PSEB 10th Class Maths Solutions Chapter 11 Constructions Ex 11.1

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 11 Constructions Ex 11.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1

In each of the questions, give the justification of the construction also.

Question 1.
Draw a line segment of length 7.6 cm and divide ¡tin the ratio 5 : 8. Measure the two parts.
Solution:
Given: A line segment of length of 7.6 cm.
Steps of construction:
1. Take a line segment AB = 7.6 cm.
2. Draw any ray AX, making an acute angle ∠BAX.
3. Locate 5 + 8 = 13 (given ratio 5: 8) points A₁, A₂, A₃, A₄, A₅, ………….., A₁₀, A₁₁, A₁₂, A₁₃ on ray AX such that A₁A₂ = A₂A₃ = A₃A₄ = …………. = A₁₁A₁₂ = A₁₂ A₁₃.
4. Join BA₁₃.
5. Through point A5, draw a line A₅C || A₁₃B (by making an angle equal to ∠A₁₃B) at A₅ intersecting AB at ‘C’. Then AC : CB = 5 : 8;

Justification:
Let us see how this method gives us the required division.
In ∆AA₁₃B,
Since A₅C || A₁₃B
∴ By Basic Proportionality Theorem
\(\frac{\mathrm{AA}_{5}}{\mathrm{~A}_{5} \mathrm{~A}_{13}}=\frac{\mathrm{AC}}{\mathrm{CB}}\)

By construction, \(\frac{\mathrm{AA}_{5}}{\mathrm{~A}_{5} \mathrm{~A}_{13}}=\frac{5}{8}\)

∴ \(\frac{\mathrm{AC}}{\mathrm{CB}}=\frac{5}{8}\)
This shows that ‘C’ divides AB in the ratio 5 : 8.
On measuring the two parts, AC = 2.9 cm and CB = 4.7 cm.

Alternative Method:
Steps of construction:
1. Take a line segment AB = 7.6 cm
2. Draw any acute angle ∠BAX
3. Draw angle ∠ABY such that ∠ABY = ∠BAX.
4. Locate the points A₁, A₂, A₃, A₄, A₅ on ray AX such that A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅.
5. Locate the points B₁, B₂, B₃, B₄, B₅, B₆, B₇, B₈ on ray BY such that B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇ = B₇B₈
6. Join A₅B₈ let it intersects AB at point Then AC : CB = 5 : 8.

justification:
In ∆ACA₅ and ∆BCB₈,
∠ACA₅ = ∠BCB₈ [vertically opp. ∠s]
∠BAA₅ = ∠ABB₈ [construction]
∴ AACA₅ ~ ABCB₈ [AA-similarity cond.]
∴ Their corresponding sides must be in the same ratio. ,
\(\frac{A C}{B C}=\frac{C A_{5}}{C B_{8}}=\frac{A_{5} A}{B_{8} B}\)
(I)(II) (III)
From I and III, \(\frac{A C}{B C}=\frac{A_{5} A}{B_{8} B}\)

But, \(\frac{A_{5} A}{B_{8} B}=\frac{5}{8}\) [construction]

\(\frac{A C}{C B}=\frac{5}{8}\).

Question 2.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of corresponding sides of the first triangle.
Solution:
Steps of construction:
1. Construct a triangle ABC with given measurements. AB = 5 cm, AC = 4 cm and BC = 6 cm.
2. Make any acute angle ∠CBX below the side BC.

3. Locate three points (greater of 2 and 3 in \(\frac{2}{3}\))B₁, B₂, B₃ on BX such that BB₁ = B₁B₂ = B₂B₃.
4. Join B₃C.
5. Through B₂ (smaller of 2 and 3 in \(\frac{2}{3}\) draw a line parallel to B₃C, which intersect BC in C’.
6. Through C’, draw a line parallel to CA meeting BA is A’.
Thus ∆A’BC’ is the required triangle whose sides are of corresponding sides of ∆ABC.

Justification of construction :
First we will show that first triangle and constructed triangle are similar.
i.e. ∆A’BC’ ~ ∆ABC.
Consider ∆A’BC’ and ∆ABC.
∠B = ∠B [Common]
∠A’C’B= ∠ACB [By construction]
∆A’C’B ~ ∆ACB [AA – similarity]
∴ Their corresponding sides must be in the same ratio.
\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}\) …………….(1)
Now, consider ∆B₂BC’ and ∆B₃BC,
∠B = ∠B [common]
∠B₂C’B = ∠B₂CB [construction]
∴ ∆B₂BC’ ~ ∆B₃BC [AA -similarity]
∴ Their corresponding sides must be in the same ratio.

