Area of a segment

Question

Two circles of radii 5cm and 12cm are drawn, partly overlapping as shown. Their centres are 13cm apart. Find the area common to the circles?

Answer 1

HINT: Draw a vertical line where the two legs of the triangle meet. Then solve for the partial area of the sector by subtracting the area of the triangle from the area of the sector.

Answer 2

We can split the area into two sectors minus the equaliteral triangles formed inside their sectors.

In math terms:

So we must first find the angles inside the right triangle:

$$\theta = \sin^{-1}(\frac{12}{13})$$

$$\theta = 67.4\ \text{degrees}$$

Similarly we can find the other angles:

$$67.4 + 90 + x = 180$$

$$x = 22.6\ \text{degrees}$$

Now, we can use the sectors:

$$22.6*2 = 45.2$$ $$144\pi*\frac{45.2}{360} = 56.8$$ Now, we have to subtract the triangle to get the remaining area: $$A = \frac{1}{2}*12*12*\sin(45.2) = 51.1$$ Area that remains: $$56.8 - 51.1 = 5.7$$

Now we do the other triangle: $$25\pi*\frac{134.8}{360} = 29.4$$

Area of triangle: $$A = \frac{1}{2}*5*5*\sin(134.8) = 8.9$$

Subtracting: $$20.5$$

Adding that to our other area:

Our final answer becomes: $$20.5 + 5.7 = 26.2$$

Comment if you have any questions. This was a lot to take in.