Question

Find the area of the shaded region in Fig.15.72, if AC=24cm, BC=10cm and O is the centre of the circle.(Use π=3.14)

Answer 1

Answer:

Area of shaded region = 145.57 cm²

Step-by-step explanation:

Angle ACB = 90°,

therefore, ∆ABC is a right angle ∆

in ∆ABC, By using Pythagoras theorem,

AB² = AC² + Bc²

=> AB² = (24)² + (10)²

=> AB² = 576 + 100

=> AB² = 676

=> AB = √676

=> AB = 26 cm

AB is the diameter,

AB = OA + OB

and OA = OB ( radius of circle )

=> OA = OB = AB/2

=> OA = OB = 26/2

=> OA = OB = 13 cm

Area ∆ABC = 1/2 × Base × height

=> Area ∆ABC = 1/2 × 10 × 24

=> Area ∆ABC = 120 cm²

Area of semi circle = 1/2 × π × r²

=> Area of semi circle = 1/2 × 22/7 × 13 × 13

=> Area of semi circle = 265.57 cm²

Area of shaded region = area of semi circle - area of ∆ABC

=> Area of shaded region = 265.57 - 120

=> Area of shaded region = 145.57 cm² ( required answer )

Answer 2

Given

• AC=24cm BC=10cm and O is the center of the circle

To find

Area of the shaded region

$\mid \overline \mathbf \red{answer}$

By pythagors theorem,

AB² = AC² + BC²

AB² = 24² + 10²

AB² = 576 + 100

AB² = 676

AB = 26 cm

diameter of the circle = 26 cm

2r = 26

r = 13 cm

Now,

Area of the shaded region

=(Area of the semicircle) - (Area of the

ΔABC)

= π r² - (1/2) (BC)(AC)

= (3.14)(13×13) - (1/2) (10)(24)

= (3.14)(169) - (10)(12)

= 530.66 - 120

=410.66cm²