Criterion:
This special case for right-angled triangles requires the length of the hypotenuse and one leg (side). Begin by drawing the given leg. At one endpoint, construct a perpendicular line segment, forming a right angle ($90^\circ$). Now, set the compass radius to the length of the hypotenuse. Place the compass point at the other endpoint of the leg and draw an arc that intersects the perpendicular line. This intersection point is the third vertex. Joining this point to the endpoints completes the right-angled triangle.
Beyond triangle constructions, this chapter's solutions also typically cover another fundamental construction:
- Constructing a line parallel to a given line through an external point: Solutions demonstrate methods, often relying on creating equal corresponding angles or equal alternate interior angles using a transversal and compass-based angle copying techniques, ensuring the resulting line is parallel to the original.
Each construction detailed in the solutions is presented as a clear, sequential algorithm. Following these steps precisely, maintaining neatness in drawing, and ensuring the accurate use of the ruler and compasses are paramount. These practical exercises serve as a powerful reinforcement of the theoretical concepts of triangle properties and the conditions required for congruence, solidifying geometric understanding through active creation.
Exercise 10.1
Question 1. Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using ruler and compasses only.
Answer:
To Construct:
A line through a given point C, parallel to a given line AB.
Steps of Construction:
1. Draw a line and label it AB. Mark a point C outside this line.
2. Choose any point D on the line AB and join C to D. This forms a transversal line.
3. With D as the centre and a convenient radius, draw an arc that cuts AB at point E and CD at point F.
4. With C as the centre and the same radius as in the previous step, draw an arc, say arc GHI, that cuts the line segment CD at G.
5. Place the compass point at F and adjust the opening to the width of the arc EF.
6. With G as the centre and the same opening as in the previous step, draw an arc to cut arc GHI at point J.
7. Draw a line passing through points C and J. This line, let's call it 'm', is the required line parallel to AB.
Justification:
In this construction, we have copied the angle $\angle CDF$ to the position of $\angle GCJ$. This makes $\angle CDF = \angle GCJ$. These two angles are a pair of alternate interior angles. When alternate interior angles are equal, the lines are parallel. Therefore, the constructed line m is parallel to the given line AB.
Question 2. Draw a line l. Draw a perpendicular to l at any point on l. On this perpendicular choose a point X, 4 cm away from l. Through X, draw a line m parallel to l.
Answer:
To Construct:
A line m parallel to a given line l at a distance of 4 cm from it.
Steps of Construction:
1. Draw a line and label it 'l'.
2. Choose any point P on line l.
3. Construct a perpendicular to l at P: With P as the centre, draw an arc that intersects line l at two points, A and B. With A and B as centres and a radius greater than AP, draw two arcs that intersect at a point Q. Join P and Q. The line PQ is perpendicular to l.
4. Locate point X: Using a ruler, measure 4 cm on your compass. Place the compass point at P and draw an arc on the perpendicular line PQ. Label this point X. Now, PX = 4 cm.
5. Construct a line m through X parallel to l: Now, construct another perpendicular at point X to the line PX. With X as the centre, draw an arc intersecting PX. Using this arc, construct a 90° angle at X (similar to step 3). Draw the line passing through X at this 90° angle. This is the required line 'm'.
Justification:
We have constructed line l and line m such that both are perpendicular to the same line, PQ. Lines that are perpendicular to the same transversal line are parallel to each other. Therefore, $m \parallel l$. The distance between the two lines is the length of the perpendicular segment PX, which is 4 cm.
Question 3. Let l be a line and P be a point not on l. Through P, draw a line m parallel to l. Now join P to any point Q on l. Choose any other point R on m. Through R, draw a line parallel to PQ. Let this meet l at S. What shape do the two sets of parallel lines enclose?
Answer:
Construction and Analysis:
Steps of Construction:
1. Draw a line l and a point P not on it.
2. Draw a line m through P such that $m \parallel l$.
3. Choose any point Q on line l and draw the line segment PQ.
4. Choose another point R on line m.
5. Through R, draw a line parallel to PQ. This is done by copying the angle $\angle PQA$ (where A is a point on l) to the position of $\angle R S B$ (where B is on m) or by copying the angle $\angle Q P C$ (where C is on m) to the position of $\angle S R D$ (where D is on l), ensuring they are corresponding or alternate interior angles. Let this new line intersect line l at point S.