\(\frac{\mathrm{B}_{2} \mathrm{~B}}{\mathrm{~B}_{3} \mathrm{~B}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{B}_{2}}{\mathrm{CB}_{3}}\)

I II III

Taking (I) and (II).
\(\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{B}_{2} \mathrm{~B}}{\mathrm{~B}_{3} \mathrm{~B}}\)

But, \(\frac{\mathrm{B}_{2} \mathrm{~B}}{\mathrm{~B}_{3} \mathrm{~B}}=\frac{2}{3}\) [construction]

\(\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{2}{3}\) ……………(2)

From (1) & (2),
\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}=\frac{2}{3}\)

⇒ A’B = \(\frac{2}{3}\) AB and BC’ = \(\frac{2}{3}\) BC; C’A’ = \(\frac{2}{3}\) CA.
Hence, the construction is Justified.

Question 3.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are of the corresponding sides of the first triangle.
Solution:

Steps of construction :
1. Construct a triangle ABC in which AB = 7 cm, BC 6 cm and AC =5 cm.
2. Make any acute angle ∠BAX below the base AB.
3. Locate seven points A₁, A₂, A₃, A₄, A₅, A₆, A₇ on the ray AX such that
AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇.
4. Join BA₅.
5. Through A₇, draw a line parallel A₅B. Let it meets AB at B’ on being produced such that AB’= \(\frac{7}{5}\) AB.
6. Through B’, draw a line parallel to BC which meets AC at C’ on being produced.
∆AB’C’ is the required triangle.

Justification of the construction.
In ∆ABC and ∆AB’C’,
∠A = ∠A [common]
∠ABC = ∠AB’C’ [corresponding ∠s]
∴ ∠ABC – ∠AB’C’ [AA-similarity]
∴ Their corresponding sides must be in the same ratio.
\(\frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{\mathrm{BC}}{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}=\frac{\mathrm{CA}}{\mathrm{C}^{\prime} \mathrm{A}}\) ……………..(1)

Again, in ∆AA₅B and AA₇B’
∠A = ∠A [common]
∠AA5B = ∠AA7 B’ [corresponding ∠s]
∴ ∆AA5B ~ ∆AA7B’ [AA – similarity]
∴ Their corresponding sides must be in the same ratio.
\(\frac{\mathrm{AA}_{5}}{\mathrm{AA}_{7}}=\frac{\mathrm{A}_{5} \mathrm{~B}}{\mathrm{~A}_{7} \mathrm{~B}^{\prime}}=\frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}\)

⇒ \(\frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{\mathrm{AA}_{5}}{\mathrm{AA}_{7}}\) [construction]

But, \(\frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{5}{7}\) …………….(2)

From (1) and (2),

\(\frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{\mathrm{BC}}{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}=\frac{\mathrm{CA}}{\mathrm{C}^{\prime} \mathrm{A}}=\frac{5}{7}\)

or \(\frac{A B^{\prime}}{A B}=\frac{B^{\prime} C^{\prime}}{B C}=\frac{C^{\prime} A}{C A}=\frac{7}{5}\)

⇒ AB’ = \(\frac{7}{5}\) AB; B’C’ = \(\frac{7}{5}\) BC and C’A’ = \(\frac{7}{5}\) CA

Hence, the sides of ∆AB’C’ are \(\frac{4}{4}\) of ∆ABC.

Question 4.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 12 times
the corresponding sides of the isosceles triangle.
Solution:
Given: Base of isosceles triangle is 8 cm and Altitude = 4 cm
To construct: A triangle whose sides are times the sides of isosceles triangle.
Steps of construction:
1. Take base AB = 8 cm.
2. Draw perpendicular bisector of AB. Let it intersect AB at ‘M’.
3. With M as centre and radius = 4 cm, draw an arc which intersects the perpendicular bisector at ‘C’
4. Join CA and CB.
5. ∆ABC is an isosceles with CA = CB.
6. Make any acute angle ∠BAX below the side BC.
7. Locate three (greater of ‘2’ & ‘3’ in 1\(\frac{1}{2}\) or \(\frac{3}{2}\))
A₁, A₂, A₃ on ‘AX’ such that A A₁ = A₁ A₂ = A₂ A₃.

8. Join A₂ (2nd point smaller of ‘2 and ‘3’ in ) and B.
9. Through A₃, draw a line parallel to A₂B meet AB is B’ cm being produced.
10. Through B’, draw a line parallel to BC which meets AC in C’ on being produced. ∆AB’C’ is the required triangle whose sides are 1\(\frac{1}{2}\) times the corresponding sides of ∆ABC.

