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| Exercise 11.1 | Example 1 (Before Exercise 11.2) | Exercise 11.2 |
Chapter 11 Construction
This guide delves into the solutions for Chapter 11: Constructions, a practical and foundational area of geometry. Unlike previous chapters focused on theoretical proofs and properties, this chapter emphasizes the skill of creating accurate geometric figures using only the most basic, classical tools: an ungraduated ruler (a straightedge without markings) and a pair of compasses. This limitation forces a reliance on pure geometric principles rather than measurement. Building upon constructions learned in earlier classes, this chapter introduces more intricate procedures, demanding both precision in execution and, crucially, a clear understanding of the underlying justification for why each sequence of steps produces the desired figure.
The solutions provided offer meticulous, step-by-step instructions for each construction. Equally important, they often include the geometric reasoning or proof that validates the method. This justification typically relies on congruence criteria (like SSS) or established geometric properties previously studied. Mastering these constructions enhances spatial reasoning and provides a tangible connection to geometric theorems. Precision is key – neat arcs and clearly drawn lines are essential for successful constructions.
The key constructions covered in this chapter, along with their justifications as explained in the solutions, include:
- Constructing the bisector of a given angle: The solutions detail how to draw arcs from the vertex and then intersecting arcs from points on the angle's arms. The justification usually involves proving two triangles congruent using the SSS criterion, thereby showing the equality of the angles created by the constructed line.
- Constructing the perpendicular bisector of a given line segment: This involves drawing intersecting arcs of equal radius from both endpoints of the segment. The solutions explain the steps and justify the construction, again often using SSS congruence to show that the line joining the intersection points forms right angles and bisects the original segment (connecting to properties of rhombuses or kites).
- Constructing specific angles without a protractor: Starting with the fundamental construction of a $60^\circ$ angle (by creating an equilateral triangle), the solutions demonstrate how to derive other key angles through bisection or addition/subtraction. This includes constructing angles such as $30^\circ$ (bisecting $60^\circ$), $90^\circ$ (bisecting the angle between $60^\circ$ and $120^\circ$, or constructing a perpendicular), $45^\circ$ (bisecting $90^\circ$), $120^\circ$ (adjacent $60^\circ$ angles), $75^\circ$ ($60^\circ + 15^\circ$ or $90^\circ - 15^\circ$), $105^\circ$ ($90^\circ + 15^\circ$), and $135^\circ$ ($90^\circ + 45^\circ$). Each sequence of steps is clearly outlined.
- Constructing a triangle given its base, a base angle, and the sum of the other two sides: Solutions guide drawing the base, constructing the given angle, marking a point on the angle ray corresponding to the sum of the other two sides, connecting this point to the other end of the base, and finally constructing the perpendicular bisector of this connecting line segment. The intersection of the bisector with the initial ray gives the third vertex. Justification hinges on the properties of isosceles triangles created by the perpendicular bisector.
- Constructing a triangle given its base, a base angle, and the difference of the other two sides: This construction presents two cases, depending on which of the other two sides is longer. The process is similar to the 'sum' case but involves marking the difference and requires careful attention to whether the difference is marked on the angle ray itself or its extension. The justification again relies on constructing a perpendicular bisector and the resulting isosceles triangle properties.
- Constructing a triangle given its perimeter and its two base angles: The solutions detail drawing a line segment equal to the perimeter, constructing the given base angles at the ends of this segment, bisecting these angles, and finding the intersection point of the angle bisectors (which becomes the third vertex of the required triangle). The actual base vertices are then found by constructing perpendicular bisectors of the segments connecting this vertex to the ends of the perimeter line.
For every construction, the solutions emphasize the importance of providing the justification, explicitly linking the drawing steps back to established geometric theorems and ensuring the construction is mathematically sound, not just visually approximate. This reinforces the connection between theoretical geometry and practical application.
Exercise 11.1
Question 1. Construct an angle of 90° at the initial point of a given ray and justify the construction.
Answer:
Given:
A ray OA with initial point O.
To Construct:
An angle of $90^\circ$ at the initial point O of the ray OA.
