Bihar Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1

Bihar Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1 Textbook Questions and Answers.

BSEB Bihar Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1

Question 1.
Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
Solution:
Steps of Construction :

1. Draw a line segment AB = 7.6 cm.

2. Draw a ray AC making any acute angle with AB, as shown in the figure.

3. On ray AC, starting from A, mark 5 + 8 = 13 equal line segments : AA1, A1A2, A2A3, A3A4, A4A5, A5A6, A6A7, A7A8, A8A9, A9A10, A10A11, A11A12 and A12A13.

4. Join A13B.

5. From A5, draw A5P || A13B, meeting AB at P.

6. Thus, P divides AB in the ratio 5 : 8.
On measuring the two parts, we find that AP = 2.9 cm and PB = 4.7 cm (approx.).

Justification :
In ∆ ABA13, PA5 || BA13.
So, $$\frac { AP }{ PB }$$ = $$\frac{\mathrm{AA}_{5}}{\mathrm{~A}_{5} \mathrm{~A}_{13}}$$ = $$\frac { 5 }{ 8 }$$ [By BPT]
Thus, AP : PB = 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 the corresponding sides of it.
Solution:
Steps of Construction :

1. Draw a line segment BC = 6 cm.

2. With B as centre and radius equal to 5 cm, draw an arc.

3. With C as centre and radius equal to 4 cm, draw an arc intersecting the previously drawn arc at A.

4. Join AB and AC. Then ∆ ABC is the required triangle.

5. Below BC, make an acute angle CBX.

6. Along BX, mark off three (bigger of the numerator and denominator) points : B1, B1 and B3 such that BB1 = B1B2 = B2B3

7. Join B3C. [Note that 3 is denominator]

8. From B2 draw B2D || B3C, meeting BC at D. [Note that 2 is numerator]

9. From D, draw DE || CA, meeting BA at E. Then, ∆ EBD is the required triangle whose sides are $$\frac { 2 }{ 3 }$$ of the corresponding sides of ∆ ABC.

Justification :
Since DE || CA, therefore
∆ ABC ~ ∆ EBD
and $$\frac { EB }{ AB }$$ = $$\frac { DE }{ CA }$$ = $$\frac { BD }{ BC }$$ = $$\frac { 2 }{ 3 }$$ [Since $$\frac { BD }{ BC }$$ = $$\frac { 2 }{ 3 }$$ ,by construction]
Hence, we get the new triangle similar to the given triangle whose sides are equal to $$\frac { 2 }{ 3 }$$ of the corresponding sides of ∆ ABC.

Question 3.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are $$\frac { 7 }{ 5}$$ of the corresponding sides of the first triangle.
Solution:
Steps of Construction :
1. With the given data, construct ∆ ABC in which BC = 7 cm, CA = 4 cm and AB = 5 cm.

2. Below BC, make an acute angle CB1.

3. Along BX, mark off seven (bigger of the numerator and denominator) points : B1, B2, B3, B4, B5, B6 and B7 such that BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7

4. Join B5C. [Note that 5 is denominator]

5. From B7, draw B7D || B5C, meeting BC produced at D. [Note that 7 is numerator]

6. From D, draw DE || CA, meeting BA produced at E. Then, ∆ EBD is the required triangle whose sides are $$\frac { 7 }{ 5}$$ of the corresponding sides of ∆ ABC.

Justification :
Since DE || CA, therefore
∆ ABC ~ ∆ EBD and $$\frac { EB }{ AB }$$ = $$\frac { DE }{ CA }$$ = $$\frac { BD }{ BC }$$ = $$\frac { 7 }{ 5 }$$
Hence, we get the new triangle similar to the given triangle whose sides are equal to $$\frac { 7 }{ 5 }$$ of the corresponding sides of ∆ ABC.

