ABCD is a rectangle of length 20cm and breadth 10cm. OAPB is a sector of circle of radius 10✓2. Calculate the area of the shaded region.
Answers
Given : ABCD is a rectangle of length 20 cm and breadth 10 cm. OAPB is a sector of a circle of radius 10√2 cm
To find : area of the shaded region.
Solution:
Area of Rectangle = 20 * 10 = 200 cm²
OA = OB =10√2
AB = 20 cm
AB² = OA² + OB²
=> OAB is right angle triangle at O
=> ∠AOB = 90°
Area of Sector OAB = (90/360) π (10√2)²
= (1/4)(3.14) 200
= 157 cm²
Area of Δ OAB = (1/2) * OA * OB = (1/2) * 10√2 * 10√2 = 100 cm²
area of the shaded region. = area of rectangle + area of Triangle OAB - area of Sector OAPB
= 200 + 100 - 157
= 143 cm²
area of the shaded region. = 143 cm²
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Answer:
143 cm^2
Step-by-step explanation:
Construct/complete the square(of side length 20 cm).
Notice/Assuming that O is on the line of perpendicular bisector of DC, which means O is the mid point of A'B' ∴ OA' = OB' = 10 cm
Moreover, as DCB'A' is a square, side length = 20 cm
∴ AD + AA' = 20 & BC + BB' = 20
∴ AA' = 10 & BB = 10
Now,
As in Δ AA'O: ∵ AA' = A'O = 10 cm
∴ ∠A'AO = ∠AOA' = x(say)
And, ∠AA'O = 90°
∴ 90° + x + x = 180° ⇒ x = 45°
∴ ∠AOA' = 45°
Similarly in triangle BB'O: ∠BOB' = 45°
Hence, ∠AOB = 180° - 45° - 45°
= 90°
which clearly means, AOBP forms a quarter circle and area should be 1/4 of πr². Hence, area of sector = 1/4 * 3.14 * (10√2)² = 157 cm².
Observing,
Shaded region = area of square - area of sector - area of ΔAA'O - area of ΔBB'O
= (20)² - 157 - (1/2)(10)(10) - (1/2)(10)(10)
= 400 - 157 - 50 - 50
= 143 cm^2
[units have been ignored (wherever possible ; area of triangle = 1/2 * base * height ; area of square = side^2 ; area of full circle = πr^2]
