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Answers
Answer:
Step-by-step explanation:
perimeter of region = 47cm
270/360 2pi r + OA + OB = 47
3/2 pi r + r + r = 47 ( OA=OB=r )
33r/7 + 2r = 47
33r+14r /7 = 47
47r /7 = 47
hence, r = 7cm
area of shaded region = 270/360 pi r²
=3/4 * 22/7 * 49
=3/2 * 11 * 7
=3*77/2
=115.5 cm²
area of shaded region = 115.5 cm²
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Solution :
Question : 22 ( a)
In this question , We have provided with Cubical block and hemisphere
According to the question ,
Cubical block of side 10 cm surmounted by a hemisphere . It means hemisphere touching the sides of cubical block . Therefore , the largest diameter which hemisphere can have is equal to 10 cm .
Radius = Diameter /2
⇒ 10/2
⇒ 5 = Radius
Finding the Surface area of the Cubical block
Surface area of the Cubical block = Surface area of the Cube - area of circular part of the hemisphere + surface area of the hemisphere
⇒ 6a^2 - πr^2 + 2πr^2
⇒ 6 * 10 * 10 - 22/7 * 5 * 5 + 2 * 22/7 * 5 * 5
⇒ 600 - 22/7 * 25 + 44/7 * 25
⇒ 600 - (22*25)/7 + ( 44 * 25)/7
⇒ 600 - 550/7 + 1100/7
⇒ 600 - 78.5 + 157.5
⇒ 600 + 79
⇒ 679 cm^2
Therefore , 679 cm^2 is our required surface area
Finding the Cost of painting
⇒ (679/100) × 5
⇒ 6.79 × 5
⇒ 33.95
Therefore , the Cost of painting is equal to Rs. 33.95
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Question : 22 ( b)
Area of Shaded region = Area of Semi-Perimeter APB + Area of Semi - Circle AQO
We have Given Perimeter of the figure is 47 cm
Let Radius of Semi-Perimeter APB be r then radius of AQO will be R/2
Finding the radius of Semi-Perimeter APB and Semi- Circle AQO
Perimeter = Arc length of APB + Arc length of AQO + length of OB
⇒ 47 = πr + π(R/2) + R
⇒ 47 = R( 3π/2 + 1)
⇒ 47 = R( ( 3 * 22/7)/(2) + 1 )
⇒ 47 = R( (3.14)/(2) * 3 + 1 )
⇒ 47 = R( 1.57 * 3 + 1)
⇒ 47 = R(5.71)
⇒ R = (47)/(5.71)
Therefore , the Radius of APB = 47/5.71 cm
Radius of AQO = 27/5.71 cm
Finding the area of Shaded Region
Area of Shaded Region = Area of Semi-Perimeter APB + Area of Semi-Perimeter AQO
⇒ 1/2πr^2 + 1/2πr^2
⇒ 1/2 * 3.14 * (47/5.71)^2 + 1/2 * 3.14 * ( 27/5.71)^2
⇒ 141.4 cm^2
Therefore , the area of the shaded region is equal to 141.4 cm^2