The areas of two similar triangles are in respectively $9 mathrm{~cm}^{2}$ and $16 mathrm{~cm}^{2}$. The ratio of their corresponding sides is
Question:

The areas of two similar triangles are in respectively $9 \mathrm{~cm}^{2}$ and $16 \mathrm{~cm}^{2}$. The ratio of their corresponding sides is

(a) 3 : 4
(b) 4 : 3
(c) 2 : 3
(d) 4 : 5

Solution:

Given: Areas of two similar triangles are 9cm2 and 16cm2.

To find: Ratio of their corresponding sides.

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\frac{\operatorname{ar}(\text { triangle } 1)}{\operatorname{ar}(\text { triangle } 2)}=\left(\frac{\text { side } 1}{\text { side } 2}\right)^{2}$

$\frac{9}{16}=\left(\frac{\text { side 1 }}{\text { side 2 }}\right)^{2}$

Taking square root on both sides, we get

side1side2=34

So, the ratio of their corresponding sides is 3 : 4.

Hence the correct answer is $(a)$.

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