Construction:
1. Draw a ray OA.
2. With O as centre and a convenient radius, draw an arc of a circle which intersects OA at a point P.
3. With P as centre and the same radius as before, draw an arc intersecting the previously drawn arc at point Q.
4. With Q as centre and the same radius as before, draw an arc intersecting the arc at point R.
5. With Q and R as centres and a radius greater than half of QR, draw two arcs intersecting each other at point S.
6. Draw the ray OS.
Then $\angle$AOS is the required angle of $90^\circ$.
Justification:
Join OQ, OP, PQ, OR, QR, QS, and RS.
By construction, OP = PQ = OQ (Radii of arcs drawn with the same radius).
Therefore, $\triangle$OPQ is an equilateral triangle.
$\angle$POQ = $60^\circ$
Similarly, by construction, OQ = QR = OR (Radii of arcs drawn with the same radius).
Therefore, $\triangle$OQR is an equilateral triangle.
$\angle$QOR = $60^\circ$
Now, consider $\triangle$OQS and $\triangle$ORS.
OQ = OR
(Radii of the same arc)
OS = OS
(Common side)
QS = RS
(By construction)
Therefore, by SSS congruence rule:
$\triangle$OQS $\cong$ $\triangle$ORS
By Corresponding Parts of Congruent Triangles (CPCT):
$\angle$QOS = $\angle$ROS
We know that $\angle$QOR = $\angle$QOS + $\angle$ROS.
$60^\circ$ = $\angle$QOS + $\angle$QOS
$60^\circ$ = 2$\angle$QOS
$\angle$QOS = $\frac{60^\circ}{2} = 30^\circ$
Now, consider the angle $\angle$AOS (which is the same as $\angle$POS).
$\angle$AOS = $\angle$POQ + $\angle$QOS
$\angle$AOS = $60^\circ + 30^\circ = 90^\circ$
Thus, the angle constructed is indeed $90^\circ$.
Question 2. Construct an angle of 45° at the initial point of a given ray and justify the construction.
Answer:
Given:
A ray OA with initial point O.
To Construct:
An angle of $45^\circ$ at the initial point O of the ray OA.
Construction:
1. Draw a ray OA.
2. At the initial point O, construct an angle of $90^\circ$. Let the ray forming the $90^\circ$ angle be OB, so that $\angle$AOB = $90^\circ$. (Follow the steps from Question 1 to construct the $90^\circ$ angle).
3. With O as centre and a convenient radius, draw an arc which intersects OA at P and OB at Q.
4. With P as centre and a radius greater than half of PQ, draw an arc.
5. With Q as centre and the same radius, draw another arc intersecting the previously drawn arc at point R.
6. Draw the ray OR.
Then $\angle$AOR is the required angle of $45^\circ$.
Justification:
We have constructed $\angle$AOB = $90^\circ$.
By construction, the ray OR is the bisector of $\angle$AOB.
We can justify this by joining PR and QR.
In $\triangle$OPR and $\triangle$OQR:
OP = OQ
(Radii of the same arc)
OR = OR
(Common side)
PR = QR
(By construction)
Therefore, by SSS congruence rule:
$\triangle$OPR $\cong$ $\triangle$OQR
By Corresponding Parts of Congruent Triangles (CPCT):
$\angle$AOR = $\angle$BOR
Since OR bisects $\angle$AOB, we have:
$\angle$AOR = $\frac{1}{2}\angle$AOB
We know that $\angle$AOB = $90^\circ$ from the first part of the construction.
$\angle$AOR = $\frac{1}{2} \times 90^\circ = 45^\circ$
Thus, the angle constructed is indeed $45^\circ$.
Question 3. Construct the angles of the following measurements:
(i) 30°
(ii) $22\frac{1}{2}^\circ$ °
(iii) 15°
Answer:
(i) Construction of $30^\circ$
Given:
A ray OA with initial point O.
To Construct:
An angle of $30^\circ$ at the initial point O of the ray OA.