Justification of construction :
First we will prove ∆AB’C’ are ∆ABC and similar.
Consider ∆ AB’C’ and ∆ ABC
∠A = ∠A [Common]
∠AB’C’ = ∠ABC [By construction]
∠AB’C’ ~ ∠ABC [By AA – similarityj
∴ Their corresponding sides must be in the same ratio
\(\frac{\mathrm{AB}^{\prime}}{\mathrm{AB}}=\frac{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}}{\mathrm{CA}}\) ……………(1)

Now consider ∆ A₃AB’ and ∆ A,AB
∠A = ∠A [common]
∠B’A₃A = ∠BA₂A [By construction]
∴∆ A₃A B’ – ∆A₂AB [AA – similarity]
∴ Their corresponding sides must be in the same ratio
\(\frac{A_{3} A}{A_{2} A}=\frac{A B^{\prime}}{A B}=\frac{B^{\prime} A_{3}}{B A_{2}}\)
I II III
Taking (I) & (II),
\(\frac{A B^{\prime}}{A B}=\frac{A_{3} A}{A_{2} A}\)

But, \(\frac{A_{3} A}{A_{2} A}=\frac{3}{2}\) [construction]
⇒ \(\frac{\mathrm{AB}^{\prime}}{\mathrm{AB}}=\frac{3}{2}\) ……………..(2)
From (1) & (2)m
\(\frac{\mathrm{AB}^{\prime}}{\mathrm{AB}}=\frac{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}}{\mathrm{CA}}=\frac{3}{2}\left(1 \frac{1}{2}\right)\)

⇒ AB’ = 1\(\frac{1}{2}\) (AB); B’C’ = 1\(\frac{1}{2}\) BC and C’A’ = 1\(\frac{1}{2}\) (CA)
Hence, given result is justified.

Question 5.
Draw a triangle ABC with side BC = 6 cm, AB =5 cm and ¿ABC = 60°. Then construct a triangle whose sides are of the corresponding sides of the triangle ABC.
Solution:
Steps of construction :
1. Take a line segment BC = 6 cm
2. Construct an angle of measure 60° at point B. i.e., ∠CBX = 60°.
3. With B as centre and radius 5 cm draw an arc intersecting BX at ‘A’

4. Join A and C.
5. At B, make any acute angle ∠CBY below the side BC.
6. Locate four points (greater of 3 and 4 in \(\frac{3}{4}\)) B₁, B₂, B₃, B₄ on BY such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄. .
7. Join B₄ and C.
8. Draw a line through B₃ (smaller of 3 and 4 in ) parallel to B₄C making corresponding angles. Let the line through B₃ intersects BC in C’.
9. Through C’, draw a line parallel to CA which intersects BA at A’.
The ∆A’BC’ is the required triangle whose sides are \(\frac{3}{4}\) of sides of ∆ABC.

Justification of the construction:
Consider ∆A’BC’ and ∆ABC
∠B = ∠B [commoni
∠A’C’B = ∠ACB [corresponding ∠s]
∴ ∆A’BC’ ~ ∆ABC [AA – similarity]
∴ Their corresponding sides must be in the same ratio.

∴ \(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}\) ……………..(1)

Now consider ∆B₃B C’ and ∆B₄BC.
∠B = ∠B [common]
∠ C’ B₃B = ∠CB₄B [corresponding ∠s]
∆B₃BC’ ~ ∆B₄BC [AA – similarity con.]
Their corresponding sides must be in the same ratio.
\(\frac{B_{3} B}{B_{4} B}=\frac{B C^{\prime}}{B C}=\frac{C^{\prime} B_{3}}{C B_{4}}\)
(I) (II) (III)

From (I) and (II),
\(\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{B}_{3} \mathrm{~B}}{\mathrm{~B}_{4} \mathrm{~B}}\)

But, = \(\frac{\mathrm{B}_{3} \mathrm{~B}}{\mathrm{~B}_{4} \mathrm{~B}}=\frac{3}{4}\) [construction]

\(\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{3}{4}\) ………..(3)

From (1) and (3)
\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}=\frac{3}{4}\)

and C’A’= \(\frac{3}{4}\) CA.
∆A’BC’ is the required triangle whose sides are \(\frac{3}{4}\) sides of ∆ABC.

