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Areas Related to Circles — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsAreas Related to Circles4 Marks
Question
In the following figure, ABC is a right-angled triangle,$\angle\text{B}=90^\circ,$ $AB = 28\ cm$ and $BC = 21\ cm$.
With $AC$ as diameter a semicircle is drawn and with $BC$ as radius a quarter circle is drawn.
Find the area of the shaded region correct to two decimal places.
✓

Answer

In right $\triangle\text{ABC},$ $AB = 28\ cm, BC = 21\ cm$

$\therefore$ $AC^2 = AB^2 + BC^2$ (Pythagoras theorem)
$= (28)^2+ (21)^2 = 784 + 441 = 1225 = (35)^2$
$\therefore$ $AC = 35\ cm$
 $\therefore$ Radius of quadrant $BCD = 21\ cm$ and radius of semicircle on AC as diameter $=\frac{35}{2}\text{cm}$
Area of $\triangle\text{ABC}=\frac{1}{2}\text{ AB}\times\text{BC}=\frac{1}{2}\times28\times21\text{cm}^2=294\text{cm}^2$
Area of quadrant BCD $=\frac{1}{4}\pi\text{r}^2$
$=\frac{1}{4}\times\frac{22}{7}\times21\times21\text{cm}^2$
$=\frac{693}{2}=346.5\text{cm}^2$
and area of semicircle on AC as diameter
$=\frac{1}{2}\pi\text{R}^2=\frac{1}{2}\times\frac{22}{7}\times\frac{35}{2}\times\frac{35}{2}\text{cm}^2$
$=\frac{1925}{4}=481.25\text{cm}^2$
$\therefore$ Area of shaded region = Area of $\triangle\text{ABC}$ + area of semicircle - area of quadrant
$=294+481.25-346.50\text{cm}^2$
$=428.75\text{cm}^2$

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