RBSE Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 11 Constructions Exercise 11.1

In each of the following constructions, give the justification of the construction also :

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. First of all draw a line segment AB = 7.6 cm.
2. Now draw a ray AC which makes an acute angle with AB.
3. Locate 5 + 8 = 13 equal line segments on ray AC starting from A : AA₁, A₁A₂, A₂A₃, A₃A₄, A₄A₅, A₅A₆, A₆A₇, A₇A₈, A₈A₉, A₉A₁₀, A₁₀A₁₁, A₁₁A₁₂ and A₁₂A₁₃
4. Now join A₁₃B
5. Through A₅ draw A₅P || A₁₃B which meets AB at R
6. Thus, P divides AB in the ratio 5 : 8.
On measuring the two parts AP = 2.9 cm. and PB = 4.7 cm.
Justification—
In △ABA₁₃
PA₅ || BA₁₃
∴ △APA₅ ~ △ABA₁₃
⇒ \(\frac{AP}{PB}\) = \(\frac{\mathrm{AA}_{5}}{\mathbf{A}_{5} \mathbf{A}_{13}}\) = \(\frac{5}{8}\)
⇒ 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 the first triangle.
Solution:
Steps of Construction :
1. First of all draw a line segment BC = 7 cm.
2. Now taking B as centre and radius of 5 cm, draw an arc.
3. With C as centre and radius 4 cm, draw an arc to intersect the former arc at A.
4. Join AB and AC.
Then DABC is the required triangle.
5. Make an acute angle below BC.
6. On BX, mark three points B₁, B₂ and B₃ such that
BB₁ = B₁B₂ = B₂B₃
7. Now join B₃C
8. Through B₂ draw B₂D || B₃C which meets BC at D.
9. Through D draw ED || AC which meets BA at E.
Thus △EBD is the required triangle whose sides are \(\frac{2}{3}\) times the corresponding sides of △ABC.

Justification—Since DE || CA
∴ △ABC ~ △EBD
and \(\frac{EB}{AB}\) = \(\frac{BD}{BC}\) = \(\frac{DE}{CA}\) = \(\frac{2}{3}\)
Thus, we get a new triangle which is similar to the given triangle, whose sides are \(\frac{2}{3}\) times the corresponding sides of △ABC.

Question 3.
Construct a triangle with sides 5 cm., 6 cm. and 7 cm. and then another triangle where sides are \(\frac{7}{5}\) of the corresponding sides of the first triangle.
Solution:
Steps of Construction :
1. First of all construct a triangle ABC with given values in which BC = 7 cm., CA = 6 cm. and AB = 5 cm.
2. Now below BC make an acute angle CBX,
3. Where 7 points on BX :
B₁,B₂, B₃, B₄, B₅, B₆ and B₇ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇
4. Now join B₅C.
5. Through B₇ draw B₇D || B₅C. Join BC producing upto D.
6. Through D draw DE || CA, which meets BA produced at E.
Thus △EBD is the required triangle whose sides are \(\frac{7}{5}\) times the corresponding sides of △ABC.

Justification—
Since DE || CA
∴ △ABC ~ △EBD
and \(\frac{EB}{AB}\) = \(\frac{BD}{BC}\) = \(\frac{DE}{CA}\) = \(\frac{7}{5}\)
Thus we get a new triangle similar to the given triangle, where sides are \(\frac{7}{5}\) times the
corresponding side of △ABC.

Question 4.
Construct an isosceles triangle where 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. First of all draw BC = 8 cm.
2. Draw the perpendicular bisector PQ of line segment BC which meets BC at M.
3. From MP, cut MA = 4 cm.
4. Join BA and CA.
Now the DABC obtained is the required triangle.
5. Produce BC to D such that
BD = 12 cm.
6. Draw DE || CA which meets BA produced at E.
Then, △EBD will be the required triangle.

Justification—Since DE || CA
∴ △ABC ~ △EBD
and, \(\frac{EB}{AB}\) = \(\frac{DE}{CA}\) = \(\frac{BD}{BC}\) = \(\frac{12}{8}\) = \(\frac{3}{2}\)
Thus, we get a new triangle similar to the given triangle, whose sides are \(\frac{3}{2}\) times, i.e. 1\(\frac{1}{2}\) the corresponding sides of △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. First of all construct DABC with given values in which BC = 6 cm., DABC = 60° and AB = 5 cm.
2. Now make an acute angle DCBX below
3. Locate four points B₁, B₂, B₃ and B₄ on BX. Thus BB₁ = B₁B₂ = B₂B₃ = B₃B₄
4. Then join B₄C and draw B₃D || B₄C.
5. Now through D, draw ED || AC which meets BA at E.
Then △EBD will be the required triangle whose sides will be \(\frac{3}{4}\) times the corresponding sides of △ABC.

Justification—
Since DE || CA
∴ △ABC ~ △EBD
and, \(\frac{EB}{AB}\) = \(\frac{BD}{BC}\) = \(\frac{DE}{CA}\) = \(\frac{3}{2}\)
Thus, we get a new triangle similar to the given triangle, where sides are \(\frac{3}{2}\) times the 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. First of all construct △ABC with the given values in which BC = 7 cm.,
∠B = 45°, ∠C = 180° – (∠A + ∠B)
⇒ ∠C = 180° – (105° + 45°)
= 180° – 150° = 30°
2. Now make aft acute angle ∠CBX below BC.
3. Locate four points on BX :
B₁, B₂, B₃ and B₄
Thus BB₁ = B₁B₂ = B₂B₃ = B₃B₄
4. Then join B₃C.
5. Through B₄, draw B₄D || B₃C which meets BC produced at D.
6. Through D, draw ED || AC which meets BA produced at E.
Then △EBD will be the required triangle where sides will be \(\frac{4}{3}\) times the corresponding sides of △ABC.

Justification—
Since, DE || CA
∴ △ABC ~ △EBD
and, \(\frac{EB}{AB}\) = \(\frac{BD}{BC}\) = \(\frac{DE}{CA}\) = \(\frac{4}{3}\)
Thus, we get a new triangle similar to the given triangle, whose sides are \(\frac{4}{3}\) times the corresponding 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 where sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle.
Solution:
Steps of Construction :
1. First of all construct △ABC with given values in which BC = 4 cm., ∠B = 90° and BA = 3 cm.
2. Now make an acute angle CBX below BC.
3. Locate five points on BX :
B₁, B₂, B₃, B₄ and B₅ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅
4. Join B₃C.
5. Through B₅, draw B₅D || B₃C which meets BC produced at D.
6. Through D, draw ED || AC which meets BA produced at E. Thus, △EBD is the required triangle, whose sides are \(\frac{5}{3}\) times the corresponding sides of △ABC.

Justification—
Since, DE || CA
∴ △ABC ~ △EBD
and, \(\frac{EB}{AB}\) = \(\frac{BD}{BC}\) = \(\frac{DE}{CA}\) = \(\frac{5}{3}\)
Thus, we get a new triangle similar to the given triangle, whose sides are \(\frac{5}{3}\) times the corresponding sides of △ABC.