In Fig $31$ $A B C D$ is a rectangle with diameter $32 m$ by $18 m$ . $A D E$ is a triangle such that $E F ⊥ A D$ and $E F = 14 c m$ . Calculate the area of the shaded region.

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Solution

area of shaded portion = area of rectangle - area of triangle

$= 32 \times 18 - \frac{1}{2} \times 18 \times 14$

$= 576 - 126$

$= 450 m^{2}$

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In Fig 31 ABCD is a rectangle with diameter 32 m by 18 m. ADE is a triangle such that EF⊥AD and EF=14 cm. Calculate the area of the shaded region.

area of shaded portion = area of rectangle - area of triangle=32×18−12×18×14=576−126=450m2

In Fig $31$ $A B C D$ is a rectangle with diameter $32 m$ by $18 m$ . $A D E$ is a triangle such that $E F ⊥ A D$ and $E F = 14 c m$ . Calculate the area of the shaded region.

area of shaded portion = area of rectangle - area of triangle

$= 32 \times 18 - \frac{1}{2} \times 18 \times 14$

$= 576 - 126$

$= 450 m^{2}$