In fig, ABC is a triangle right angled at A. Semi circles are drawn on AB, AC and BC as diameters. Find the area of the shaded region.
Answers
AB = 6 cm , AC = 8 cm
Diameter AB = 6 cm
Diameter AC = 8 cm
In right angled ∆ABC,
BC² = AB² + AC² [by Pythagoras theorem]
BC² = 6² + 8²
BC² = 36 +64 = 100
BC = √ 100 = 10
BC= 10 cm
Area of semicircle = ½(πr²)
Area of semicircle with AB as diameter = ½(π) (6/2)²= 1/2π (3)² = 9π/2 cm²
Area of semicircle with AC as diameter = ½(π) (8/2)²= 1/2π (4)² = 1/2π (16) =8π cm²
Area of ∆ABC = ½ × AC × AB = 1/2× 8 ×6 = 4×6 =24 cm²
Area of semicircle with BC as diameter= ½(π) (10/2)²= 1/2π (5)² = 1/2π (25) =25π/2 cm²
Area of shaded region= Area of semicircle with AB as diameter + Area of semicircle with AC as diameter + Area of ∆ABC - Area of semicircle with BC as diameter
= 9π/2 + 8π + 24 - 25π/2
= 9π/2 + 8π - 25π/2+ (24)
= π(9/2 +8 -25/2) + 24
= π(4.5 - 12.5 +8) +24
= π (-8 +8) + 24
= π (0)+24
= 0 + 24 = 24
Area of shaded region= 24 cm²
Hence, the Area of shaded region = 24 cm²
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Answer:
Step-by-step explanation:
Given :
AB = 6 cm , AC = 8 cm
Diameter AB = 6 cm
Diameter AC = 8 cm
In right angled ∆ABC,
BC² = AB² + AC² [by Pythagoras theorem]
BC² = 6² + 8²
BC² = 36 +64 = 100
BC = √ 100 = 10
BC= 10 cm
Area of semicircle = ½(πr²)
Area of semicircle with AB as diameter = ½(π) (6/2)²= 1/2π (3)² = 9π/2 cm²
Area of semicircle with AC as diameter = ½(π) (8/2)²= 1/2π (4)² = 1/2π (16) =8π cm²
Area of ∆ABC = ½ × AC × AB = 1/2× 8 ×6 = 4×6 =24 cm²
Area of semicircle with BC as diameter= ½(π) (10/2)²= 1/2π (5)² = 1/2π (25) =25π/2 cm²
Area of shaded region= Area of semicircle with AB as diameter + Area of semicircle with AC as diameter + Area of ∆ABC - Area of semicircle with BC as diameter
= 9π/2 + 8π + 24 - 25π/2
= 9π/2 + 8π - 25π/2+ (24)
= π(9/2 +8 -25/2) + 24
= π(4.5 - 12.5 +8) +24
= π (-8 +8) + 24
= π (0)+24
= 0 + 24 = 24
Area of shaded region= 24 cm²
Hence, the Area of shaded region = 24 cm²
HOPE THIS WILL HELP YOU...
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