Construction:
1. Draw a ray OA.
2. With O as centre and a convenient radius, draw an arc of a circle which intersects OA at a point P.
3. With P as centre and the same radius, draw an arc intersecting the previously drawn arc at point Q.
4. With O and Q as centres and a radius greater than half of OQ, draw two arcs intersecting each other at point R.
5. Draw the ray OR.
Then $\angle$AOR is the required angle of $30^\circ$.
Justification:
Join OQ and PQ.
By construction, OP = PQ = OQ (Radii of arcs drawn with the same radius).
Therefore, $\triangle$OPQ is an equilateral triangle.
$\angle$POQ = $60^\circ$
By construction, the ray OR is the angle bisector of $\angle$POQ.
Therefore, $\angle$POR = $\angle$QOR = $\frac{1}{2} \angle$POQ.
$\angle$AOR = $\angle$POR
(Same angle)
$\angle$AOR = $\frac{1}{2} \times 60^\circ = 30^\circ$
Thus, the angle constructed is indeed $30^\circ$.
(ii) Construction of $22\frac{1}{2}^\circ$
Given:
A ray OA with initial point O.
To Construct:
An angle of $22\frac{1}{2}^\circ$ at the initial point O of the ray OA.
Construction:
1. Draw a ray OA.
2. At the initial point O, construct an angle of $90^\circ$. Let the ray forming the $90^\circ$ angle be OB, so that $\angle$AOB = $90^\circ$. (Follow the steps from Question 1 to construct the $90^\circ$ angle).
3. With O as centre and a convenient radius, draw an arc which intersects OA at P and OB at Q.
4. With P and Q as centres and a radius greater than half of PQ, draw two arcs intersecting each other at point R. Draw the ray OR. $\angle$AOR = $45^\circ$.
5. With O as centre and a convenient radius, draw an arc intersecting OA at P and OR at S.
6. With P and S as centres and a radius greater than half of PS, draw two arcs intersecting each other at point T.
7. Draw the ray OT.
Then $\angle$AOT is the required angle of $22\frac{1}{2}^\circ$.
Justification:
We have constructed $\angle$AOB = $90^\circ$.
By bisecting $\angle$AOB, we constructed ray OR such that $\angle$AOR = $\frac{1}{2}\angle$AOB = $\frac{1}{2} \times 90^\circ = 45^\circ$.
Now, we bisected $\angle$AOR using ray OT.
Therefore, $\angle$AOT = $\frac{1}{2}\angle$AOR.
$\angle$AOT = $\frac{1}{2} \times 45^\circ = 22.5^\circ = 22\frac{1}{2}^\circ$
Thus, the angle constructed is indeed $22\frac{1}{2}^\circ$.
(iii) Construction of $15^\circ$
Given:
A ray OA with initial point O.
To Construct:
An angle of $15^\circ$ at the initial point O of the ray OA.
Construction:
1. Draw a ray OA.
2. At the initial point O, construct an angle of $60^\circ$. Let the ray forming the $60^\circ$ angle be OB, so that $\angle$AOB = $60^\circ$. (Follow steps 2 and 3 from construction of $30^\circ$). Let the arc intersect OA at P and the new ray OB at Q.
3. Construct the angle bisector of $\angle$AOB. Let the bisecting ray be OC, so that $\angle$AOC = $30^\circ$. (Follow steps 4 and 5 from construction of $30^\circ$). Let the bisector OC intersect the initial arc at point R.
4. With O as centre and a convenient radius, draw an arc intersecting OA at P and OC at R.
5. With P and R as centres and a radius greater than half of PR, draw two arcs intersecting each other at point S.
6. Draw the ray OS.
Then $\angle$AOS is the required angle of $15^\circ$.
Justification:
We have constructed $\angle$AOB = $60^\circ$.
By bisecting $\angle$AOB, we constructed ray OC such that $\angle$AOC = $\frac{1}{2}\angle$AOB = $\frac{1}{2} \times 60^\circ = 30^\circ$.
Now, we bisected $\angle$AOC using ray OS.
Therefore, $\angle$AOS = $\frac{1}{2}\angle$AOC.
$\angle$AOS = $\frac{1}{2} \times 30^\circ = 15^\circ$
Thus, the angle constructed is indeed $15^\circ$.
