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:3

# Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle.

Share

Get Hindi Medium and English Medium NCERT Solution for Class 10 Maths to download.

Please follow the link to visit website for first and second term exams solutions.

https://www.tiwariacademy.com/ncert-solutions/class-10/maths/chapter-11/

Step 1

Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 5 cm radius respectively.

Let these arcs intersect each other at point C. ΔABC is the required triangle having length of sides as 5 cm, 6 cm, and 7 cm respectively.

Step 2

Draw a ray AX making acute angle with line AB on the opposite side of vertex C.

Step 3

Locate 7 points, A₁, A₂, A₃, A₄, A₅, A₆, A₇, (as 7 is greater between 5 and 7), on line AX such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇.

Step 4

Join BA₅ and draw a line through A₇ parallel to BA₅ to intersect extended line segment AB at point B’.

Step 5

Draw a line through B’ parallel to BC intersecting the extended line segment AC at C’ ΔAB’C’ is the required triangle.

Justification

The construction can be justified by proving that

AB’ = 7/5 AB, B’C’ = 7/5 BC, AC’ = 7/5 AC

In ΔABC and Δ AB’C’,

∠ABC = ∠AB’C’ (Corresponding angles)

∠BAC = ∠B’AC’ (common)

∴ ΔABC ∼ ΔAB’C’ (AA similarity Criterion)

⇒ AB’/AB = B’C’/BC = AC’/AC …(1)

In ΔAA₅B/AA₇B’,

∠A₅AB = ∠A₇AB’ (common)

∠AA₅B = ∠AA₇B'(Corresponding angles)

∴ ΔAA₅B ∼ ΔAA₇B’ (AA similarity criterion)

⇒ AB’/AB = AA₅/AA₇ ⇒ AB’/AB = 5/7 …(2)

On comparing equations (1) and (2), we obtain.

AB’/AB = B’C’/BC = AC’/AC = 7/5

⇒ AB’ = 7/5 AB, B’C’ = 7/5 BC, AC’ = 7/5 AC

This justifies the construction.