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Mensuration — MATHS STD 7 — Question

Gujarat BoardEnglish MediumSTD 7MATHSMensuration3 Marks
Question
In the given figure, $ABCD$ is a rectangle with length $= 36m$ and breadth $= 24m$. In $\triangle\text{ADE},\text{EF}\perp\text{AD}$ and $EF = 15\ m$. Calculate the area of the shaded region.
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Answer

$A B C D$ is a rectangle in which $A B=36 m$
and $B C=24 m$
In $\triangle AED$,
$EF =15 m$
$A D=B C=24 m$
Now area of rectangle $A B C D=l \times b$
$=36 \times 24 cm^2=864 cm^2$
Area of $\triangle AED =\frac{1}{2} \times AD \times EF$
$=\frac{1}{2} \times 24 \times 15 cm^2=180 cm^2$
Area of shaded portion $=864-180=684 m^2$

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