Question 4. Construct the following angles and verify by measuring them by a protractor:
(i) $75^\circ$
(ii) $105^\circ$
(iii) $135^\circ$
Answer:
(i) Construction of $75^\circ$
To Construct:
An angle of $75^\circ$ at the initial point O of a given ray OA.
Construction:
1. Draw a ray OA.
2. With O as centre and a convenient radius, draw an arc which intersects OA at point P.
3. With P as centre and the same radius, draw an arc intersecting the first arc at Q. $\angle$POQ = $60^\circ$.
4. With Q as centre and the same radius, draw an arc intersecting the first arc at R. $\angle$POR = $120^\circ$.
5. With Q and R as centres and a radius greater than half of QR, draw two arcs intersecting each other at S. Draw ray OS. $\angle$POS = $90^\circ$.
6. The arc intersects OS at a point (let's call it T). With Q and T as centres and a radius greater than half of QT, draw two arcs intersecting each other at U.
7. Draw the ray OU.
Then $\angle$AOU is the required angle of $75^\circ$. This is because $\angle$POQ = $60^\circ$ and $\angle$QOT = $90^\circ - 60^\circ = 30^\circ$. Ray OU bisects $\angle$QOT, so $\angle$QOU = $\frac{1}{2}\angle$QOT = $\frac{1}{2} \times 30^\circ = 15^\circ$. Therefore, $\angle$AOU = $\angle$POQ + $\angle$QOU = $60^\circ + 15^\circ = 75^\circ$.
Verification:
Place the protractor with its centre on O and the base line along OA. Read the measure of $\angle$AOU. It should be $75^\circ$.
(ii) Construction of $105^\circ$
To Construct:
An angle of $105^\circ$ at the initial point O of a given ray OA.
Construction:
1. Draw a ray OA.
2. With O as centre and a convenient radius, draw an arc which intersects OA at point P.
3. With P as centre and the same radius, draw an arc intersecting the first arc at Q ($60^\circ$).
4. With Q as centre and the same radius, draw an arc intersecting the first arc at R ($120^\circ$).
5. With Q and R as centres and a radius greater than half of QR, draw two arcs intersecting each other at S. Draw ray OS ($90^\circ$).
6. The arc intersects OS at a point (let's call it T). With T and R as centres and a radius greater than half of TR, draw two arcs intersecting each other at U.
7. Draw the ray OU.
Then $\angle$AOU is the required angle of $105^\circ$. This is because $\angle$AOS = $90^\circ$ and $\angle$AOR = $120^\circ$. Ray OU bisects the angle between the $90^\circ$ line OS and the $120^\circ$ line OR, which is $\angle$TOR = $120^\circ - 90^\circ = 30^\circ$. So $\angle$TOU = $\frac{1}{2} \times 30^\circ = 15^\circ$. Therefore, $\angle$AOU = $\angle$AOS + $\angle$SOU = $90^\circ + 15^\circ = 105^\circ$. (Note: T lies on OS).
Verification:
Place the protractor with its centre on O and the base line along OA. Read the measure of $\angle$AOU. It should be $105^\circ$.
(iii) Construction of $135^\circ$
To Construct:
An angle of $135^\circ$ at the initial point O of a given ray OA.
Construction:
1. Draw a ray OA. Extend OA backwards through O to form a line POQ, where P is on the other side of O.
2. With O as centre and a convenient radius, draw a semicircle intersecting OP at R and OA at P.
3. Construct a $90^\circ$ angle at O on line PQ. With P as centre and the same radius, draw an arc on the semicircle at Q. With Q as centre and the same radius, draw an arc on the semicircle at R. With Q and R as centres, draw arcs intersecting at S. Draw ray OS. $\angle$POS = $90^\circ$ and $\angle$AOS = $90^\circ$.
4. The semicircle intersects OS at a point (let's call it T). With R and T as centres and a radius greater than half of RT, draw two arcs intersecting each other at U.
