Question
In Fig below, ABCD is a rectangle with dimensions 32m by 18m. ADE is a triangle such that $\text{EF}\bot\text{AD}$ and EF = 14cm. Calculate the area of the shaded region.
✓
Answer
We have,
Area of the rectangle $=A B \times B C=32 m \times 18 m=576 m^2$
Area of the triangle $=\frac{1}{2}( AD \times FE )$
$=\frac{1}{2}(BC \times FE)[\text { Since } AD=BC]$
$=\frac{1}{2}(18 m \times 14 m)$
$=9 m \times 14 m$
$=126 m^2$
Area of the shaded region = Area of the rectangle - Area of the triangle
$=(576-126) m^2$
$=450 m^2$
Area of the rectangle $=A B \times B C=32 m \times 18 m=576 m^2$
Area of the triangle $=\frac{1}{2}( AD \times FE )$
$=\frac{1}{2}(BC \times FE)[\text { Since } AD=BC]$
$=\frac{1}{2}(18 m \times 14 m)$
$=9 m \times 14 m$
$=126 m^2$
Area of the shaded region = Area of the rectangle - Area of the triangle
$=(576-126) m^2$
$=450 m^2$
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