In the given Fig. 2,O is the centre of the circle with AC =24 cm,AB =7 cm and angle BOD =90°. Find the area of the shaded region [Take π = 3.14]
Answers
Answer:
Area of shaded region = 284 cm² (approx)
Step-by-step explanation:
In the figure, area of shaded region is as follows
= area of semicircle ABC + area of quadrant BOD - area triangle ABC
As, BC is the diameter so, angle BAD = 90 degree
=> triangle ABC is right angled at A .
According to Pythagorus Theorem -
BC² = AB² + AC²
=> BC² = 7² + 24²
= 49 + 576
= 625
=> BC = 25 cm
Thus, area of semicircle ABC = (pi * r * r)/2
= (3.14 * 25/2 * 25/2)/2
= 245.3125 cm²
area of quadrant BOD = (pi * r * r)/4
= (3.14 * 25/2 * 25/2)/4
= 122.656 cm²
area of right triangle ABC = 1/2 * base * height
= 1/2 * 7 * 24
= 84 cm²
Thus, area of shaded region = 245.3125 + 122.656 - 84
= 283.9685 cm²
= 284 cm² (approx)
Answer:
283.97 cm²
Step-by-step explanation:
In the given Fig. 2,O is the centre of the circle with AC =24 cm,AB =7 cm and angle BOD =90°. Find the area of the shaded region [Take π = 3.14]
BC is diameter as it is passing through center so
∠CAB = 90°
now using Pythagoras theorem
BC² = AC² + AB²
=> BC² = 24² + 7²
=> BC² = 576 + 49
=> BC² = 625
=> BC = 25 cm
BC = 25 cm = Diameter
so Radius = 25/2 = 12.5cm
Area of shaded region = Area of semi circle - Area of Triangle + Area of region with 90 deg angle
= ((1/2) * 3.14 * 12.5²) - ((1/2) * 24 * 7 )+ ((90/360)*3.14*12.5²)
= 245.31 - 84 + 122.66
= 283.97 cm²