Question
$ABC$ is an isosceles right-angled triangle with $\angle ABC = 90^\circ $. A semi-circle is drawn with $AC$ as the diameter. If $AB = BC = 7 cm$, find the area of the shaded region. [Take $\pi = 22/7]$
✓
Answer
In ΔABC, using Pythagoras theorem $AC^2 = AB^2 + BC^2$
$AC^2= (7)^2 + (7)^2$
$AC^2= 49 + 49$
$AC^2= 98$
$\Rightarrow AC =7 \sqrt{2}$
Radius of semi-circle $=\frac{ AC }{2}=\frac{7 \sqrt{2}}{2}$
$\therefore$ Area of the shaded region = Area of semi-circle – Area of ΔABC
Area of semi-circle = $\frac{1}{2} \pi r^2=\frac{1}{2} \pi\left(\frac{7 \sqrt{2}}{2}\right)^2$
$=\frac{98 \pi}{8}=\frac{49 \pi}{4}$
$=\frac{49}{4}$
$=\frac{49}{4} \times \frac{22}{7}$
$=\frac{77}{2} cm ^2$
Area of $\triangle ABC =\frac{1}{2} \times BC \times AB$
$=\frac{1}{2} \times 7 \times 7$
$=\frac{49}{2} cm ^2$
Thus, Area of the shaded region = $\frac{77}{2} cm ^2-\frac{49}{2} cm ^2=\frac{28}{2} cm ^2=14 cm ^2$.
$AC^2= (7)^2 + (7)^2$
$AC^2= 49 + 49$
$AC^2= 98$
$\Rightarrow AC =7 \sqrt{2}$
Radius of semi-circle $=\frac{ AC }{2}=\frac{7 \sqrt{2}}{2}$
$\therefore$ Area of the shaded region = Area of semi-circle – Area of ΔABC
Area of semi-circle = $\frac{1}{2} \pi r^2=\frac{1}{2} \pi\left(\frac{7 \sqrt{2}}{2}\right)^2$
$=\frac{98 \pi}{8}=\frac{49 \pi}{4}$
$=\frac{49}{4}$
$=\frac{49}{4} \times \frac{22}{7}$
$=\frac{77}{2} cm ^2$
Area of $\triangle ABC =\frac{1}{2} \times BC \times AB$
$=\frac{1}{2} \times 7 \times 7$
$=\frac{49}{2} cm ^2$
Thus, Area of the shaded region = $\frac{77}{2} cm ^2-\frac{49}{2} cm ^2=\frac{28}{2} cm ^2=14 cm ^2$.
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