$ \mathrm{X}, \mathrm{Y} $ and $ \mathrm{Z} $ are the midpoints of the sides of $ \Delta \mathrm{PQR} . \mathrm{A}, \mathrm{B} $ and $ \mathrm{C} $ are the midpoints of the sides of $ \triangle \mathrm{XYZ} $. If $ \mathrm{PQR}=240 \mathrm{~cm}^{2} $, find $ \mathrm{XYZ} $ and $ \mathrm{ABC} $.

Given:

\( \mathrm{X}, \mathrm{Y} \) and \( \mathrm{Z} \) are the midpoints of the sides of \( \Delta \mathrm{PQR} . \mathrm{A}, \mathrm{B} \) and \( \mathrm{C} \) are the midpoints of the sides of \( \triangle \mathrm{XYZ} \).

\( \mathrm{PQR}=240 \mathrm{~cm}^{2} \)

To do:

We have to find the area of XYZ and ABC.

Solution:

We know that,

Area of the triangle formed by joining the mid points of the sides of a triangle is equal to one-fourth the area of the given triangle.

This implies,

Area of triangle XYZ $=\frac{1}{4}\times$ Area of triangle PQR

Similarly,

Area of triangle ABC $=\frac{1}{4}\times$ Area of triangle XYZ

$=\frac{1}{4}\times\frac{1}{4}\times$ Area of triangle PQR

$=\frac{1}{16}$ Area of triangle PQR

Therefore,

Area of triangle ABC $=\frac{1}{16}\times$ Area of triangle PQR

$=\frac{1}{16}\times240$

$=15\ cm^2$

Area of triangle XYZ $=\frac{1}{4}\times$ Area of triangle PQR

$=\frac{1}{4}\times240$

$=60\ cm^2$