Question

Question for Only:- ❏ Moderators ❏ Brainly Stars ❏ Best users ✰ 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

Given : Figure

To find : The ratio between the area of the remaining cardboard after cutting circular pieces and area of cardboard.

Solution :

Let's say the radius of the circle be, r. Then the breadth of cardboard is 2r and also the length of cardboard = 5r.

Area of cardboard is given by Length × Breadth.

Area of cardboard = 2r × 5r = 10r²

Now the area of circular pieces is given by πr²

Area of 2 circles and 1 semi circle = πr²+πr²+πr²/2

Area of 2 circles and 1 semi circle = 5πr²/2

Area of cardboard after cutting out circular pieces,

Area = 10r² - 5πr²/2

Area = (20r² - 5πr²)/2

Area = 5r²(4-π)/2

Required ratio is given by,

Ratio = 5r²(4-π)/2 : 10r²

Ratio = (4-π)/2 : 2

Ratio = (4-π)/4 : 1

Ratio = (4 - 22/7)/4 : 1

Ratio = ((28-22)/7)/4 : 1

Ratio = 6/28 : 1

Ratio = 3/14 : 1

Ratio = 3 : 14

So the required ratio is 3 : 14.

Answer 2

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

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,

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

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

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

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