Question

3. In the figure, ABCD is a rectangle of sides 3 m and 4 m and BCD is a semicircle drawn on the diagonal BD of the rectangle. Find: (a) the perimeter; and (b) the area of the shaded region.​

Answer 1

Answer:

$\dashrightarrow \large \sf \blue{question}$

In the figure, ABCD is a rectangle of sides 3 m and 4 m and BCD is a semicircle drawn on the diagonal BD of the rectangle. Find: (a) the perimeter; and (b) the area of the shaded region.

$\large \sf \green{answer} \downarrow$

(a) BD is the diameter of semicircle BCD.

Then,

$\sf {bd}^{2} = {4}^{2} + {3}^{2} \\ \\ {bd}^{2} = 25 \\ \\ bd = 5$

So radius of semicircle is

$\frac{5}{2} = 2.5m$

perimeter

$= 3 + 4 + \pi \times 25 \\ \\ = 7 + \frac{22}{ 7} \times 2.5 \\ \\ = \frac{104}{7} \\ \\ = \boxed{ \sf\green{12.85m}}$

(b) BCD is the semicircle on the diagonal BD of the rectangle.

Now,

In semicircle BOC,

base=3cm

so radius=

$\frac{3}{2} = 1.5cm \\ \\ \sf{area} = \frac{1}{2} {\pi r}^{2} \\ \\ = \frac{1}{ \cancel2} \times \frac{ { \cancel{22}} \: \: ^{11} }{7} \times \frac{1.5}{10} \times \frac{1.5}{10} \\ \\ = \frac{2475}{700} \\ \\ = {3.53 \sf \: cm}^{2}$

In semicircle C×D

Base=4cm

So radius=

$\frac{ { \cancel4}^{2} }{ \cancel2} \\ \\ = 2cm$

Area=

$\frac{1}{2} {\pi r}^{2} \\ \\ = \frac{1}{ \cancel2} \times \frac{ { \cancel{22}} \: ^{11} }{7} \times 2 \times 2 \\ \\ = \frac{44}{7} \\ \\ = {6.28cm}^{2}$

Now area of shaded region

$= {3.53}^{2} + {6.28}^{2} \\ \\ \boxed{ \sf\red{ = {9.81cm}^{2}}}$

$\Large\textsf{Hope \: It \: Helped}$