NCERT Solutions for Class 10 Maths Chapter 11

Important NCERT Questions

Constructions

NCERT Books for Session 2022-2023

CBSE Board and UP Board Others state Board

EXERCISE 11.1

Page No:220

Questions No:5

# Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and angle B = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.

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A ΔA’BC’ whose sides are 3/4 of the corresponding sides of Δ ABC can be drawn as follows.

Step 1

Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.

Step 2

Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

Step 3

Locat 4 points (as 4 is greater in 3 and 4), B₁, B₂, B₃, B₄, on line segment BX.

Step 4

Join B₄C and Draw a line through B₃, parallel to B₄C intersecting BC AT C’.

Step 5

Draw a line through C’ parallel to AC intersecting AB at A’. ΔA’BC’ is the required triangle.

Justification

The construction can be justified by proving A’B = 3/4 AB, BC’ = 3/4 BC, A’C’ = 3/4 AC

In ΔA’BC’ and ΔABC,

∠A’C’B = ∠ACB (Corresponding angles)

∠A’BC’ = ∠ABC (Common)

∴ ΔA’BC’ ∼ ΔABC (AA similarity criterion)

⇒ A’B/AB = BC’/BC = A’C’/AC …(1)

In ΔBB₃C’ and ΔBB₄C,

∠B₃BC’ = ∠B₄BC (Common)

∠BB₃C’ = ∠BB₄C ( Corresponding angles)

∴ ΔBB₃C ∼ ΔBB₄C (AA similarity criterion)

⇒ BC’/BC = BB₃/BB₄ ⇒ BC’/BC = 3/4 …(2)

From equations (1) and (2), we obtain

A’B/AB = BC’/BC = A’C’/AC = 3/4

⇒ A’B = 3/4 AB, BC’ = 3/4 BC A’C’ = 3/4 AC

This justifies the construction.