Question

in given fig BC is diameter of AB = 3 AC =4 and angle A = 90 find the area of shaded region

Answer 1

Answer:

in triangle ABC. ab^2 + ac^2 =bc^2

BC = 5cm

Area of shaded = ar(circle -triangle)

= 19.63 - 6

=13.63 cm^2

Answer 2

In the given figure we first find area of circle and triangle after that we minus area of triangle from area of circle then we have the area of shaded region.

Let's do it,

First of all we find area of triangle,

In ∆ABC angle A is 90 therefore ∆ABC is right triangle.

In ∆ABC,

BC^2 = AB^2 + AC^2

$= {(3)}^{2} + {(4)}^{2} \\ = 9 + 16 \\ = 25 \\ = \sqrt{25} \\ = 5$

we find BC = 5cm

area of ∆ABC = 1/2 × base × hight

$= \frac{1}{2} \times 3 \times 4 \\ = 3 \times 2 \\ = 6 {cm}^{2}$

Find area of circle.

BC is diameter and OC is radius,

$2r = d \\ r = \frac{1}{2} \times d \\ r = \frac{1}{2} \times 5 \\ r = 2.5$

area of circle,

$= \pi {r}^{2} \\ = \frac{22}{7} \times {(2.5)}^{2} \\ = \frac{22}{7} \times \frac{25}{10} \times \frac{25}{10} \\ = 19.64 {cm}^{2}$

area of shaded region = area of circle -

area of ∆

$= 19.64 - 6\\ = 13.64 {cm}^{2}$

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