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 : AA_{1}, A_{1}A_{2}, A_{2}A_{3,} A_{3}A_{4}, A_{4}A_{5}, A_{5}A_{6}, A_{6}A_{7}, A_{7}A_{8}, A_{8}A_{9}, A_{9}A_{10}, A_{10}A_{11}, A_{11}A_{12} and A_{12}A_{13}.

4. Join A_{13}B.

5. From A_{5}, draw A_{5}P || A_{13}B, 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 ∆ ABA_{13}, PA_{5} || BA_{13}.

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 : B_{1}, B_{1} and B_{3} such that BB_{1} = B_{1}B_{2} = B_{2}B_{3}

7. Join B_{3}C. [Note that 3 is denominator]

8. From B_{2} draw B_{2}D || B_{3}C, 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 CB_{1}.

3. Along BX, mark off seven (bigger of the numerator and denominator) points : B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6} and B_{7} such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5} = B_{5}B_{6} = B_{6}B_{7}

4. Join B_{5}C. [Note that 5 is denominator]

5. From B_{7}, draw B_{7}D || B_{5}C, 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 B_{1}, B_{2} and B_{3} on BX such that BB_{1} = B_{1}B_{2} = B_{2}B_{3}

7. Join BA_{2}C.

8. From B_{3}, draw B_{3}D || B_{2}C 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 CB_{1}.

3. Along BX, mark off 4 points : B_{1}, B_{2}, B_{3} and B_{4} such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}.

4. Join B_{4}C.

5. From B_{3}, draw B_{3}D || B_{4}C 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 : B_{1}, B_{2}, B_{3} and B_{4} such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}.

4. Join B_{3}C.

5. From B_{4}, draw B_{4}D || B_{3}C, 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 : B_{1}, B_{2}, B_{3} and B_{4} such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}.

4. Join B_{3}C.

5. From B_{5}, draw B_{5}D || B_{3}C, 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.