Question

from a circular sheet of radius 14 cm two circles of 3.5 cm and a rectangle of length3cm &breadth 1cm are removedfind the area of remainingsheet (take pie=22/7)

Answer 1

r1 = 14 cm, r2 = 3.5 cm, l = 3 cm, b = 1 cm
area remaining = original area - area of 2 circles - area of rectangle
= πr1² - 2πr2² - l*b
= 22/7 * 14² - 2 * 22/7 * 3.5² - 3*1
= 616 - 77 - 3
= 536 cm²

Answer 2

$ar \: of \: circle \: = \pi {r}^{2} \\ = \frac{22 \times 14 \times 14}{7} = 44 \times 14 = 606 {cm}^{2} \\ ar \: ofrect \: = 3 {cm}^{2} \\ ar \: of \: two \: cir = 2 \times \frac{22}{7} \times \frac{35}{10} \times \frac{35}{10} \\ = 77 {cm}^{2} \\ left \: area \: = 606 - (77 + 3) = 526 {cm}^{2}$