Question

The areas of two similar triangles are 100 $c{m}^2$ and 64 $c{m}^2$. If the median of greater side of first triangle is 13 cm, find the corresponding median of the other triangle.

• A. $20.3cm$
• B. $15cm$
• C. $10.4cm$
• D. $16cm$

Answer 1

Answer: (C)

$10.4cm$

Let, $Area\left(ABC\right)=100c{m}^{2}Area\left(PQR\right)=64c{m}^2$

Since Median divides triangle into two triangles equal area, So,

$Area\left(ABD\right)=50c{m}^{2}Area\left(PQS\right)=32c{m}^2$

Now, using $BPT$ and Area theorem, we get

$\frac{AD}{PS}=\frac{\sqrt{Area\left(ABD\right)}}{\sqrt{Area\left(PQS\right)}}=\frac{\sqrt{50}}{\sqrt{32}}=\frac{5}{4}$

Let, $AD=13cm,$

So, $PS=10.4cm.$