Question

Find the area of the smaller segment whose chord is 8" long in a circle with an 8" radius. I'm not sure how to find the arc degree...

Answer 1

the triangle formed by the chord and radii is equilateral (all sides 8 inches). So the angle subtended at the centre is 60 degrees

the problems is now

Area of the sector - area of the triangle.

find the perpendicular height of the triangle



by pythagoras the height is 8^2 - 4^2

h\[h = \sqrt{48} = 4\sqrt{3}\]

area of the triangle is

\[A = \frac{1}{2} \times 8 \times 4\sqrt{3} = 16\sqrt{3}\]

next the sector

the angle at the centre is 60 so the fraction of the circle is 60/360 or 1/6

Area of the sector is

\[A = \frac{1}{6} \times \pi \times 8^2\]

then the area of the segment is

\[A = \frac{1}{6} \pi 8^2 - 16\sqrt{3}\]

I'll leave you to evaluate the answer