Question

Q. From a rectangular sheet of paper ABCD with AB = 40 cm and AD = 28 cm, a semi-circular portion with BC as diameteris cut off. Find the area of remining paper (use π = 22/7)

Answer 1

Given:

Length of paper (l) = 40 cm

Width of paper (b) = 40 cm

Area of paper:

$\implies$ $\sf{Length \times Breadth}$

$\implies$ $\sf{40 \times 28}$

$\implies$ $\sf{ {1120 \: cm}^{2}}$

Diameter of semi-circle = 28 cm

Therefore:

Radius of semi-circle:

$\implies$ $\sf{r = \frac{d}{2}}$

$\implies$ $\sf{r = \frac{28}{2}}$

$\implies$ $\sf{r = 14 \: cm}$

Thus:

Area of semi-circle:

$\implies$ $\sf{ \frac{1}{2} \pi r^{2}}$

$\implies$ $\sf{ \frac{1}{2} \times \frac{22}{7} \times 14 \times 14}$

$\implies$ $\sf{ {308 \: cm}^{2}}$

Therefore:

Area of remaining paper:

$\boxed{\sf{1120 - 308 = 812}}$

Hence:

Answer: $\sf{812 \: {cm}^{2}}$

Answer 2

Given sheet of paper ABCD AB = 40 cm, AD = 28 cm ⇒ CD = 40 cm, BC = 28 cm [since ABCD is rectangle] Semicircle be represented as BMC with BC as diameter Radius = (1/2) x BC=(1/2) x 28=14cm Area of remaining (shaded region) = (area of rectangle) – (area of semicircle)