5. Draw the ray OU.
Then $\angle$AOU is the required angle of $135^\circ$. This is because $\angle$AOS = $90^\circ$. Ray OU bisects the angle $\angle$SOP, which is $180^\circ - 90^\circ = 90^\circ$. So $\angle$SOJ = $\frac{1}{2} \times 90^\circ = 45^\circ$. Therefore, $\angle$AOU = $\angle$AOS + $\angle$SOU = $90^\circ + 45^\circ = 135^\circ$. (Note: R lies on OP, P lies on OA, and T lies on OS in the previous step description. Let's use points on the semicircle consistently). Let the semicircle intersect OA at P, and the ray extending backwards at P'. Let the $90^\circ$ ray be OB. Semicircle intersects OB at Q. Bisect the angle between ray OP' and OB. Let P' be the point on the extended ray. The semicircle intersects OP' at P'. The $90^\circ$ ray OB intersects the semicircle at Q. Bisect the angle $\angle$P'OQ. Let the bisector be OC. Then $\angle$AOC = $\angle$AOB + $\angle$BOC = $90^\circ + 45^\circ = 135^\circ$. Let's redraw the steps for clarity.
Corrected Construction for $135^\circ$:
1. Draw a line A'OA, with O as the initial point for the angle on the ray OA.
2. With O as centre, draw a semicircle that intersects OA at P and A'O at P'.
3. Construct a $90^\circ$ angle at O. Let the ray forming the $90^\circ$ angle be OB, such that $\angle$AOB = $90^\circ$. The ray OB intersects the semicircle at point Q.
4. The angle $\angle$A'OB is $180^\circ - 90^\circ = 90^\circ$. We need to bisect this angle $\angle$A'OB.
5. With Q and P' as centres and a radius greater than half of QP', draw two arcs intersecting each other at R.
6. Draw the ray OR.
Then $\angle$AOR is the required angle of $135^\circ$. This is because $\angle$AOB = $90^\circ$ and ray OR bisects $\angle$A'OB. So, $\angle$BOR = $\angle$A'OR = $\frac{1}{2}\angle$A'OB = $\frac{1}{2} \times 90^\circ = 45^\circ$. Therefore, $\angle$AOR = $\angle$AOB + $\angle$BOR = $90^\circ + 45^\circ = 135^\circ$.
Verification:
Place the protractor with its centre on O and the base line along OA. Read the measure of $\angle$AOR. It should be $135^\circ$.
Question 5. Construct an equilateral triangle, given its side and justify the construction.
Answer:
Given:
A line segment of a given length (let's call it 's').
To Construct:
An equilateral triangle with side length equal to the given segment 's'.
Construction:
1. Draw a line segment AB of the given length 's'.
2. With A as the centre and radius equal to the length of AB, draw an arc.
3. With B as the centre and the same radius (equal to the length of AB), draw another arc intersecting the previously drawn arc at point C.
4. Join A to C and B to C.
Then $\triangle$ABC is the required equilateral triangle.
Justification:
By construction, the length of the line segment AB is the given side length 's'.
In the construction, we drew an arc with centre A and radius AB. So, any point on this arc is at a distance equal to AB from A.
Point C lies on this arc, so AC = AB.
We also drew an arc with centre B and radius AB. So, any point on this arc is at a distance equal to AB from B.
Point C also lies on this second arc, so BC = AB.
Therefore, we have AC = AB and BC = AB.
By transitivity, AC = BC = AB.
Since all three sides of $\triangle$ABC are equal, it is an equilateral triangle.
Thus, the constructed triangle is an equilateral triangle with the given side length.
Example 1 (Before Exercise 11.2)
Example 1. Construct a triangle ABC, in which ∠B = 60°, ∠ C = 45° and AB + BC + CA = 11 cm.
Answer:
Given:
In $\triangle$ABC, $\angle$B = $60^\circ$, $\angle$C = $45^\circ$, and perimeter AB + BC + CA = 11 cm.
To Construct:
A triangle ABC satisfying the given conditions.
Construction:
1. Draw a line segment XY equal to the perimeter, i.e., XY = 11 cm.
2. At point X, construct an angle equal to half of $\angle$B, i.e., $\angle$YXL = $\frac{1}{2} \times 60^\circ = 30^\circ$.