Question 4.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1$$\frac { 1 }{ 2 }$$ times the corresponding sides of the isosceles triangle.
Solution:
Steps of Construction :

1. Draw BC = 8 cm.

2. Construct PQ, the perpendicular bisector of line segment BC meeting BC at M.

3. Along MP, cut off MA = 4 cm.

4. Join BA and CA. Then, ∆ ABC so obtained is the required triangle.

5. Make an acute angle CBX below BC.

6. Mark three points B1, B2 and B3 on BX such that BB1 = B1B2 = B2B3

7. Join BA2C.

8. From B3, draw B3D || B2C meeting BC produced at D.

9. From D, draw DE || CA meeting BA produced at E. Then, ∆ EBD is the required triangle.

Justification :
Since DE || CA, therefore ∆ ABC ~ ∆ EBD
and $$\frac { EB }{ AB }$$ = $$\frac { DE }{ CA }$$ = $$\frac { BD }{ BC }$$ = $$\frac { 3 }{ 2 }$$
Hence, we get the new triangle similar to the given $$\frac { 3 }{ 2 }$$ triangle whose sides are $$\frac { 3 }{ 2 }$$, i.e., 1 $$\frac { 1 }{ 2 }$$ times of the corresponding sides of the isosceles ∆ ABC.

Question 5.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are $$\frac { 3 }{ 4 }$$ of the corresponding sides of the triangle ABC.
Solution:
Steps of Construction :
1. With the given data, construct ∆ ABC in which BC 6 cm, ∠ABC = 60° and AB = 5 cm.

2. Below BC, make an acute angle CB1.

3. Along BX, mark off 4 points : B1, B2, B3 and B4 such that BB1 = B1B2 = B2B3 = B3B4.

4. Join B4C.

5. From B3, draw B3D || B4C to meet BC at D.

6. From D, draw ED || AC, meeting BA at E. Then, ∆ EBD is the required triangle whose sides are $$\frac { 3 }{ 4 }$$ th of the corresponding sides of ∆ ABC.

Justification :
Since DE || CA, therefore ∆ ABC ~ ∆ EBD
and $$\frac { EB }{ AB }$$ = $$\frac { DE }{ CA }$$ = $$\frac { BD }{ BC }$$ = $$\frac { 3 }{ 4 }$$
Hence, we get the new triangle similar to the given triangle whose sides are equal to $$\frac { 3 }{ 4 }$$ of the corresponding 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 $$\frac { 4 }{ 3 }$$ times the corresponding sides of ∆ ABC.
Solution:
Steps of Construction :
1. With the given data, construct ∆ ABC in which BC = 7 cm, ∠B = 45°, ∠C = 180° – (∠A + ∠B).

i. e., ∠C = 180° – (105° + 45°)
= 180° – 150° = 30°.

2. Below BC, make an acute ∠CBX.

3. Along BX, mark off four points : B1, B2, B3 and B4 such that BB1 = B1B2 = B2B3 = B3B4.

4. Join B3C.

5. From B4, draw B4D || B3C, meeting BC produced at D.

6. From D, draw DE || CA, meeting BA produced at E. Then, ∆ EBD is the required triangle whose sides are $$\frac { 4 }{ 3 }$$ times of the corresponding sides of ∆ ABC.

Justification :
Since DE || CA, therefore ∆ ABC ~ ∆ EBD
and $$\frac { EB }{ AB }$$ = $$\frac { DE }{ CA }$$ = $$\frac { BD }{ BC }$$ = $$\frac { 4 }{ 3 }$$
Hence, we get the new triangle similar to the given triangle whose sides are equal to $$\frac { 4 }{ 3 }$$ times of the corresponding I sides of ∆ ABC.

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. With given data, construct ∆ ABC in which BC = 4 cm, ∠B = 90° and BA = 3 cm.

2. Below BC, make an acute angle CBX.

3. Along BX, mark off four points : B1, B2, B3 and B4 such that BB1 = B1B2 = B2B3 = B3B4.

4. Join B3C.

5. From B5, draw B5D || B3C, meeting BC produced at D.

6. From D, draw DE || CA, meeting BA produced at E.
Then, EBD is the required triangle whose sides are $$\frac { 5 }{ 3 }$$ times of the corresponding sides of ∆ ABC.

Justification :
Since DE || CA, therefore ∆ ABC ~ ∆ EBD
and $$\frac { EB }{ AB }$$ = $$\frac { DE }{ CA }$$ = $$\frac { BD }{ BC }$$ = $$\frac { 5 }{ 3 }$$
Hence, we get the new triangle similar to the given triangle whose sides are equal to $$\frac { 5 }{ 3 }$$ times of the corresponding sides of ∆ ABC.