Analysis of the Enclosed Shape:
The figure enclosed by the two sets of parallel lines is the quadrilateral PQSR.
By construction, we have:
1. The line m (which contains the side PR) is parallel to the line l (which contains the side QS). Therefore, PR $\parallel$ QS.
2. The line drawn through R was constructed to be parallel to the line PQ. Therefore, RS $\parallel$ PQ.
A quadrilateral in which both pairs of opposite sides are parallel is, by definition, a parallelogram.
Conclusion:
The shape enclosed by the two sets of parallel lines is a parallelogram.
Example 1 (Before Exercise 10.2)
Example 1. Construct a triangle ABC, given that AB = 5 cm, BC = 6 cm and AC = 7 cm.
Answer:
Given:
The lengths of the three sides of a triangle ABC:
AB = 5 cm
BC = 6 cm
AC = 7 cm
To Construct:
A triangle ABC with the given side lengths.
Steps of Construction:
1. Draw a line segment BC of length 6 cm using a ruler.
2. Open the compasses to a radius of 5 cm (length of AB).
3. Place the compass pointer at point B and draw an arc.
4. Open the compasses to a radius of 7 cm (length of AC).
5. Place the compass pointer at point C and draw another arc, intersecting the arc drawn in step 3 at point A.
6. Join point A to point B using a ruler. (AB = 5 cm)
7. Join point A to point C using a ruler. (AC = 7 cm)
8. The triangle ABC obtained is the required triangle with sides AB = 5 cm, BC = 6 cm, and AC = 7 cm.
Exercise 10.2
Question 1. Construct ∆XYZ in which XY= 4.5 cm, YZ = 5 cm and ZX = 6 cm.
Answer:
Given:
The lengths of the three sides of a triangle XYZ:
XY = 4.5 cm
YZ = 5 cm
ZX = 6 cm
To Construct:
A triangle XYZ with the given side lengths.
Steps of Construction:
1. Draw a line segment YZ of length 5 cm using a ruler.
2. Open the compasses to a radius of 4.5 cm (length of XY).
3. Place the compass pointer at point Y and draw an arc.
4. Open the compasses to a radius of 6 cm (length of ZX).
5. Place the compass pointer at point Z and draw another arc, intersecting the arc drawn in step 3 at point X.
6. Join point X to point Y using a ruler. (XY = 4.5 cm)
7. Join point X to point Z using a ruler. (ZX = 6 cm)
8. The triangle XYZ obtained is the required triangle with sides XY = 4.5 cm, YZ = 5 cm, and ZX = 6 cm.
Question 2. Construct an equilateral triangle of side 5.5 cm.
Answer:
Given:
The side length of an equilateral triangle is 5.5 cm.
In an equilateral triangle, all three sides are equal. Let the triangle be ABC. Then AB = BC = AC = 5.5 cm.
To Construct:
An equilateral triangle with side length 5.5 cm.
Steps of Construction:
1. Draw a line segment BC of length 5.5 cm using a ruler.
2. Open the compasses to a radius of 5.5 cm.
3. Place the compass pointer at point B and draw an arc.
4. Place the compass pointer at point C (keeping the same radius of 5.5 cm) and draw another arc, intersecting the arc drawn in step 3 at point A.
5. Join point A to point B using a ruler. (AB = 5.5 cm)
6. Join point A to point C using a ruler. (AC = 5.5 cm)
7. The triangle ABC obtained is the required equilateral triangle with all sides equal to 5.5 cm.
Question 3. Draw ∆PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?
Answer:
Given:
The lengths of the three sides of a triangle PQR:
PQ = 4 cm
QR = 3.5 cm
PR = 4 cm
To Construct:
A triangle PQR with the given side lengths.