3. At point Y, construct an angle equal to half of $\angle$C, i.e., $\angle$XYM = $\frac{1}{2} \times 45^\circ = 22.5^\circ$.
4. Let the ray XL and the ray YM intersect at point A.
5. Draw the perpendicular bisector of the line segment AX. Let it intersect XY at point B.
6. Draw the perpendicular bisector of the line segment AY. Let it intersect XY at point C.
7. Join AB and AC.
Then $\triangle$ABC is the required triangle.
Justification:
By construction, B lies on the perpendicular bisector of AX. Therefore, any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment.
AB = BX
Since AB = BX, $\triangle$ABX is an isosceles triangle. In $\triangle$ABX, the angles opposite the equal sides are equal.
$\angle$BAX = $\angle$BXA
The angle $\angle$BXA is the same as $\angle$YXL, which we constructed as $30^\circ$.
$\angle$BXA = $30^\circ$
(By construction)
So, $\angle$BAX = $30^\circ$.
Now, consider $\angle$ABC, which is an exterior angle of $\triangle$ABX at vertex B.
The exterior angle of a triangle is equal to the sum of the two opposite interior angles.
$\angle$ABC = $\angle$BAX + $\angle$BXA
$\angle$ABC = $30^\circ + 30^\circ = 60^\circ$
Thus, $\angle$B = $60^\circ$, which is one of the required angles.
Similarly, by construction, C lies on the perpendicular bisector of AY. Therefore:
AC = CY
Since AC = CY, $\triangle$ACY is an isosceles triangle. In $\triangle$ACY, the angles opposite the equal sides are equal.
$\angle$CAY = $\angle$CYA
The angle $\angle$CYA is the same as $\angle$XYM, which we constructed as $22.5^\circ$.
$\angle$CYA = $22.5^\circ$
(By construction)
So, $\angle$CAY = $22.5^\circ$.
Now, consider $\angle$ACB, which is an exterior angle of $\triangle$ACY at vertex C.
$\angle$ACB = $\angle$CAY + $\angle$CYA
$\angle$ACB = $22.5^\circ + 22.5^\circ = 45^\circ$
Thus, $\angle$C = $45^\circ$, which is the other required angle.
Now let's check the perimeter of $\triangle$ABC. The perimeter is AB + BC + CA.
From our justification, we know AB = BX and AC = CY.
Perimeter = AB + BC + CA = BX + BC + CY.
Since B and C are points on the line segment XY, and from the construction, B lies between X and C, and C lies between B and Y (or B before C on XY), the sum of the lengths BX, BC, and CY is equal to the length of the segment XY.
Perimeter = BX + BC + CY = XY.
By construction, XY = 11 cm.
AB + BC + CA = 11 cm
All the given conditions are satisfied.
Hence Justified.
Exercise 11.2
Question 1. Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB + AC = 13 cm.
Answer:
Given:
In $\triangle$ABC, BC = 7 cm, $\angle$B = $75^\circ$, and AB + AC = 13 cm.
To Construct:
A triangle ABC satisfying the given conditions.
Construction:
1. Draw a line segment BC of length 7 cm.
2. At point B, construct an angle $\angle$CBP = $75^\circ$ using a compass and ruler. (Construct $90^\circ$ and $60^\circ$, then bisect the angle between them to get $75^\circ$).
3. Along the ray BP, cut off a line segment BD equal to AB + AC = 13 cm.
4. Join CD.
5. Construct the perpendicular bisector of the line segment CD.
6. Let the perpendicular bisector intersect BD at point A.
7. Join AC.
Then $\triangle$ABC is the required triangle.
Justification:
By construction, $\angle$B = $75^\circ$ and BC = 7 cm.
Point A lies on the perpendicular bisector of CD. A property of the perpendicular bisector is that any point on it is equidistant from the endpoints of the segment.
Therefore, AC = AD.
By construction, BD = 13 cm.
Also, from the figure, point A lies on the line segment BD. So, BD can be expressed as the sum of BA and AD.
