Question

help me out to find oit the answer​

Answer 1

See above attachment ✔️✔️

${} \huge \bold\purple{hope \: it \: helps \: you..}$

Answer 2

ANSWER:-

Given:

ABCP is a quadrant of a circle of radius 14cm, with AC as diameter a semicircle is drawn.

To find:

Find area of shaded region.

Solution:

In right angled ∆ABC,

AC² = AB² + BC²

AC² = (14)² + (14)²

AC² = 2× (14)²

AC = √2× (14)²

AC = 14√2cm.

Required area:

Area ACQA - Area ACPA

=) Semicircle ACQA - (Area of quadrant ABCPA - area of ∆ABC).........(1)

Therefore,

From equation (1), we get;

Required Area:APCQA

$required \: area = [ \frac{1}{2}\pi ( {r)}^{2} - ( \frac{1}{4} \pi( {r)}^{2} - \frac{1}{2} \times AB \times BC)] \\ \\ = > [\frac{1}{2} ( \frac{22}{7} \times (7 \sqrt{2} ) {}^{2} ) - ( \frac{1}{4} \times \frac{22}{7} \times ( {14)}^{2} - \frac{1}{2} \times 14 \times 14)]\\ \\ = > (154 - 154 + 98) {cm}^{2} \\ \\ = > 0 + 98 {cm}^{2} \\ \\ = > 98 {cm}^{2}$

Hence,

Area of shaded region is 98cm².

Hope it helps ☺️