Question
In the following figure, ABC is a right angled triangle in which $\angle\text{A}=90^\circ,$ $AB = 21\ cm$ and $AC = 28\ cm$. Semi-circles are described on AB, BC and AC as diameters. Find the area of the shaded region.
✓
Answer
In the figure, ABC is a right triangle in a semicircle $\angle\text{A}=90^\circ,$ $AB = 21\ cm$ and $AC = 28\ cm$
Semicircles are drawn on BC and AC as diameters
In right $\triangle\text{ABC}$
$BC^2 = AB^2 + AC^2$ (Pythagoras Theorem)
$= 21^2 + 28^2$
$= 441 + 784 = 1225 = (35)^2$
$\therefore$ $BC = 35\ cm$
Now radius of bigger semicircle (R) $=\frac{35}{2}\text{cm},$
of semicircle at $\text{AB}=\frac{21}{2}\text{cm}$ and of semicircle on $\text{AC}=\frac{28}{2}\text{cm}=14\text{cm}$
Now area of shaded portion
= Area of semicircle on AB as diameter + area of semicircle on AC as diameter + area of $\triangle\text{ABC}$ - area of
semicircle on BC as diameter
$=\frac{1}{2}\pi\Big(\frac{21}{2}\Big)^2+\frac{1}{2}\cdot\pi\Big(\frac{28}{2}\Big)^2+\frac{1}{2}\text{AB}\times\text{AC}-\frac{1}{2}\pi\Big(\frac{35}{2}\Big)^2$
$=\frac{\pi}{2}\bigg[\Big(\frac{21}{2}\Big)^2+\Big(\frac{28}{2}\Big)^2-\Big(\frac{35}{2}\Big)^2\bigg]+\frac{1}{2}\times21\times28$
$=\frac{\pi}{2}\Big[\frac{441}{4}+196-\frac{1225}{4}\Big]+294$
$=\frac{\pi}{2}\big[441+784-1225\big]+294$
$=\frac{\pi}{2}\times\big[1225-1225\big]+294$
$=\frac{\pi}{2}\times0+294=294\text{cm}^2$
Semicircles are drawn on BC and AC as diameters
In right $\triangle\text{ABC}$
$BC^2 = AB^2 + AC^2$ (Pythagoras Theorem)
$= 21^2 + 28^2$
$= 441 + 784 = 1225 = (35)^2$
$\therefore$ $BC = 35\ cm$
Now radius of bigger semicircle (R) $=\frac{35}{2}\text{cm},$
of semicircle at $\text{AB}=\frac{21}{2}\text{cm}$ and of semicircle on $\text{AC}=\frac{28}{2}\text{cm}=14\text{cm}$
Now area of shaded portion
= Area of semicircle on AB as diameter + area of semicircle on AC as diameter + area of $\triangle\text{ABC}$ - area of
semicircle on BC as diameter
$=\frac{1}{2}\pi\Big(\frac{21}{2}\Big)^2+\frac{1}{2}\cdot\pi\Big(\frac{28}{2}\Big)^2+\frac{1}{2}\text{AB}\times\text{AC}-\frac{1}{2}\pi\Big(\frac{35}{2}\Big)^2$
$=\frac{\pi}{2}\bigg[\Big(\frac{21}{2}\Big)^2+\Big(\frac{28}{2}\Big)^2-\Big(\frac{35}{2}\Big)^2\bigg]+\frac{1}{2}\times21\times28$
$=\frac{\pi}{2}\Big[\frac{441}{4}+196-\frac{1225}{4}\Big]+294$
$=\frac{\pi}{2}\big[441+784-1225\big]+294$
$=\frac{\pi}{2}\times\big[1225-1225\big]+294$
$=\frac{\pi}{2}\times0+294=294\text{cm}^2$
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