Steps of Construction:
1. Draw a line segment QR of length 3.5 cm using a ruler.
2. Open the compasses to a radius of 4 cm (length of PQ).
3. Place the compass pointer at point Q and draw an arc.
4. Open the compasses to a radius of 4 cm (length of PR).
5. Place the compass pointer at point R and draw another arc, intersecting the arc drawn in step 3 at point P.
6. Join point P to point Q using a ruler. (PQ = 4 cm)
7. Join point P to point R using a ruler. (PR = 4 cm)
8. The triangle PQR obtained is the required triangle.
Type of Triangle:
To determine the type of triangle, we examine the lengths of its sides:
PQ = 4 cm
QR = 3.5 cm
PR = 4 cm
Since two sides of the triangle, PQ and PR, have equal lengths (4 cm), the triangle is an isosceles triangle.
Question 4. Construct ∆ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.
Answer:
Given:
The lengths of the three sides of a triangle ABC:
AB = 2.5 cm
BC = 6 cm
AC = 6.5 cm
To Construct and Measure:
1. Construct a triangle ABC with the given side lengths.
2. Measure the angle ∠B.
Steps of Construction:
1. Draw a line segment BC of length 6 cm using a ruler.
2. Open the compasses to a radius of 2.5 cm (length of AB).
3. Place the compass pointer at point B and draw an arc.
4. Open the compasses to a radius of 6.5 cm (length of AC).
5. Place the compass pointer at point C and draw another arc, intersecting the arc drawn in step 3 at point A.
6. Join point A to point B using a ruler. (AB = 2.5 cm)
7. Join point A to point C using a ruler. (AC = 6.5 cm)
8. The triangle ABC obtained is the required triangle.
Measurement of ∠B:
1. Place the centre of the protractor at the vertex B.
2. Align the base line of the protractor with the arm BC.
3. Read the angle measure where the arm BA crosses the protractor scale.
4. On measuring ∠B with a protractor, we find that ∠B = 90°.
Note: We can verify this using the converse of the Pythagorean theorem. Check if $AC^2 = AB^2 + BC^2$.
$AC^2 = (6.5)^2 = 42.25$
$AB^2 + BC^2 = (2.5)^2 + (6)^2 = 6.25 + 36 = 42.25$
Since $AC^2 = AB^2 + BC^2$, the triangle ABC is a right-angled triangle with the right angle at B.
Example 2 (Before Exercise 10.3)
Example 2. Construct a triangle PQR, given that PQ = 3 cm, QR = 5.5 cm and ∠PQR = 60°.
Answer:
Given:
The lengths of two sides and the included angle of a triangle PQR:
PQ = 3 cm
QR = 5.5 cm
∠PQR = 60°
To Construct:
A triangle PQR with the given measurements (SAS criterion).
Steps of Construction:
1. Draw a line segment QR of length 5.5 cm using a ruler.
2. Construct an angle of 60° at point Q:
a. With Q as the centre and any convenient radius, draw an arc that intersects QR at point A.
b. With A as the centre and the same radius, draw another arc intersecting the first arc at point B.
c. Draw a ray QX passing through point B. The angle ∠XQR is 60°.
3. Open the compasses to a radius of 3 cm (length of PQ).
4. Place the compass pointer at point Q and draw an arc intersecting the ray QX at point P.
5. Join point P to point R using a ruler.
6. The triangle PQR obtained is the required triangle with PQ = 3 cm, QR = 5.5 cm, and ∠PQR = 60°.
Exercise 10.3
Question 1. Construct ∆DEF such that DE = 5 cm, DF = 3 cm and m∠EDF = 90°.
Answer:
Given:
The lengths of two sides and the included angle of a triangle DEF:
DE = 5 cm
DF = 3 cm
m∠EDF = 90°
To Construct:
A triangle DEF with the given measurements (SAS criterion).
Steps of Construction:
1. Draw a line segment DE of length 5 cm using a ruler.
2. Construct an angle of 90° at point D:
a. With D as the centre and any convenient radius, draw a semicircle that intersects DE.
b. From the intersection point on DE, draw two arcs on the semicircle to mark 60° and 120°.
c. From the 60° and 120° marks, draw two arcs that intersect above the semicircle.
d. Draw a ray DX passing through this intersection point. The angle ∠XDE is 90°.