Question 6.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°. ∠A = 105°. Then construct a triangle whose sides are j- times
the corresponding sides of ∆ABC.
Solution:
Steps of construction:
1. Construct the triangle ABC with the given measurements.
BC = 7 cm; ∠B = 45, ∠A = 105°
By angle sum property of triangle
∠A + ∠B + ∠C= 180°
105° + 45° + ∠C = 180°
∠C = 180 – 150° = 30°
2. Make any acute angle ∠CBX at point B, below the sides BC.

3. Locate four points (greater of 3 and 4 in \(\frac{4}{3}\)) B₁, B₂, B₃, B₄ on ‘BX’ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.
4. Join B₃C (smaller of 3 and 4 in \(\frac{4}{3}\)).
5. Through B₄, draw a line parallel to B₃C meeting BC in C’ on being produced.
6. Through C’, draw another line parallel to CA meeting BA in A’ on being produced.
7. ∆A’BC’ is the required triangle whose sides are times the triangle ABC.

Justification of construction:
Consider the ∆ A’BC’ and ∆ ABC,
∠B = ∠B [common]
∠A’C’B = ∠ACB [construction]
∴ ∆A’BC’ ~ ∆ABC [AA – similarity]
∴ Their corresponding sides must be in the same ratio

\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}\) ………….(1)

Again, consider ∆B₄B C’ and ∆B₃BC,
∠B = ∠B [common]
∠C’B4B = ∠CB3B [By consiruction]
∴ BB C’ AB3BC [AA-si niilarity]
∴ Their corresponding sides must be in the same ratio
\(\frac{\mathrm{B}_{4} \mathrm{~B}}{\mathrm{~B}_{3} \mathrm{~B}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{B}_{4}}{\mathrm{CB}_{3}}\)
I II III

Taking I and II members.

\(\frac{B C^{\prime}}{B C}=\frac{B_{4} B}{B_{3} B}\)

But, \(\frac{B_{4} B}{B_{3} B}=\frac{4}{3}\) (construction)

or \(\frac{B C^{\prime}}{B C}=\frac{4}{3}\) ………….(2)

From (1) and (2),

\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}=\frac{4}{3}\)

⇒ A’B = \(\frac{4}{3}\) AB; BC’ = \(\frac{4}{3}\) BC and C’A’ = \(\frac{4}{3}\) CA
Hence the construction is justified.

Question 7.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle.
Solution:
Steps of construction:
1. Draw a right triangle using given conditions. Consider the triangle as ABC in which BC = 4 cm; AB = 3 cm and
∠B = 90°.
2. Make any acute angle ∠CBX below the line BC.
3. Locate five points (greater of 5 and 3 in \(\frac{5}{3}\)) B₁, B₂, B₃, B₄. B₅ on BX such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅.
4. Join B₃ (smaller of ‘5’ and ‘3’ in \(\frac{5}{3}\)) and ‘C’.

5. Through B₅. draw a line parallel to BC meeting BC is C’ on being produced.
6. Again draw a line through C’ parallel to CA meeting BA in A’ on being produced.
∆A’BC’ is the required triangle whose sides are \(\frac{5}{3}\) times the sides of ∆ABC.

Justification of construction :
Consider ∆A’BC’ and ∆ABC
∠B = ∠B [common]
∠A’C’B = ∠ACB [By construction]
∴ ∆A’BC’ ~ ∆ABC [AA-similarity condition]
∴ Their corresponding sides must be in the same ratio
\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}\) ……………..(1)

Again, in ∆B₅C’B and ∆XB₃CB,
∠B = ∠B [common]
∠C’B₅B = ∠CB₃B [By construction]
∴ ∆B₅C’B ~ ∆B₃CB [AA-similarityj
∴ Their corresponding sidcs must be in the same ratio.

\(\frac{\mathrm{B}_{5} \mathrm{C}^{\prime}}{\mathrm{B}_{3} \mathrm{C}}=\frac{\mathrm{C}^{\prime} \mathrm{B}}{\mathrm{CB}}=\frac{\mathrm{BB}_{5}}{\mathrm{BB}_{3}}\)

I II III

Taking II and III members.
\(\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{B}_{5} \mathrm{~B}}{\mathrm{~B}_{3} \mathrm{~B}}\)

But, \(\frac{B_{5} B}{B_{3} B}=\frac{5}{3}\) [construction]

\(\frac{B C^{\prime}}{B C}=\frac{5}{3}\) ……………(2)
From (1) and (2),

\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}=\frac{5}{3}\)

⇒ A’B = \(\frac{5}{3}\) AB; BC’ = \(\frac{5}{3}\) BC and C’A’ = \(\frac{5}{3}\) CA
Hence the construction is justified.