Find the area of the shaded region in the following figure, if AC = 24 cm, BC = 10 cm and O is the centre of the circle. (Use π = 3.14)
Answers
Answer:
The area of shaded region is 145.33 cm².
Step-by-step explanation:
GIVEN :
AC = 24 cm, BC = 10 cm
Area of ∆ ABC = ½ × AC × BC
= ½ × 24 × 10
= 12 × 10
= 120 cm²
Area of ∆ ABC = 120 cm²
∠ACB = 90°
[Angle in a semicircle is a right angle]
In ∆ABC , By using pythagoras theorem,
AB² = AC² + BC²
AB² = 24² + 10²
AB² = 576 + 100
AB² = 676
AB = √676
AB = 26 cm
AB is the diameter of a circle.
Radius of a circle (OA) ,r = AB/2 = 26/2 = 13 cm
Radius of a circle,r = 13 cm
Area of a circle = πr²
= 3.14 × 13²
= 3.14 × 169
= 530.66 cm²
Area of a circle = 530.66 cm²
Area of a semi circle = ½ × πr²
= ½ × 3.14 × 13²
= ½ × 3.14 × 169
= ½ × 530.66
Area of a semi circle = 265.33 cm²
Area of shaded region = Area of circle - (Area of semicircle + Area of triangle)
= 530.66 - (265.33 + 120)
= 530.66 - 385.33
= 145.33 cm²
Area of shaded region = 145.33 cm²
Hence, the area of shaded region is 145.33 cm².
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Solution:
i ) In ABC , <C = 90°
/* Angle in a semicircle */
AB² = AB² + BC²
/* Phythogarian theorem */
=> AB² = 24² + 10²
= 576+100
= 676
=> AB = √676 = 26 cm
ii) Radius of the circle (r)
= AB/2
= 26/2
= 13 cm
ii) Area of the shaded region
= Area of the semicircle - Area of ∆ABC
= (πr²/2) - (1/2)×AC×AB
= 1/2 [ πr² - AC×AB ]
= 1/2[3.13×13² - 24 × 10 ]
= 1/2[ 3.13×169 - 240]
= 1/2[ 530.66 - 240]
=1/2 × 290.66
= 145.33 cm²
Therefore,
Area of the shaded region
= 145.33 cm²
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