3. Open the compasses to a radius of 3 cm (length of DF).
4. Place the compass pointer at point D and draw an arc intersecting the ray DX at point F.
5. Join point F to point E using a ruler.
6. The triangle DEF obtained is the required triangle with DE = 5 cm, DF = 3 cm, and m∠EDF = 90°.
Question 2. Construct an isosceles triangle in which the lengths of each of its equal sides is 6.5 cm and the angle between them is 110°.
Answer:
Given:
An isosceles triangle (let's call it ABC) with:
Length of equal sides = 6.5 cm (Let AB = BC = 6.5 cm)
Angle between the equal sides = 110° (Let ∠B = 110°)
To Construct:
An isosceles triangle ABC with the given measurements.
Steps of Construction:
1. Draw a line segment BC of length 6.5 cm using a ruler.
2. Place the protractor at point B and align its base line with BC.
3. Mark a point corresponding to 110° on the protractor.
4. Draw a ray BX passing through the marked point. The angle ∠XBC is 110°.
5. Open the compasses to a radius of 6.5 cm (length of the other equal side, AB).
6. Place the compass pointer at point B and draw an arc intersecting the ray BX at point A.
7. Join point A to point C using a ruler.
8. The triangle ABC obtained is the required isosceles triangle with AB = 6.5 cm, BC = 6.5 cm, and ∠B = 110°.
Question 3. Construct ∆ABC with BC = 7.5 cm, AC = 5 cm and m∠C = 60°.
Answer:
Given:
The lengths of two sides and the included angle of a triangle ABC:
$BC = 7.5$ cm
$AC = 5$ cm
$m\angle C = 60°$
To Construct:
A triangle ABC with the given measurements (SAS criterion).
Steps of Construction:
1. Draw a line segment BC of length 7.5 cm using a ruler.
2. Construct an angle of 60° at point C:
a. With C as the centre and any convenient radius, draw an arc that intersects BC at point D.
b. With D as the centre and the same radius, draw another arc intersecting the first arc at point E.
c. Draw a ray CX passing through point E. The angle ∠XCB is 60°.
3. Open the compasses to a radius of 5 cm (length of AC).
4. Place the compass pointer at point C and draw an arc intersecting the ray CX at point A.
5. Join point A to point B using a ruler.
6. The triangle ABC obtained is the required triangle with BC = 7.5 cm, AC = 5 cm, and m∠C = 60°.
Example 3 (Before Exercise 10.4)
Example 3. Construct ∆XYZ if it is given that XY = 6 cm, m∠ZXY = 30° and m∠XYZ = 100°.
Answer:
Given:
The length of one side and the two angles at its endpoints for triangle XYZ:
$XY = 6$ cm
$m\angle ZXY = 30°$ (Angle at vertex X)
$m\angle XYZ = 100°$ (Angle at vertex Y)
To Construct:
A triangle XYZ with the given measurements (ASA criterion).
Steps of Construction:
1. Draw a line segment XY of length 6 cm using a ruler.
2. Construct an angle of 30° at point X:
a. Construct a 60° angle at X: With X as the centre, draw an arc intersecting XY at A. With A as the centre and the same radius, cut the arc at B. The ray from X through B makes a 60° angle.
b. Bisect the 60° angle: With A and B as centres and a radius greater than half of AB, draw arcs intersecting at C.
c. Draw a ray XP passing through C. The angle ∠PXY is 30°.
3. Construct an angle of 100° at point Y:
a. Place the centre of the protractor at point Y and align its base line with YX.
b. Mark a point corresponding to 100° on the protractor.
c. Draw a ray YQ passing through the marked point. The angle ∠QYX is 100°.
4. Extend the rays XP and YQ until they intersect each other. Mark the point of intersection as Z.
5. The triangle XYZ obtained is the required triangle with XY = 6 cm, m∠ZXY = 30°, and m∠XYZ = 100°.
Exercise 10.4
Question 1. Construct ∆ABC, given m∠A = 60°, m∠B = 30° and AB = 5.8 cm.
Answer:
Given:
For triangle ABC:
$m\angle A = 60°$
$m\angle B = 30°$
$AB = 5.8$ cm
To Construct:
A triangle ABC with the given measurements (ASA criterion).
