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The areas of two similar triangles are 169 cm^2 and 121 cm^2 respectively. If the longest side of the larger triangle is 26 cm, find the longest side of the smaller triangle.

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Question

The areas of two similar triangles are 169 cm2 and 121 cm2 respectively. If the longest side of the larger triangle is 26 cm, find the longest side of the smaller triangle.   

Sum
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Solution

It is given that the triangles are similar.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Let the longest side of smaller triangle be X cm.  

`(ar("Larger triangle"))/(ar("Smaller triangle"))=(("Longest side of larger trainglle" )^2)/((L"ongest side of smaller trainglle")^2)` 

⇒ `169/121=26^2/x^2` 

⇒ `x= sqrt((26xx26xx121)/169)` 

= 22
Hence, the longest side of the smaller triangle is 22 cm. 

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Chapter 7: Triangles - EXERCISE 7C [Page 417]

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R.S. Aggarwal Mathematics [English] Class 10
Chapter 7 Triangles
EXERCISE 7C | Q 4. | Page 417
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