search
Log In
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
0 votes
101 views

The figure shows the rectangle $\text{ABCD}$ with a semicircle and a circle inscribed inside in it as shown. What is the ratio of the area of the circle to that of the semicircle?

                                                                   

  1. $\left ( \sqrt{2}-1 \right )^{2}:1$
  2. $2\left ( \sqrt{2}-1 \right )^{2}:1$
  3. $\left ( \sqrt{2}-1 \right )^{2}:2$
  4. None of these
in Quantitative Aptitude 13.3k points 306 2250 2467
edited by
101 views

1 Answer

0 votes
Given that, the figure.

DIAGRAM

Let the radius of the semicircle be $\text{‘R’}$ and the circle be $’r’.$

The triangle $\triangle \text{OMB}$ is right-angle triangle. so we can apply the Pythagorean theorem.

$\text{(Hypotenuse)}^{2} = \text{(Perpendicular)}^{2} + \text{(Base)}^{2} $

$\Rightarrow \text{(OB)}^{2} = \text{(OM)}^{2} + \text{(MB)}^{2} $

$\Rightarrow \text{(OB)}^{2} = \text{R}^{2} + \text{R}^{2} $

$\Rightarrow \text{(OB)}^{2} = \text{2R}^{2} $

$\Rightarrow \boxed{ \text{OB} = \sqrt{2} \;  \text{R} }$

Similarly  $ \boxed{ \text{O’B} = \sqrt{2} \; r }$

Now, $\text{OB} = \text{ON} + \text{NO’} + \text{O’B}$

$\Rightarrow \sqrt{2} \; \text{R} = \text{R} + r + \sqrt{2} \; r $

$\Rightarrow \sqrt{2} \; \text{R} – \text{R} =   r + \sqrt{2} \; r $

$\Rightarrow \text{R}(\sqrt{2}-1) =   r(\sqrt{2}+1) $

$\Rightarrow \frac{r}{\text{R}} = \frac{\sqrt{2}-1}{\sqrt{2}+1}$

$\Rightarrow \frac{r}{\text{R}} = \frac{(\sqrt{2}-1)}{(\sqrt{2}+1)} \times \frac{(\sqrt{2}-1)}{(\sqrt{2}-1)}$

$\Rightarrow \frac{r}{\text{R}} = \frac{(\sqrt{2}-1)^{2}}{(2-1)}$

$\Rightarrow \boxed{\frac{r}{\text{R}} = \frac{(\sqrt{2}-1)^{2}}{1}}$

$\therefore$ The ratio of the area of the circle to that of the semicircle $ = \frac{\pi \; r^{2}}{\frac{\pi}{2}\; \text{R}^{2}}$
10.1k points 4 8 30

Related questions

0 votes
1 answer
1
116 views
$\text{ABCD}$ is a square of area $4$ with diagonals $\text{AC}$ and $\text{BD}$, dividing square into $4$ congruent triangles. Figure looks like four non-over lapping triangles. Then the sum of the perimeters of the triangles is $8\left ( 2+\sqrt{2} \right )$ $8\left ( 1+\sqrt{2} \right )$ $4\left ( 1+\sqrt{2} \right )$ $4\left ( 2+\sqrt{2} \right )$
asked Mar 9, 2020 in Quantitative Aptitude go_editor 13.3k points 306 2250 2467 116 views
0 votes
1 answer
2
87 views
$\text{PQRS}$ is a square. $\text{SR}$ is a tangent (at point $\text{S})$ to the circle with centre $\text{O}$ and $\text{TR = OS}$. Then the ratio of area of the circle to the area of the square is $\pi /3$ $11/7$ $3 /\pi$ $7/11$
asked Mar 9, 2020 in Quantitative Aptitude go_editor 13.3k points 306 2250 2467 87 views
0 votes
0 answers
3
96 views
From a circular sheet of paper with a radius $20$ cm, four circles of radius $5$ each are cut out. What is the ratio of the uncut to the cut portion? $1: 3$ $4: 1$ $3: 1$ $4: 3$
asked Mar 9, 2020 in Quantitative Aptitude go_editor 13.3k points 306 2250 2467 96 views
0 votes
1 answer
4
131 views
$\text{ABCD}$ is a rhombus with the diagonals $\text{AC}$ and $\text{BD}$ intersecting at the origin on the $x\text{-}y$ plane. The equation of the straight line $\text{AD}$ is $x + y = 1$. What is the equation of $\text{BC}?$ $x + y = -1$ $x - y = -1$ $x + y = 1$ None of these
asked Mar 9, 2020 in Quantitative Aptitude go_editor 13.3k points 306 2250 2467 131 views
0 votes
0 answers
5
125 views
In the figure above, $\text{AB = BC = CD = DE = EF = FG = GA}$. Then $\angle \text{DAE}$ is approximately $15^{\circ}$ $20^{\circ}$ $30^{\circ}$ $25^{\circ}$
asked Mar 9, 2020 in Quantitative Aptitude go_editor 13.3k points 306 2250 2467 125 views
...