Steps of Construction:
1. Draw a line segment AB of length 5.8 cm using a ruler.
2. Construct an angle of 60° at point A:
a. With A as the centre and any convenient radius, draw an arc intersecting AB at point D.
b. With D as the centre and the same radius, draw another arc intersecting the first arc at point E.
c. Draw a ray AX passing through point E. The angle ∠XAB is 60°.
3. Construct an angle of 30° at point B:
a. Construct a 60° angle at B first. Then, bisect this angle to get 30°.
b. Draw a ray BY making an angle of 30° with BA. The angle ∠YBA is 30°.
4. Let the rays AX and BY intersect at point C.
5. The triangle ABC obtained is the required triangle with m∠A = 60°, m∠B = 30°, and AB = 5.8 cm.
Question 2. Construct ∆PQR if PQ = 5 cm, m∠PQR = 105° and m∠QRP = 40°.
(Hint: Recall angle-sum property of a triangle).
Answer:
Given:
For triangle PQR:
$PQ = 5$ cm
$m\angle PQR = 105°$ (Angle at Q)
$m\angle QRP = 40°$ (Angle at R)
To Construct:
A triangle PQR with the given measurements.
Analysis using Angle Sum Property:
We are given one side (PQ) and the angle at one endpoint (∠PQR at Q). To construct the triangle using the ASA criterion, we need the angle at the other endpoint of the side PQ, which is ∠QPR (angle at P).
We can find m∠QPR using the angle sum property of a triangle:
$m\angle QPR + m\angle PQR + m\angle QRP = 180°$
$m\angle QPR + 105° + 40° = 180°$
$m\angle QPR + 145° = 180°$
$m\angle QPR = 180° - 145°$
$m\angle QPR = 35°$
So, we need to construct a triangle PQR with PQ = 5 cm, m∠QPR = 35°, and m∠PQR = 105°.
Steps of Construction:
1. Draw a line segment PQ of length 5 cm using a ruler.
2. Construct an angle of 35° at point P:
a. Place the centre of the protractor at point P and align its base line with PQ.
b. Mark a point corresponding to 35° on the protractor.
c. Draw a ray PX passing through the marked point. The angle ∠XPQ is 35°.
3. Construct an angle of 105° at point Q:
a. Place the centre of the protractor at point Q and align its base line with QP.
b. Mark a point corresponding to 105° on the protractor.
c. Draw a ray QY passing through the marked point. The angle ∠YQP is 105°.
4. Let the rays PX and QY intersect at point R.
5. The triangle PQR obtained is the required triangle. By construction, PQ = 5 cm, m∠QPR = 35°, m∠PQR = 105°. The third angle, m∠QRP, will automatically be $180° - (35° + 105°) = 180° - 140° = 40°$, as required.
Question 3. Examine whether you can construct ∆DEF such that EF = 7.2 cm, m∠E = 110° and m∠F = 80°. Justify your answer.
Answer:
Given:
Potential measurements for a triangle DEF:
$EF = 7.2$ cm
$m\angle E = 110°$
$m\angle F = 80°$
Examination and Justification:
To determine if a triangle can be constructed with these measurements, we need to check if they satisfy the fundamental properties of a triangle.
One fundamental property is the Angle Sum Property, which states that the sum of the interior angles of any triangle must be exactly 180°.
Let's find the sum of the two given angles, m∠E and m∠F:
$m\angle E + m\angle F = 110° + 80° = 190°$
The sum of just these two angles is 190°, which is already greater than the required total sum of 180° for all three angles of a triangle.
According to the angle sum property:
$m\angle D + m\angle E + m\angle F = 180°$
If we substitute the given values:
$m\angle D + 110° + 80° = 180°$
$m\angle D + 190° = 180°$
$m\angle D = 180° - 190° = -10°$
This implies that the third angle, m∠D, would have to be -10°, which is impossible as angles in a triangle must be positive.
Conclusion:
No, we cannot construct a triangle ∆DEF with the given measurements because the sum of the two given angles (m∠E + m∠F = 190°) exceeds 180°, which violates the angle sum property of a triangle.
Example 4 (Before Exercise 10.5)
Example 4. Construct ∆LMN, right-angled at M, given that LN = 5 cm and MN = 3 cm.
