Circle Formulae
1. Area: $$\pi\ r^2$$
2. Circumference: $$2\pi r$$
3. Length of Arc: $$\left(\frac{\theta}{360}\right)\left(2\pi r\right)$$
4. Area of Sector: $$\left(\frac{\theta}{360}\right)\left(\pi r^2\right)$$
Circle - Formulae
Rarely Tested
Question 1
In a triangle ABC, AB = AC = 8 cm. A circle drawn with BC as diameter passes through A. Another circle drawn with center at A passes through Band C. Then the area, in sq. cm, of the overlapping region between the two circles is
Solution
BC is the diameter of circle C2 so we can say that $$\angle BAC=90^{\circ\ }$$ as angle in the semi circle is $$90^{\circ\ }$$
Therefore overlapping area = $$\frac{1}{2}$$(Area of circle C2) + Area of the minor sector made be BC in C1
AB= AC = 8 cm and as $$\angle BAC=90^{\circ\ }$$, so we can conclude that BC= $$8\sqrt{2}$$ cm
Radius of C2 = Half of length of BC = $$4\sqrt{2}$$ cm
Area of C2 = $$\pi\left(4\sqrt{2}\right)^2=32\pi$$ $$cm^2$$
A is the centre of C1 and C1 passes through B, so AB is the radius of C1 and is equal to 8 cm
Area of the minor sector made be BC in C1 = $$\frac{1}{4}$$(Area of circle C1) - Area of triangle ABC = $$\frac{1}{4}\pi\left(8\right)^2-\left(\frac{1}{2}\times8\times8\right)=16\pi-32$$ $$cm^2$$
Therefore,
Overlapping area between the two circles= $$\frac{1}{2}$$(Area of circle C2) + Area of the minor sector made be BC in C1
= $$\frac{1}{2}\left(32\pi\right)\ +\left(16\pi-32\right)=32\left(\pi-1\right)$$ $$cm^2$$
Question 2
A circular plot of land is divided into two regions by a chord of length $$10\sqrt{3}$$ meters such that the chord subtends an angle of 120° at the center. Then, the area, in square meters, of the smaller region is
Solution
This is the situation that is described in the question above, Angle AOB is 120 degrees and the chord AB is of length 10cm.
Using Cosine rule we can find the length of AO and AB
$$\cos\left(AOB\right)=\frac{\left(r^2+r^2-300\right)}{2r^2}$$
$$-\frac{1}{2}=\frac{\left(2r^2-300\right)}{2r^2}$$
$$3r^2=300$$
$$r=10$$
Area of the Sector AOB is $$\frac{120}{360}\times\ \pi\ \times\ r^2$$
$$\frac{1}{3}\times\ \pi\ \times\ \frac{100}{1}$$ which is $$\frac{100\pi}{3}\ $$
Area of triangle AOB is $$\frac{1}{2}\times\ r^2\times\ \sin\left(120\right)$$
$$\frac{1}{2}\times\ \frac{100}{1}\times\ \frac{\sqrt{3}}{2}$$
Area of triangle AOB is $$\frac{75}{\sqrt{3}}$$
The smaller region will be, $$\frac{100\pi\ }{3}-\frac{75}{\sqrt{3}}$$
Taking 25 common we will get,
$$25\left(\cfrac{4 \pi}{3} - \sqrt{3}\right)$$
Question 3
Three concentric circles A, B and C have radii in the ratio 1:2:4. What is the ratio of area enclosed between A and B and the area enclosed between B and C
Solution
Area enclosed between A and B(region 1) = area of circle B - area of circle A = $$\pi\ \left(2\right)^2-\pi\ \left(1\right)^2=3\pi\ $$
Area enclosed between B and C(region 2) = area of circle C - area of circle B = $$\pi\ \left(4\right)^2-\pi\ \left(2\right)^2=12\pi\ $$
Required ratio = $$\frac{3\pi}{12\pi\ }=\frac{1}{4}\ $$
The answer is option C.
Question 4
The sum of the areas of two circles touching each other externally is 170$$\pi$$ sq cm and the distance between the centres of the two circles is 18 cm. Find the ratio of the radii of the two circles?
Solution
$$x^{2}$$ + $$y^{2}$$ = 170 and x + y = 18. So, x = 11 cm and y = 7 cm
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