BD = BA + AD
Substitute BD = 13 cm and AD = AC into the equation:
13 cm = BA + AC
AB + AC = 13 cm
Thus, the triangle ABC has BC = 7 cm, $\angle$B = $75^\circ$, and AB + AC = 13 cm, which matches the given conditions.
Hence Justified.
Question 2. Construct a triangle ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 3.5 cm.
Answer:
Given:
In $\triangle$ABC, BC = 8 cm, $\angle$B = $45^\circ$, and AB – AC = 3.5 cm (where AB > AC).
To Construct:
A triangle ABC satisfying the given conditions.
Construction:
1. Draw a line segment BC of length 8 cm.
2. At point B, construct an angle $\angle$CBP = $45^\circ$ using a compass and ruler. (Construct $90^\circ$ and then bisect it).
3. Along the ray BP, cut off a line segment BD equal to AB – AC = 3.5 cm.
4. Join CD.
5. Construct the perpendicular bisector of the line segment CD.
6. Let the perpendicular bisector intersect the ray BP at point A.
7. Join AC.
Then $\triangle$ABC is the required triangle.
Justification:
By construction, $\angle$B = $45^\circ$ and BC = 8 cm.
Point A lies on the perpendicular bisector of CD. By the property of a perpendicular bisector, any point on it is equidistant from the endpoints of the segment.
Therefore, AC = AD.
By construction, BD = 3.5 cm.
From the figure, point A lies on the ray BD. Since A is the intersection of the perpendicular bisector and the ray BP, and D is between B and A (because AB > AC implies BD < AB), we have:
BA = BD + DA
Substitute BD = 3.5 cm and DA = AC into the equation:
BA = 3.5 cm + AC
BA - AC = 3.5 cm
AB - AC = 3.5 cm
Thus, the triangle ABC has BC = 8 cm, $\angle$B = $45^\circ$, and AB – AC = 3.5 cm, which matches the given conditions.
Hence Justified.
Question 3. Construct a triangle PQR in which QR = 6cm, ∠Q = 60° and PR – PQ = 2cm.
Answer:
Given:
In $\triangle$PQR, QR = 6 cm, $\angle$Q = $60^\circ$, and PR – PQ = 2 cm (where PR > PQ).
To Construct:
A triangle PQR satisfying the given conditions.
Construction:
1. Draw a line segment QR of length 6 cm.
2. At point Q, construct an angle $\angle$RQX = $60^\circ$ using a compass and ruler.
3. Extend the ray XQ backwards to a point Y (forming a line YQX).
4. Along the ray QY (the extended part of XQ), cut off a line segment QS equal to PR – PQ = 2 cm.
5. Join SR.
6. Construct the perpendicular bisector of the line segment SR.
7. Let the perpendicular bisector intersect the ray QX at point P.
8. Join PR.
Then $\triangle$PQR is the required triangle.
Justification:
By construction, QR = 6 cm and $\angle$PQR = $60^\circ$ (since P lies on ray QX).
Point P lies on the perpendicular bisector of SR. By the property of a perpendicular bisector, any point on it is equidistant from the endpoints of the segment.
Therefore, PR = PS.
By construction, QS = 2 cm.
From the figure, point Q lies on the line segment PS. So, PS can be expressed as the sum of PQ and QS.
PS = PQ + QS
Substitute PS = PR and QS = 2 cm into the equation:
PR = PQ + 2 cm
PR - PQ = 2 cm
Thus, the triangle PQR has QR = 6 cm, $\angle$Q = $60^\circ$, and PR – PQ = 2 cm, which matches the given conditions.
Hence Justified.
Question 4. Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.
Answer:
Given:
In $\triangle$XYZ, $\angle$Y = $30^\circ$, $\angle$Z = $90^\circ$, and perimeter XY + YZ + ZX = 11 cm.
To Construct:
A triangle XYZ satisfying the given conditions.