Answer:
Given:
For triangle LMN:
Right-angled at M ($m\angle LMN = 90°$)
Length of hypotenuse LN = 5 cm
Length of leg MN = 3 cm
To Construct:
A right-angled triangle LMN with the given measurements (RHS criterion).
Steps of Construction:
1. Draw a line segment MN of length 3 cm using a ruler.
2. Construct a right angle (90°) at point M:
a. With M as the centre and any convenient radius, draw an arc intersecting MN at point A.
b. With A as the centre and the same radius, draw an arc intersecting the first arc at point B.
c. With B as the centre and the same radius, draw another arc intersecting the first arc (further along) at point C.
d. With B and C as centres and the same radius (or any radius greater than half the distance BC), draw arcs intersecting each other at point P.
e. Draw a ray MX passing through point P. The angle ∠XMN is 90°.
3. Open the compasses to a radius of 5 cm (length of the hypotenuse LN).
4. Place the compass pointer at point N and draw an arc intersecting the ray MX.
5. Mark the point of intersection as L.
6. Join point L to point N using a ruler.
7. The triangle LMN obtained is the required right-angled triangle with MN = 3 cm, LN = 5 cm, and m∠LMN = 90°.
Exercise 10.5
Question 1. Construct the right angled ∆PQR, where m∠Q = 90°, QR = 8cm and PR = 10 cm.
Answer:
Given:
For triangle PQR:
$m\angle Q = 90°$ (Right-angled at Q)
$QR = 8$ cm (Length of one leg)
$PR = 10$ cm (Length of the hypotenuse)
To Construct:
A right-angled triangle PQR with the given measurements (RHS criterion).
Steps of Construction:
1. Draw a line segment QR of length 8 cm using a ruler.
2. Construct a right angle (90°) at point Q:
a. With Q as the centre and any convenient radius, draw an arc intersecting QR at point A.
b. With A as the centre and the same radius, draw an arc intersecting the first arc at point B.
c. With B as the centre and the same radius, draw another arc intersecting the first arc (further along) at point C.
d. With B and C as centres and the same radius (or any radius greater than half the distance BC), draw arcs intersecting each other at point D.
e. Draw a ray QX passing through point D. The angle ∠XQR is 90°.
3. Open the compasses to a radius of 10 cm (length of the hypotenuse PR).
4. Place the compass pointer at point R and draw an arc intersecting the ray QX.
5. Mark the point of intersection as P.
6. Join point P to point R using a ruler.
7. The triangle PQR obtained is the required right-angled triangle with QR = 8 cm, PR = 10 cm, and m∠Q = 90°.
Question 2. Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.
Answer:
Given:
A right-angled triangle (let's call it ABC) with:
Hypotenuse length = 6 cm (Let AC = 6 cm)
Length of one leg = 4 cm (Let BC = 4 cm)
Let the right angle be at vertex B ($m\angle B = 90°$).
To Construct:
A right-angled triangle ABC with the given measurements (RHS criterion).
Steps of Construction:
1. Draw a line segment BC of length 4 cm using a ruler.
2. Construct a right angle (90°) at point B:
a. With B as the centre and any convenient radius, draw an arc intersecting BC at point D.
b. With D as the centre and the same radius, draw an arc intersecting the first arc at point E.
c. With E as the centre and the same radius, draw another arc intersecting the first arc (further along) at point F.
d. With E and F as centres and the same radius (or any radius greater than half the distance EF), draw arcs intersecting each other at point G.
e. Draw a ray BX passing through point G. The angle ∠XBC is 90°.
3. Open the compasses to a radius of 6 cm (length of the hypotenuse AC).
4. Place the compass pointer at point C and draw an arc intersecting the ray BX.
5. Mark the point of intersection as A.
6. Join point A to point C using a ruler.
7. The triangle ABC obtained is the required right-angled triangle with leg BC = 4 cm, hypotenuse AC = 6 cm, and m∠B = 90°.
Question 3. Construct an isosceles right-angled triangle ABC, where m∠ACB = 90° and AC = 6 cm.
