n the figure, ABC is a right angled triangle with B = 90 , BC = 21 cm and - Maths - Areas Related to Circles

Question

n the figure, ABC is a right angled triangle with ∠B =
90°, BC = 21 cm and AB = 28 cm With AC as diameter
of a semicircle and with BC as radius, a quadratic circle is drawn
Find the area of the shaded portion (Approximately)

(A)

428 75 cm2

(B)

857 50 cm2

C)

214 37 cm3

(D)

638 50 cm2

Gursheen kaur. · Accepted Answer

Area of shaded region= Area of semi-circle with AC as diameter + Area of ΔABC – Area of quadrant of circle with BC as radius

In right ΔABC,

By Pythagoras theorem,

AC² = AB² + BC²

⇒AC² = (28 cm )² + (21 cm)²

⇒ AC² = 1225 cm²

⇒ AC = 35 cm

∴ Area of the shaded region,

$= \frac{1}{2} \times π (\frac{AC}{2})^{2} + \frac{1}{2} \times BC × AB - \frac{1}{4} \times π (BC)^{2} = \frac{1}{2} \times \frac{22}{7} \times (\frac{35 cm}{2})^{2} + \frac{1}{2} \times 21 cm × 28 cm - \frac{1}{4} \times \frac{22}{7} \times (21 cm)^{2} = 481.25 {cm}^{2} + 294 {cm}^{2} - 346.5 {cm}^{2} = 428.75 {cm}^{2}$

Thus, the area of the shaded region = 428.75 cm².

28

n the figure, ABC is a right angled triangle with ∠B = 90°, BC = 21 cm and AB = 28 cm. With AC as diameter of a semicircle and with BC as radius, a quadratic circle is drawn. Find the area of the shaded portion (Approximately) (A) 428.75 cm2 (B) 857.50 cm2 C) 214.37 cm3 (D) 638.50 cm2

n the figure, ABC is a right angled triangle with ∠B = 90^°, BC = 21 cm and AB = 28 cm. With AC as diameter of a semicircle and with BC as radius, a quadratic circle is drawn. Find the area of the shaded portion (Approximately)

(A)
428.75 cm²
(B)
857.50 cm²

C)
214.37 cm³
(D)
638.50 cm²