Area and perimeter Questions and answers

Given that, a = 6

According to the formula,

Percentage increase in area

= \( \Large \left(2a+\frac{a^{2}}{100} \right) \% \)

= \( \Large \left(2\times 6+\frac{86}{100} \right)% \)

= (12 + 0.36)% = 12.36%

Distance covered in 1 round

= \( \Large \frac{Total \ Distance}{Total \ round} \)

= \( \Large \frac{1.76\times 1000}{350} \)

= \( \Large \frac{176}{35}m \)

=\( \Large 2\pi r=\frac{1.76\times 100}{35}cm \)

2r = Diameter=\( \Large \frac{17600\times 7}{22\times 35} \)

= 160 cm

Ratio of the areas of the circumcircle and incircle of a square

=\( \Large \frac{(Diagonal)^{2}\pi}{(Side)^{2}\pi}=\frac{(Side\times \sqrt{2})^{2}}{(Side)^{2}}=\frac{2}{1} \)or 2:1

Let the radii of two circles are \( \Large r_{1} \) and \( \Large r_{2} \) respectively.

Given,

\( \Large \frac{Circumference \ of \ 1st \ circle}{Circumference \ of \ 2nd \ circle}=\frac{2}{3} \)

=> \( \Large \frac{2\pi r_{1}}{2\pi r_{2}}=\frac{2}{3} \)=> \( \Large \frac{r_{1}}{r_{2}}=\frac{2}{3} \) => \( \Large \left(\frac{r_{1}}{r_{2}}\right)^{2}=\frac{4}{9} \)

\( \Large \frac{Area \ of \ 1st \ circle}{Area \ of \ 2nd \ circle} \)= \( \Large \left(\frac{r_{1}}{r_{2}}\right)^{2}=\frac{4}{9} \)

We know that, area of segment (PRQP) = Area of sector (OPRQO)- Area of \( \Large \triangle OPQ\)

=\( \Large \frac{\pi r^{2}\theta}{360}-\frac{1}{2}r^{2}sin\theta \)
 

so, the area of a segment of a circle is always less than area of its corresponding sector.

II. Distance travelled by a circular wheel of diameter 2d cm in one revolution

= \( \Large 2\times 3.14\times d \)=6.28d

which is greater than 6d cm.

Increase in Circumference of circle = 5%
Increase in radius is also 5%
Now, increase in area of circle
=\( \Large \left(2a+\frac{a^{2}}{100} \right)% \)

where,a=increase in radius

=\( \Large \left(2\times 5+\frac{5\times 5}{100} \right)\% \)=10.25%

Let original radius be r.

Then, according to the question,

\( \Large \pi (r+1)^{2}-\pi r^{2} \)=22

=> \( \Large \pi\times [(r+1)^{2}-r^{2}] \)=22

=> \( \Large \frac{22}{7}\times (r+1+r)(r+1-r) \)=22

=> 2r+1=7 => 2r=6

r=\( \Large \frac{6}{2} \)=3 cm

In \( \Large \triangle AOB \),AO=OB=r[radius of circle]

By Pythagoras theorem,

\( \Large AB^{2}=OA^{2}+OB^{2} => (5)^{2}=r^{2}+r^{2} \)

\( \Large r^{2}=\frac{25}{2} \)cm

Now, area of sector AOB = \( \Large \frac{\theta}{360 degree}\times \pi r^{2} \)

=\( \Large \frac{90 degree}{360 degree}\times \pi\times 25/2=\frac{25\pi}{8}cm^{2} \)

Now, area of minor segment = area of sector - area of triangle

=\( \Large \frac{25\pi}{8}-\frac{r^{2}}{2}=\frac{25\pi}{8}-\frac{25}{4}=\left(\frac{25\pi-50}{8}\right) \)

Area of major segment = Area of circle - Area of minor segment

=\( \Large \pi r^{2}-\left(\frac{25\pi-50}{8}\right)=\frac{25\pi}{2}-\frac{(25\pi-50)}{8} \)

=\( \Large \frac{100\pi-25\pi+50}{8}=\frac{75\pi+50}{8}\)

=\( \Large \frac{25}{8}(3\pi+2)=\frac{25}{4}\left(\frac{3\pi}{2}+1\right)cm^{2} \)

Let the radius of circular field = r m.

Speed of person in m/s =\( \Large \frac{30}{60}=\frac{1}{2} \)m/s

According to the question,

\( \Large \frac{2\pi r}{1/2}-\frac{2r}{1/2} \)=30 => \( \Large 4\pi r \)-4r=30

=> \( \Large \left(4\times 22/7 - 4 \right)r \)=30

=> (12.5-4)r=30 => (8.5)r=30

=> r=\( \Large \frac{30}{8.5} \)=3.5 m.