Answer:
Given:
An isosceles right-angled triangle ABC with:
$m\angle ACB = 90°$ (Right angle at C)
$AC = 6$ cm (Length of one leg)
Since the triangle is isosceles and right-angled at C, the two legs adjacent to the right angle must be equal. Therefore, $AC = BC$.
So, we also have $BC = 6$ cm.
To Construct:
An isosceles right-angled triangle ABC with AC = 6 cm, BC = 6 cm, and m∠ACB = 90° (SAS criterion).
Steps of Construction:
1. Draw a line segment BC of length 6 cm using a ruler.
2. Construct a right angle (90°) at point C:
a. With C as the centre and any convenient radius, draw an arc intersecting BC at point D.
b. With D as the centre and the same radius, draw an arc intersecting the first arc at point E.
c. With E as the centre and the same radius, draw another arc intersecting the first arc (further along) at point F.
d. With E and F as centres and the same radius (or any radius greater than half the distance EF), draw arcs intersecting each other at point G.
e. Draw a ray CX passing through point G. The angle ∠XCB is 90°.
3. Open the compasses to a radius of 6 cm (length of the other leg AC).
4. Place the compass pointer at point C and draw an arc intersecting the ray CX.
5. Mark the point of intersection as A.
6. Join point A to point B using a ruler.
7. The triangle ABC obtained is the required isosceles right-angled triangle with AC = 6 cm, BC = 6 cm, and m∠ACB = 90°.
Miscellaneous questions
Below are given the measures of certain sides and angles of triangles. Identify those which cannot be constructed and, say why you cannot construct them. Construct rest of the triangles.
Triangle
1. ∆ABC
2. ∆PQR
3. ∆ABC
4. ∆LMN
5. ∆ABC
6. ∆PQR
7. ∆XYZ
8. ∆DEF
Given measurements
m∠A = 85°; m∠B = 115°; AB = 5 cm.
m∠Q = 30°; m∠R = 60°; QR = 4.7 cm.
m∠A = 70°; m∠B = 50°; AC = 3 cm.
m∠L = 60°; m∠N = 120°; LM = 5 cm.
BC = 2 cm; AB = 4 cm; AC = 2 cm.
PQ = 3.5 cm.; QR = 4 cm.; PR = 3.5 cm.
XY = 3 cm; YZ = 4 cm; XZ = 5 cm
DE = 4.5cm; EF = 5.5cm; DF = 4 cm.
Answer:
1. ∆ABC: m∠A = 85°; m∠B = 115°; AB = 5 cm.
Analysis: Check the angle sum property.
$m\angle A + m\angle B = 85° + 115° = 200°$
The sum of two angles is 200°, which is greater than 180°. This violates the angle sum property of a triangle.
Conclusion: ∆ABC cannot be constructed.
2. ∆PQR: m∠Q = 30°; m∠R = 60°; QR = 4.7 cm.
Analysis: Check the angle sum property.
$m\angle Q + m\angle R = 30° + 60° = 90°$
The sum is less than 180°, so a triangle is possible. $m\angle P = 180° - 90° = 90°$. The measurements satisfy the ASA criterion (Angle-Side-Angle) using side QR and angles ∠Q and ∠R.
Conclusion: ∆PQR can be constructed.
Steps of Construction:
1. Draw a line segment QR of length 4.7 cm.
2. At point Q, construct an angle of 30° (e.g., construct 60° and bisect it). Draw a ray QX making this angle with QR.
3. At point R, construct an angle of 60°. Draw a ray RY making this angle with RQ.
4. Let the rays QX and RY intersect at point P.
5. ∆PQR is the required triangle.
3. ∆ABC: m∠A = 70°; m∠B = 50°; AC = 3 cm.
Analysis: Check the angle sum property.
$m\angle A + m\angle B = 70° + 50° = 120°$
The sum is less than 180°, so a triangle is possible. $m\angle C = 180° - 120° = 60°$. The measurements satisfy the ASA criterion (Angle-Side-Angle) using side AC and angles ∠A and ∠C.
Conclusion: ∆ABC can be constructed.
Steps of Construction:
1. Draw a line segment AC of length 3 cm.
2. At point A, construct an angle of 70° using a protractor. Draw a ray AX making this angle with AC.
3. At point C, construct an angle of 60°. Draw a ray CY making this angle with CA.
4. Let the rays AX and CY intersect at point B.
5. ∆ABC is the required triangle.
