Question

heya !!!!!!!
solve this​

Answer 1

Given :

• Δ ABC is a right angled triangle.
• Semicircles are drawn on AB, AC and BC as diameter.

To Find :

• Area of the shaded region.

Solution :

Let's find the length of hypotenuse, BC of triangle ΔABC.

Since ΔABC is a right angled triangle,

•°• By pythagoras theorem,

$\sf{\longrightarrow{BC^2=AB^2+AC^2}}$

$\sf{\longrightarrow{BC^2=3^2\:+\:4^2}}$

$\sf{\longrightarrow{BC^2=9+16}}$

$\sf{\longrightarrow{BC^2=25}}$

$\sf{\longrightarrow{BC=\sqrt{25}}}$

$\sf{\longrightarrow{BC=5\:cm}}$

Now, let's calculate the area of the triangle ABC.

Formula :

$\large{\boxed{\bold{\dfrac{1}{2}\:\times\:b\:\times\:h}}}$

Here,

• b = base = 3 cm
• h = height = 4 cm

Block in the data,

$\sf{\longrightarrow{Area_{triangle}\:=\:\dfrac{1}{2}\:\times\:3\:\times\:4}}$

$\sf{\longrightarrow{Area_{triangle}\:=\:\dfrac{1}{2}\:\times\:12}}$

$\boxed{\sf{Area_{triangle}\:=\:6\:cm^2}\:\:\:\:(1)}$

Now, calculate the area of semicircle with diameter AB.

Formula :

$\large{\boxed{\bold{Area_{semicircle}\:=\:\dfrac{\pi\:r^2}{2}}}}$

AB = Diameter = 3 cm.

$\sf{\longrightarrow{Radius,r\:=\:\dfrac{Diameter}{2}}}$

$\sf{\longrightarrow{r=\dfrac{3}{2}}}$

$\sf{\longrightarrow{r=1.5\:cm}}$

Now, using the formula.

$\sf{\longrightarrow{Area_{semicircle}\:=\:\dfrac{\pi\:(1.5)^2}{2}}}$

$\sf{\longrightarrow{Area_{semicircle}\:=\:\dfrac{\pi\:2.25}{2}}}$

$\boxed{\sf{Area_{semicircle}\:=\:\dfrac{2.25\:\pi}{2}}\:\:\:\:(2)}$

Now, consider area of semicircle with diameter AC.

AC = 4 cm

$\sf{\longrightarrow{r\:=\:\dfrac{4}{2}}}$

$\sf{\longrightarrow{r=2\:cm}}$

$\sf{\longrightarrow{Area_{semicircle}\:=\:\:\dfrac{\pi\:r^2}{2}}}$

$\sf{\longrightarrow{Area_{semicircle}\:=\:\dfrac{\pi\:2^2}{2}}}$

$\sf{\longrightarrow{Area_{semicircle}\:=\:\dfrac{4\:\pi}{2}}}$

$\boxed{\sf{Area_{semicircle}\:=\:2\:\pi\:\:\:\:(3)}}$

Now, area of semicircle with diameter BC.

BC = 5 cm.

$\sf{\longrightarrow{r=\dfrac{5}{2}}}$

$\sf{\longrightarrow{r=2.5\:cm}}$

$\sf{\longrightarrow{Area_{semicircle}\:=\:\dfrac{\:\pi\:r^2}{2}}}$

$\sf{\longrightarrow{Area_{semicircle}\:=\:\dfrac{\pi\:2.5^2}{2}}}$

$\boxed{\sf{Area_{semicircle}\:=\:\dfrac{6.25\:\pi}{2}}\:\:\:\:\:(4)}$

Now, area of shaded region will be area of triangle + area of two semi circle - area of semi circle on diameter BC.

$\sf{\longrightarrow{Shaded\:region\:=\:6\:+\:\dfrac{2.25\pi}{2}\:+\:2\:\pi\:-\:\dfrac{6.25\pi}{2}}}$

$\bold{\big[Using\:equations\:(1),(2),(3)\:and\:(4)\big]}$

$\sf{\longrightarrow{Shade\:region\:=\:\dfrac{12+2.25\:\pi}{2}+2\:\pi-\:\dfrac{6.25\:\pi}{2}}}$

$\sf{\longrightarrow{Shaded\:region\:=\:\dfrac{12+2.25\:\pi\:+\:4\:\pi-\:6.25\:\pi}{2}}}$

$\sf{\longrightarrow{Shaded\:region\:=\:\dfrac{12\:\:\cancel{+6.25\:\pi-\:6.25\pi}}{2}}}$

$\sf{\longrightarrow{Shaded\:region\:=\:\dfrac{12}{2}}}$

$\sf{\longrightarrow{Shaded\:region\:=\:6}}$

$\large{\boxed{\bold{\purple{Area\:of\:shaded\:region\:=\:6\:cm^2}}}}$