answer my 33rd question plz
Answers
In the given figure:-
r of big circle=42
r of small circle=21
the area of outer ring:-
area of big circle - area of small circle
=22/7×42 - 22/7×21
=22/7(42²-21²)
=22/7×1323
=4158
area of sector ring:-
area of big sector - area of small sector
22/7×60(42² - 21²)/360
249480/360
693
area of shaded region:-
4158-693
3465
thats all.......
Answer:
3465 cm²
Step-by-step explanation:
Method - 1:
Given:
Radius of inner circle r = 21 cm
Radius of outer circle R = 42 cm
Angle of sector, θ = ∠AOB = ∠COD = 60°
Also, area of a circle with radius 'r' is πr²
area of a sector with angle θ and radius r is : (θ/360°) πr²
Now,
Required area = πR² - πr² - (θ/360 πR² - θ/360 πr²)
Place the values, we get
⇒ π(42)² - π(21)² - (60/360 * π(42)² - 60/360 * π(21)²)
⇒ π(1764 - 441 - (1764/6 - 441/6))
⇒ π(1323 - 220.5)
⇒ (22/7) * 1102.5
⇒ 3465 cm²
Therefore, Area of shaded region = 3465 cm²
Method - 2:
Area of region ABCD :
∴ Area of sector AOB - Area of sector COD
= (60/360 πR² - 60/360 πr²)
= (60/360 * 22/7 * 42² - 60/360 * 22/7 * 21²)
= (22 * 42 - 11 * 21)
= 693 cm²
Now,
Area of circular ring = (πR² - πr²)
= (22/7 * 42² - 22/7 * 21²)
= (5544 - 1386)
= 4158 cm²
Thus,
Area of shaded region :
Area of circular ring - Area of region ABCD
⇒ 4158 - 693
⇒ 3465 cm²
Therefore, Area of shaded region = 3465 cm²
Hope it helps!