4. ∆LMN: m∠L = 60°; m∠N = 120°; LM = 5 cm.
Analysis: Check the angle sum property.
$m\angle L + m\angle N = 60° + 120° = 180°$
The sum of two angles is 180°. This means the third angle, m∠M, would be $180° - 180° = 0°$. A triangle cannot have an angle of 0°.
Conclusion: ∆LMN cannot be constructed.
5. ∆ABC: BC = 2 cm; AB = 4 cm; AC = 2 cm.
Analysis: Check the Triangle Inequality Theorem.
The sum of the lengths of any two sides must be greater than the length of the third side.
$BC + AC > AB$? $\implies 2 + 2 > 4$? $\implies 4 > 4$? This is false (4 is not greater than 4).
Since the sum of two sides (BC + AC) is equal to the third side (AB), the points A, B, and C are collinear and do not form a triangle.
Conclusion: ∆ABC cannot be constructed.
6. ∆PQR: PQ = 3.5 cm; QR = 4 cm; PR = 3.5 cm.
Analysis: Check the Triangle Inequality Theorem.
$PQ + QR > PR$? $\implies 3.5 + 4 > 3.5$? $\implies 7.5 > 3.5$ (True).
$PQ + PR > QR$? $\implies 3.5 + 3.5 > 4$? $\implies 7 > 4$ (True).
$QR + PR > PQ$? $\implies 4 + 3.5 > 3.5$? $\implies 7.5 > 3.5$ (True).
All conditions are met. The measurements satisfy the SSS criterion.
Conclusion: ∆PQR can be constructed (It is an isosceles triangle).
Steps of Construction:
1. Draw a line segment QR of length 4 cm.
2. With Q as the centre, draw an arc with radius 3.5 cm (length of PQ).
3. With R as the centre, draw an arc with radius 3.5 cm (length of PR).
4. Let the two arcs intersect at point P.
5. Join P to Q and P to R.
6. ∆PQR is the required triangle.
7. ∆XYZ: XY = 3 cm; YZ = 4 cm; XZ = 5 cm.
Analysis: Check the Triangle Inequality Theorem.
$XY + YZ > XZ$? $\implies 3 + 4 > 5$? $\implies 7 > 5$ (True).
$XY + XZ > YZ$? $\implies 3 + 5 > 4$? $\implies 8 > 4$ (True).
$YZ + XZ > XY$? $\implies 4 + 5 > 3$? $\implies 9 > 3$ (True).
All conditions are met. The measurements satisfy the SSS criterion. (Note: $3^2 + 4^2 = 9 + 16 = 25 = 5^2$, so it's a right-angled triangle).
Conclusion: ∆XYZ can be constructed.
Steps of Construction:
1. Draw a line segment YZ of length 4 cm.
2. With Y as the centre, draw an arc with radius 3 cm (length of XY).
3. With Z as the centre, draw an arc with radius 5 cm (length of XZ).
4. Let the two arcs intersect at point X.
5. Join X to Y and X to Z.
6. ∆XYZ is the required triangle.
8. ∆DEF: DE = 4.5 cm; EF = 5.5 cm; DF = 4 cm.
Analysis: Check the Triangle Inequality Theorem.
$DE + EF > DF$? $\implies 4.5 + 5.5 > 4$? $\implies 10 > 4$ (True).
$DE + DF > EF$? $\implies 4.5 + 4 > 5.5$? $\implies 8.5 > 5.5$ (True).
$EF + DF > DE$? $\implies 5.5 + 4 > 4.5$? $\implies 9.5 > 4.5$ (True).
All conditions are met. The measurements satisfy the SSS criterion.
Conclusion: ∆DEF can be constructed.
Steps of Construction:
1. Draw a line segment EF of length 5.5 cm.
2. With E as the centre, draw an arc with radius 4.5 cm (length of DE).
3. With F as the centre, draw an arc with radius 4 cm (length of DF).
4. Let the two arcs intersect at point D.
5. Join D to E and D to F.
6. ∆DEF is the required triangle.