# $A, B$ 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

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

$=6\ cm^2$

Updated on: 10-Oct-2022

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