\( \mathrm{X}, \mathrm{Y} \) 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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