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Areas of Circle, Sector and Segment — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsAreas of Circle, Sector and Segment5 Marks
Question
In the given figure, from a rectangular region ABCD with AB = 20cm, a right triangle AED with AE = 9cm and DE = 12cm, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside the region. Find the area of the shaded region. $\big[\text{Use }\pi=3.14\big]$
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Answer

In right-angled $\triangle\text{AED},$
$AD^2 = DE^2 + AE^2 = 12^2 + 9^2 = 144 + 81 = 225$
$\Rightarrow\text{AD}=\sqrt{225}=15\text{cm}$
Now, Area of $\triangle\text{AED}=\frac12\times\text{DE}\times\text{AE}=\frac12\times12\times9=54\text{cm}^2$
Length of rectangle ABCD = AB = 20cm
Breadth of rectangle ABCD = AD = 15cm
$\therefore$ Area of rectangle $ABCD = AB \times BC = 20 \times 15 = 300cm^2$​​​​​​​
Area of semi-circle $=\frac{1}{2}\pi\times\Big(\frac{15}{2}\Big)^2=\Big\{\frac12\times3.14\times7.5\times7.5\Big\}\text{cm}^2=88.3125\text{cm}^2$
Thus, Area of rectangle ABCD + Area of semi-circle - Area of $\triangle\text{AED}$
$= 300 + 88.31 - 54$
$= 334.31cm^2​​​​​​​$​​​​​​​

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