 # ML Aggarwal Solutions Class 10 Mathematics Solutions for Similarity Exercise 13.3 in Chapter 13 - Similarity

The area of two similar triangles are 36 cm² and 25 cm². If an altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other triangle.

Solution:-

The square of any ratio between any two corresponding sides of two similar triangles is equal to the ratio of their areas.

From the question it is given that,

The area of two similar triangles are 36 cm² and 25 cm².

Let us assume ∆PQR ~ ∆XYZ, PM and XN are their altitudes.

So, area of ∆PQR = 36 cm2

Area of ∆XYZ = 25 cm2

PM = 2.4 cm

Assume XN = a

We know that,

area of ∆PQR/area of ∆XYZ = PM2/XN2

36/25 = (2.4)2/a2

By cross multiplication we get,

36a2 = 25 (2.4)2

a2 = 5.76 × 25/36

a2 = 144/36

a2 = 4

a = √4

a = 2 cm

Therefore, altitude of the other triangle XN = 2 cm.

Related Questions

Lido

Courses

Teachers

Book a Demo with us

Syllabus

Maths
CBSE
Maths
ICSE
Science
CBSE

Science
ICSE
English
CBSE
English
ICSE
Coding

Terms & Policies

Selina Question Bank

Maths
Physics
Biology

Allied Question Bank

Chemistry
Connect with us on social media!