The areas of two similar triangles are 121 cm2 and 64 cm2 respectively.

Question:

The areas of two similar triangles are $121 \mathrm{~cm}^{2}$ and $64 \mathrm{~cm}^{2}$ respectively. If the median of the first triangle is $12.1 \mathrm{~cm}$, then the corresponding median of the other triangle is

(a) 11 cm
(b) 8.8 cm
(c) 11.1 cm
(d) 8.1 cm

Solution:

Given: The area of two similar triangles is $121 \mathrm{~cm}^{2}$ and $64 \mathrm{~cm}^{2}$ respectively. The median of the first triangle is $2.1 \mathrm{~cm}$.

To find: Corresponding medians of the other triangle

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

$\frac{\operatorname{ar}(\text { trianglel })}{\operatorname{ar}(\text { triangle2 })}=\left(\frac{\text { median } 1}{\text { median } 2}\right)^{2}$

$\frac{121}{64}=\left(\frac{12.1}{\text { median2 } 2}\right)^{2}$

Taking square root on both side, we get

118=12.1cmmedian2⇒median2= 8.8 cm

Hence the correct answer is $(b)$

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