# $\mathrm{X}, \mathrm{Y}$ 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

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

$=60\ cm^2$

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Updated on: 10-Oct-2022

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