Question
Two circles each of radius 7 cm, intersect each other. The distance between their centres is 7√2 cm. Find the area common to both the circles.
✓
Answer
Let $O$ and $O_1$ be the centres of two circles intersecting each other at $A$ and $B$.
Then $O A=O B=O_1 A=O_1 B=7 cm$
and $OO _1=7 \sqrt{2} cm$
$OO _1^2=98$
Since $OA ^2+ O _1 A ^2=7^2$
$ =98$
$= OO _1{ }^2 \ldots \ldots[\text { from (i)] }$
$m _{\angle O_1}=90^{\circ}$
$\square OAO _1 B \text { is a square. }$
$m \angle AOB = m \angle AO O _1 B =90^{\circ}$
$=\left(90 \times \frac{\pi}{180}\right)^{\circ}=\left(\frac{\pi}{2}\right)^c $
Now, $A ($ sector $OAB )=\frac{1}{2} r ^2 \theta$
$ =\frac{1}{2} \times 7^2 \times \frac{\pi}{2}$
$=\frac{49 \pi}{4} \text { sq.cm } $
$\frac{4}{4}$ sq.cm
$ \text { and } A \left(\text { sector } O _1 AB \right)=\frac{1}{2} r ^2 \theta$
$=\frac{1}{2} \times 7^2 \times \frac{\pi}{2}$
$=\frac{49 \pi}{4} sq . cm$
$A (\square OAO , B )=(\text { side })^2=(7)^2=49 \text { sq.cm }$
$\therefore$ Required area $=$ area of shaded portion $= A ($ sector $OAB )+ A \left(\right.$ sector $\left.\left.O _1 AB \right)\right)- A (\square$ $\left.OAO _1 B \right)$
$ =\frac{49 \pi}{4}+\frac{49 \pi}{4}-49$
$=\frac{49 \pi}{2}-49$
$=49\left(\frac{\pi}{2}-1\right) \text { sq.cm } $
Then $O A=O B=O_1 A=O_1 B=7 cm$
and $OO _1=7 \sqrt{2} cm$
$OO _1^2=98$
Since $OA ^2+ O _1 A ^2=7^2$
$ =98$
$= OO _1{ }^2 \ldots \ldots[\text { from (i)] }$
$m _{\angle O_1}=90^{\circ}$
$\square OAO _1 B \text { is a square. }$
$m \angle AOB = m \angle AO O _1 B =90^{\circ}$
$=\left(90 \times \frac{\pi}{180}\right)^{\circ}=\left(\frac{\pi}{2}\right)^c $
Now, $A ($ sector $OAB )=\frac{1}{2} r ^2 \theta$
$ =\frac{1}{2} \times 7^2 \times \frac{\pi}{2}$
$=\frac{49 \pi}{4} \text { sq.cm } $
$\frac{4}{4}$ sq.cm
$ \text { and } A \left(\text { sector } O _1 AB \right)=\frac{1}{2} r ^2 \theta$
$=\frac{1}{2} \times 7^2 \times \frac{\pi}{2}$
$=\frac{49 \pi}{4} sq . cm$
$A (\square OAO , B )=(\text { side })^2=(7)^2=49 \text { sq.cm }$
$\therefore$ Required area $=$ area of shaded portion $= A ($ sector $OAB )+ A \left(\right.$ sector $\left.\left.O _1 AB \right)\right)- A (\square$ $\left.OAO _1 B \right)$
$ =\frac{49 \pi}{4}+\frac{49 \pi}{4}-49$
$=\frac{49 \pi}{2}-49$
$=49\left(\frac{\pi}{2}-1\right) \text { sq.cm } $
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
A box contains 2 blue and 3 pink balls and another box contains 4 blue and 5 pink balls. One ball is drawn at random from one of the two boxes and it is found to be pink. Find the probability that it was drawn from (i) first box. (ii)second box
Prove the following by the principle of mathematical induction:
$2 + 5 + 8 + 11 + ... +(3\text{n} - 1) =\frac{1}{2}\text{n}(3\text{n}+1)$ $$
→$2 + 5 + 8 + 11 + ... +(3\text{n} - 1) =\frac{1}{2}\text{n}(3\text{n}+1)$ $$
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\sqrt{2\sin\text{x}}-1}{\big(\frac{\pi}{2}-\text{x}\big)}$
→$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\sqrt{2\sin\text{x}}-1}{\big(\frac{\pi}{2}-\text{x}\big)}$
If $\sin\text{x}=\frac{12}{13}$ and x lies in the second quadrant, find the value of $\sec\text{x}+\tan\text{x}.$
→Sketch the graphs of the following functions:
$\text{f(x)}=\cot2\text{x}$
→$\text{f(x)}=\cot2\text{x}$
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin(\text{a}+\text{x})+\sin(\text{a}-\text{x})-2\sin\text{a}}{\text{x}\sin\text{x}}$
→$\lim\limits_{\text{x}\rightarrow0}\frac{\sin(\text{a}+\text{x})+\sin(\text{a}-\text{x})-2\sin\text{a}}{\text{x}\sin\text{x}}$
In a group of 50 persons, 14 drink tea but not coffee and 30 drink tea. Find:
How may drink tea and coffee both.
How may drink tea and coffee both.
If α and β are the complex cube roots of unity, show that
→1.$\alpha^2+\beta^2+\alpha \beta=0$
(ii) $\alpha^4+\beta^4+\alpha^{-1} \beta^{-1}=0$
2.If x = a + b, y = αa + βb and z = aβ + bα, where α and β are complex cube roots of unity,
show that $x y z=a^3+b^3$.
Prove the following by the principle of mathematical induction:
$5^{2 n+2}+24 n-25$ is divisible by 576 for all $n \in N$
→$5^{2 n+2}+24 n-25$ is divisible by 576 for all $n \in N$
Find the general solutions of the following equations:
$\cos4\text{x}=\cos2\text{x}$
→$\cos4\text{x}=\cos2\text{x}$