Quiz 4

Original rectangle: Shorter side = 2 Longer side = x
Second rectangle: Longer side = 2 Shorter side = x/2 (As it is folded into half at midpoint)
Ratio of 1st rectangle = x/2.
2nd rectangle =( 2/x)/2 = 4/x.
Ratio is same. x/2 = 4/x.
x2 = 8, x = 2√2.
So, area of smaller rectangle = (x/2) * 2 => x = 2√2.

Hence, option 2 is correct.

∆ PRO and ∆ QSO are similar

OQ = ⇒ OQ = 21
∴ PQ = 7 ⇒ PQ
QO = = 7/21
1/3 . Ans.(2)

4x + 3x = PQ = 7
⇒ x = 1. So, the radius of circle II = 3 cm. Ans.(2)

32+ $SO^{2}$
= 212
⇒ $SO^2$
= 212
– 32
⇒ $SO^2$
= 432 ⇒ SO = 12 root 3 . Ans.(3)

Let the diameter of circle C

AB=x

Now $P_{o}B$=x/2

As P is the mid-point

$P_{o}P_{1}$=x/4=P_{1}B

Radius of C1=x/8 

And $P_{1}P_{2}= x/8 = P_{2}B$

Radius of C2= $P_{1}P_{2}$ =x/16

And radius of C 3=$P_{2}P_{3}$=x/32

Now area of $c_{1}+c_{2}+c_{_{3}}$

$\pi \left ( \frac{x}{8} \right )^{2}+\pi \left ( \frac{x}{16} \right )^{2}+\pi \left ( \frac{x}{32} \right )^{2}+....\infty$

This is an infinte GP with common ration 1/4
$=\pi x^{2}\times \frac{\frac{1}{64}}{1-\frac{1}{4}}=\frac{\pi x^{2}}{48}$

Area of Unshaded Region= $\frac{\pi x^{2}}{4}-\frac{\pi x^{2}}{48}=\frac{11\times \pi x^{2}}{48}$

Ratio of Unshaded Portion to Original Circle
$\frac{11\times \pi x^{2}}{48}:\frac{\pi x^{2}}{4}$=11:12

 

 

 

area of square=1 cm^2
area of one sector=90/360*3.14*1 cm^2 =0.785 cm^2
area of remaining portion=area of sq.-area of one sector=.215cm^2
common area of both the circle= 1-(2*0.215)=0.572cm^2