Question
In the given figure, $ABCD$ is a rectangle with $\text{AB}=80\text{cm}$ and $\text{BC}=70\text{cm},$ $\angle\text{AED}=90^\circ$ and $\text{DE}=42\text{cm}$ A semicircle is dn wn, taking BC as diameter. Find the area o the shaded region.
✓
Answer
Length of rectangle $ABCD = AB = 80cm$
Breadth of rectangle $ABCD = BC = 70cm$
$\therefore$ Area of rectangle $ABCD = AB \times BC = 80 \times 70 = 5600cm^2$
In right-angled $\triangle\text{AED},$
$AE^2 = (AD^2 - DE^2) = (70^2 - 42^2) = (70 + 42) = 112 \times 28 = 4 \times 28 \times 28$
$\Rightarrow AE = 2 \times 28 = 56cm$
$\therefore$ Area of $\triangle\text{AED}=\frac12\times\text{DE}\times\text{AE}=\frac12\times42\times56=1176\text{cm}^2$
Area of semi-circle $=\frac12\pi\times\Big(\frac{70}{2}\Big)^2$
$=\Big\{\frac{1}{2}\times\frac{22}{7}\times35\times35\Big\}\text{cm}^2=1925\text{cm}^2$
Thus, Area of the shaded region
= Area of rectangle ABCD $- ($Area of $\triangle\text{AED}$ + Area of semi-circle$)$
$= 5600 - (1176 + 1925)$
$= 5600 - 3101$
$= 2499cm^2$
Breadth of rectangle $ABCD = BC = 70cm$
$\therefore$ Area of rectangle $ABCD = AB \times BC = 80 \times 70 = 5600cm^2$
In right-angled $\triangle\text{AED},$
$AE^2 = (AD^2 - DE^2) = (70^2 - 42^2) = (70 + 42) = 112 \times 28 = 4 \times 28 \times 28$
$\Rightarrow AE = 2 \times 28 = 56cm$
$\therefore$ Area of $\triangle\text{AED}=\frac12\times\text{DE}\times\text{AE}=\frac12\times42\times56=1176\text{cm}^2$
Area of semi-circle $=\frac12\pi\times\Big(\frac{70}{2}\Big)^2$
$=\Big\{\frac{1}{2}\times\frac{22}{7}\times35\times35\Big\}\text{cm}^2=1925\text{cm}^2$
Thus, Area of the shaded region
= Area of rectangle ABCD $- ($Area of $\triangle\text{AED}$ + Area of semi-circle$)$
$= 5600 - (1176 + 1925)$
$= 5600 - 3101$
$= 2499cm^2$
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