Question

the figure below shows a circle centre of radius 10 cm the chord PQ=16cm calculate the area of the shaded region​

Answer 1

Step-by-step explanation:

OPQ originally forms a sector, the formula for sector is

$\frac{x}{360} \pi {r}^{2}$

where x is the degrees of rotation between the two radii.

We know three sides length and is trying to find an angle between the radii so we can use law of cosines which states that

$16 = \sqrt{10 {}^{2} + 10 {}^{2} - 2(100) \times \cos(o) }$

This isn't the standard formula, it's for this problem

$16 = \sqrt{200 - 200 \times \cos(o) }$

$16 = \sqrt{200 - 200 \cos(o) }$

$256 = 200 - 200 \cos(o)$

$56 = - 200 \cos(o)$

$- \frac{7}{25} = \cos(o)$

$\cos {}^{ - 1} ( - \frac{7}{25} ) = \cos {}^{ - 1} ( \cos(o) )$

$106.26 = o$

So we found our angle of rotation, which is 106.26

Now, we do the sector formula.

$\frac{106.26}{360} (100)\pi$

$\frac{10626}{360} \pi$

is the area of sector.

Now let find the area of triangle, we can use Heron formula,

The area of a triangle is

$\sqrt{s(s - a)(s - b)(s - c)}$

where s is the semi-perimeter.

To find s, add all the side lengths up, 10,10,16 and divide it by 2.

Which is

$\frac{36}{2} = 18$

$\sqrt{18(18 - 10)(18 - 10)(18 - 16)}$

$\sqrt{18(8)(8)(2)}$

$\sqrt{(36)(64)}$

$6 \times 8 = 48$

So our area of the triangle is 48. Now, to find the shaded area subtract the main area,( the sector of the circle) by the area of the triangle so we get

$\frac{10626}{360} - 48$

Which is an approximate or

44.73