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