In $ \triangle \mathrm{ABC}, \mathrm{M} $ and $ \mathrm{N} $ are the midpoints of $ \mathrm{AB} $ and $ \mathrm{AC} $ respectively. If the area of $ \triangle \mathrm{ABC} $ is $ 90 \mathrm{~cm}^{2} $, find the area of $ \triangle \mathrm{AMN} $.
Given:
In \( \triangle \mathrm{ABC}, \mathrm{M} \) and \( \mathrm{N} \) are the midpoints of \( \mathrm{AB} \) and \( \mathrm{AC} \) respectively.
The area of \( \triangle \mathrm{ABC} \) is \( 90 \mathrm{~cm}^{2} \).
To do:
We have to find the area of \( \triangle \mathrm{AMN} \).
Solution:
We know that,
Area of the triangle formed by joining the midpoints of the sides of a triangle is equal to one-fourth the area of the given triangle.
Similarly,
Area of the triangle formed by a vertex and the midpoints of the adjacent sides is equal to one-fourth the area of the given triangle.
Therefore,
Area of triangle AMN $=\frac{1}{4}\times$ Area of triangle ABC
Area of triangle AMN $=\frac{1}{4}\times 90\ cm^2$
Area of triangle AMN $=22.5\ cm^2$
The area of \( \triangle \mathrm{AMN} \) is $22.5\ cm^2$.
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