Construction:
1. Draw a line segment AB of length 11 cm (equal to the perimeter).
2. At point A, construct an angle $\angle$PAB = $15^\circ$ ($\frac{1}{2} \times 30^\circ$) using compass and ruler.
3. At point B, construct an angle $\angle$QBA = $45^\circ$ ($\frac{1}{2} \times 90^\circ$) using compass and ruler.
4. Let the rays AP and BQ intersect at point X. This point X is the third vertex of the triangle.
5. Draw the perpendicular bisector of the line segment AX. Let this perpendicular bisector intersect AB at point Y.
6. Draw the perpendicular bisector of the line segment BX. Let this perpendicular bisector intersect AB at point Z.
7. Join XY and XZ.
Then $\triangle$XYZ is the required triangle.
Justification:
By construction, point Y lies on the perpendicular bisector of AX.
Therefore, by the property of the perpendicular bisector, any point on it is equidistant from the endpoints of the segment.
XY = AY
Since XY = AY, $\triangle$AXY is an isosceles triangle.
In an isosceles triangle, the angles opposite the equal sides are equal.
$\angle$XAY = $\angle$AXY
By construction, $\angle$XAY = $\angle$PAB = $15^\circ$.
$\angle$AXY = $15^\circ$
Now, consider the angle $\angle$XYZ, which is an exterior angle of $\triangle$AXY at vertex Y.
The exterior angle of a triangle is equal to the sum of the two opposite interior angles.
$\angle$XYZ = $\angle$XAY + $\angle$AXY
$\angle$XYZ = $15^\circ + 15^\circ = 30^\circ$
Thus, $\angle$Y = $30^\circ$, which is one of the required angles.
Similarly, by construction, point Z lies on the perpendicular bisector of BX.
XZ = BZ
Since XZ = BZ, $\triangle$BXZ is an isosceles triangle.
$\angle$XBZ = $\angle$BXZ
By construction, $\angle$XBZ = $\angle$QBA = $45^\circ$.
$\angle$BXZ = $45^\circ$
Now, consider the angle $\angle$XZY, which is an exterior angle of $\triangle$BXZ at vertex Z.
$\angle$XZY = $\angle$XBZ + $\angle$BXZ
$\angle$XZY = $45^\circ + 45^\circ = 90^\circ$
Thus, $\angle$Z = $90^\circ$, which is the other required angle.
Finally, let's check the perimeter of $\triangle$XYZ.
Perimeter = XY + YZ + ZX.
From our justification, we know XY = AY and XZ = BZ.
Perimeter = AY + YZ + BZ.
Since Y and Z are points on the line segment AB, and Y and Z lie between A and B, the sum of the lengths AY, YZ, and ZB is equal to the length of the segment AB.
AY + YZ + BZ = AB
By construction, AB = 11 cm.
XY + YZ + ZX = 11 cm
All the given conditions are satisfied.
Hence Justified.
Question 5. Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.
Answer:
Given:
A right triangle with base = 12 cm, and the sum of its hypotenuse and the other side = 18 cm.
To Construct:
A right triangle, say $\triangle$ABC, where the base BC = 12 cm, $\angle$B = $90^\circ$, and AB + AC = 18 cm.
Construction:
1. Draw a ray BX and cut off a line segment BC = 12 cm from it.
2. At point B, construct an angle $\angle$XBY = $90^\circ$ using a compass and ruler.
3. Along the ray BY, cut off a line segment BD = 18 cm (equal to the sum of the hypotenuse and the other side).
4. Join CD.
5. Construct the perpendicular bisector of the line segment CD.
6. Let the perpendicular bisector intersect the ray BY at point A.
7. Join AC.
Then $\triangle$ABC is the required right triangle.
Justification:
By construction, BC = 12 cm and $\angle$B = $90^\circ$.
Point A lies on the perpendicular bisector of CD. By the property of a perpendicular bisector, any point on it is equidistant from the endpoints of the segment.
Therefore, AC = AD.
By construction, BD = 18 cm.
From the figure, point A lies on the ray BD. Since A is the intersection of the perpendicular bisector and the ray BY, and A lies on BD, we have:
BD = BA + AD
Substitute BD = 18 cm and AD = AC into the equation:
18 cm = BA + AC
AB + AC = 18 cm
Thus, the triangle ABC constructed has base BC = 12 cm, $\angle$B = $90^\circ$, and the sum of the other two sides AB + AC = 18 cm, which matches the given conditions.
Hence Justified.