# finding areas perimeters and segments in circle

• May 25th 2009, 10:39 AM
llkkjj24
finding areas perimeters and segments in circle
Hi, i really need help on working this one out as it is very important for school, thanks in advance.

A circle has centre o and radius 1cm, points A,B,C,D,E and F lie on the cirumference, so that A,B,C,D,E,F is a regular hexagone(6 equal sides)

(a): Show that triangle OAB is equilateral, Find its area and the area of the hexagone. Find the area of the circle

(b) : deduce the area of the minor segment defined on AB

(c): Find perimeter of this segment
• May 25th 2009, 11:59 AM
Hello llkkjj24

Welcome to Math Help Forum!
Quote:

Originally Posted by llkkjj24
Hi, i really need help on working this one out as it is very important for school, thanks in advance.

A circle has centre o and radius 1cm, points A,B,C,D,E and F lie on the cirumference, so that A,B,C,D,E,F is a regular hexagone(6 equal sides)

(a): Show that triangle OAB is equilateral, Find its area and the area of the hexagone. Find the area of the circle

(b) : deduce the area of the minor segment defined on AB

(c): Find perimeter of this segment

(a) All angles at O are equal, because the hexagon is regular. Therefore $\angle AOB = 60^o$

$AO = OB$, radii

Therefore $\angle OAB = \angle ABO$ (isosceles triangle)

Therefore $\angle OAB = 60^o$ (angle sum of triangle)

Therefore $\triangle AOB$ is equilateral.

Using $\triangle = \tfrac12 bc\sin A$, its area = $\tfrac12.1.1.\sin 60^o = 0.433\, cm^2$

So area of hexagon $= 0.433 \times 6 = 2.598\, cm^2$.

Area of circle $= \pi.1^2 = 3.142\, cm^2$

(b) Therefore the area of the minor segment $AB = \tfrac16(3.142 - 2.598) = 0.088\, cm^2$

(c) Arc $AB = \tfrac162\pi.1 = 1.047 \,cm$

$AB = 1\, cm$

So the perimeter of this segment $= 2.047 \,cm$