Question

plz solve question number 39.....
NO SPAMMING PLZ....
NEED A FULL SOLUTION...

Answer 1

$\underline{\bold{Given:-}}$

∆ABC is a right angled triangle which is right angled at B.

AB = 28 cm

BC = 21 cm

$\underline{\bold{To\: find:-}}$

Area of shaded region

$\underline{\bold{Solution:-}}$

Using Pythagoras theorem in ∆ABC

${AC}^{2} = {AB }^{2} + {BC}^{2}$

${AC}^{2} = {28}^{2} + {21}^{2} \\ \\ {AC}^{2} = 784 + 441 \\ \\ {AC}^{2} = 1225 \\ \\ AC = \sqrt{1225} \\ \\ AC = 35cm$

Area of ∆ABC

$= \frac{1}{2} \times 21 \times 28 \\ \\ = 21 \times 14 \\ \\ = 294 {cm}^{2}$

Area of given part of circle of radius 21 cm

$= \frac{1}{4} \times \pi \times {21}^{2} \\ \\ = \frac{1}{4} \times 3.14 \times 441 \\ \\ = 346.185 {cm}^{2}$

Radius of circle of diameter 35 cm =35/2 = 17.5 cm

Area of given part of circle of diameter 35 cm

$= \frac{1}{2} \times \pi \times {17.5}^{2} \\ \\ = \frac{1}{2} \times 3.14 \times 306.25 \\ \\ = 480.81 {cm}^{2}$

Area of shaded region = Area of ∆ABC + Area of given part of circle of diameter 35 cm -
Area of given part of circle of radius 21 cm

$= 294 + 480.81 - 346.18 \\ \\ = 774.81 - 346.18 \\ \\\boxed{ = 428.63 {cm}^{2} }$