Question

from a rectangular cardboard ABCD 2 circles and 1 semicircle of a largest side are cut. calculate the ratio between the area of the remaining cardboard and area of cardboard.​

Answer 1

$\large\underline{\sf{Solution-}}$

Let us assume that

↝ The diameter of the circle be 2x units.

According to given figure,

2 circles and one semi-circle of largest side is cut out.

It implies,

Length of Rectangular card - board = 5x units

Breadth of Rectangular card - board = 2x units

So,

$\boxed{ \rm{ \: Area_{(Rectangular \: cardboard)} = (5x)(2x) = {10x}^{2} \: sq. \: units}}$

Now, we calculate the area of 2 circles and one semi - circle.

$\rm :\longmapsto\:Area_{(circles)} = \pi {(x)}^{2} + \pi {(x)}^{2} + \dfrac{1}{2}\pi {(x)}^{2}$

$\rm :\longmapsto\:Area_{(circles)} = 2\pi {(x)}^{2} + \dfrac{1}{2}\pi {(x)}^{2}$

$\rm :\longmapsto\:Area_{(circles)} = \dfrac{1}{2}[\pi {(x)}^{2} + 4\pi {(x)}^{2}]$

$\rm \implies\:\boxed{ \tt{ \: Area_{(circles)} = \dfrac{5\pi {x}^{2}}{2}}}$

So,

Remaining Area of Rectangular Card - board

$\rm \: = \:Area_{(Rectangular \: cardboard)} - Area_{(circles)}$

$\rm \: = \: {10x}^{2} - \dfrac{5\pi {x}^{2}}{2}$

$\rm \: = \: {10x}^{2} - \dfrac{5 \times 22 \times {x}^{2}}{2 \times 7}$

$\rm \: = \: {10x}^{2} - \dfrac{5 \times 11\times {x}^{2}}{7}$

$\rm \: = \: {10x}^{2} - \dfrac{55{x}^{2}}{7}$

$\rm \: = \: \dfrac{70 {x}^{2} - 55{x}^{2}}{7}$

$\rm \: = \:\dfrac{15 {x}^{2} }{7}$

$\rm \implies\:\boxed{ \tt{ \: Area_{(Remaining \: cardboard)} = \frac{15 {x}^{2} }{7} \: sq. \: units}}$

Hence,

$\red{\rm :\longmapsto\:Area_{(Remaining \: cardboard)} : Area_{(Rectangular \: cardboard)}}$

$\rm \: = \:\dfrac{15 {x}^{2} }{7} : 10 {x}^{2}$

$\rm \: = \:\dfrac{3}{7} : 2$

$\rm \: = \:3 : 14$