Question

Q. 18. In the given figure, AABC is a right angled triangle
in which A = 90°. Semicircles are drawn on
AB, AC and BC as diameters. Find the area of the
shaded region. ​

Answer 1

Gɪᴠᴇɴ :-

• ∆ABC is a right angle ∆ , Right Angle at A.
• Semicircles are drawn on AB, AC and BC as diameters.
• AB = 3cm.
• AC = 4cm.

Tᴏ Fɪɴᴅ :-

• Shaded Area . ?

Fᴏʀᴍᴜʟᴀ ᴜsᴇᴅ :-

• Shaded Area = Area of Right Angle ∆ABC = (1/2) * Base * Perpendicular.

Sᴏʟᴜᴛɪᴏɴ :-

First we will Prove This Direct Result , that, How we can conclude That , Area of Shaded Region is Equal to Area of Right ∆.

Refer To Image For Proof Now.

Use of :-

→ Area of semi-circle = (1/2) * (πr²)

→ Area of Right Angle ∆ABC = (1/2) * Base * Perpendicular.

Hence, we can conclude That,

→ Shaded Area = Area of Right ∆ABC .

→ Shaded Area = (1/2) * 4 * 3

→ Shaded Area = 6cm². (Ans.)

Therefore, Required Shaded Area is 6cm².

(Excellent Question).

Answer 2

Aɴꜱᴡᴇʀ

➜ Area of the shaded region is 6 cm²

_________________

Gɪᴠᴇɴ

✪ ABC is a right angle triangle,right angled at A

✪ There are 2 semi-circles being drawn on the sides BA ,AC and BC( they are their diameter)

_________________

ᴛᴏ ꜰɪɴᴅ

☞ The area of the shaded region

_________________

Sᴛᴇᴘꜱ

So here the area of the triangle is actually equal to the area of the right angle triangle because,

✭ The Area of semi-circle with base AB is given by,

$\large \leadsto \tt \frac{\pi{} {a}^{2} }{2}........1$

✭ So then the area of the semi-circle with base AC is given by,

$\large \tt \leadsto{} \frac{\pi {b}^{2} }{2} ........2$

✭ And the area of the semi-circle with base BC is given by,

$\large \tt \leadsto{} \frac{\pi {c}^{2} }{2}........3$

❍ So now 1+2-3+∆ABC

$\large \mapsto \tt \frac{\pi {a}^{2} }{2} + \frac{\pi {b}^{2} }{2} - \frac{\pi {c}^{2} }{2} + abc \\ \\ \large \tt \mapsto \frac{\pi}{2} ( {a}^{2} + {b}^{2} - {c}^{2} ) + abc$

As per Pythagoras theorem a²+b² = c²

So,

$\huge{\underline{ \boxed{ \sf{}0 + abc}}}$

✴ So this its enough if we find the area of the triangle to find the area of the shaded region

So the area of the shaded region is given by,

$\large \tt \dashrightarrow{} \frac{1}{2} \times base \times height \\ \\ \large \tt \dashrightarrow{} \frac{1}{ \cancel2} \times \cancel4 \times 3 \\ \\ \large \tt \pink{ \dashrightarrow{}6 \: c {m}^{2} }$

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