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 = 15m. Calculate the area of the shaded region.
✓
Answer
$A B C D$ is a rectangle in which $A B=36 m$
and $B C=24 m$
In $\triangle AED$,
$E F=15 m$
$A D=B C=24 m$
Now area of rectangle $A B C D=1 \times b$
$=36 \times 24 cm^2=864 cm^2$
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 $=864-180=684 m^2$
and $B C=24 m$
In $\triangle AED$,
$E F=15 m$
$A D=B C=24 m$
Now area of rectangle $A B C D=1 \times b$
$=36 \times 24 cm^2=864 cm^2$
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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