$ A, B $ and $ C $ are the midpoints of the sides of $ \Delta X Y Z $. $ P, Q $ and $ R $ are the midpoints of the sides of $ \triangle \mathrm{ABC} $. If $ \mathrm{ABC}=24 \mathrm{~cm}^{2} $, find $XYZ$ and $PQR$.

Given:

A, \( B \) and \( C \) are the midpoints of the sides of \( \Delta X Y Z \). \( P, Q \) and \( R \) are the midpoints of the sides of \( \triangle \mathrm{ABC} \). \( \mathrm{ABC}=24 \mathrm{~cm}^{2} \).

To do:

We have to find the area of XYZ and PQR.

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

Similarly,

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

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

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

Therefore,

$24=\frac{1}{4}\times$ Area of triangle XYZ

Area of triangle XYZ $=4\times24$

$=96\ cm^2$

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

$=\frac{1}{4}\times24$

$=6\ cm^2$