Thread: Area and Perimeter

I need some help with the following questions concerning area and perimeter:

1. Calculate the area of a regular hexagon with a side length of 10 cm.
Would I use A triangle=0.5 X 10 X 5 and then multiply that again by 6?

2. Calculate the area and perimeter of the following segments:
a. A segment with a diameter of 20 cm.
b. A segment, when it is attached to a right angled triangle which has two side lengths of 12 cm.

3. Calculate the area and perimeter of one part of the yin-yang symbol (so, I guess just the yin part).

Thank you! Please reply when you can, as soon as possible if possible!

Originally Posted by JadeKiara

I need some help with the following questions concerning area and perimeter:

1. Calculate the area of a regular hexagon with a side length of 10 cm.
Would I use A triangle=0.5 X 10 X 5 and then multiply that again by 6?

A hexagon is six sided so the area would be the sum of the

areas of six identical triangles. If we call this area A:

Area of hexagon = 6*A

We just need to find the area A of one of these triangles.


A = (1/2)base*height

The base is the side of the hexagon or 10 inches

The height is the distance from the center of the hexagon to

the center of the side and this forms a right triangle whose

one side is 10/2 = 5 and has a 60 degrees angle so that

tan(60) = height/5











There is a formula as well


(no. of sides)(length of a side)^2
-------------------------------- = AREA of Regular POLYGONS
(4)tan(180/no. of sides)

Originally Posted by JadeKiara

I need some help with the following questions concerning area and perimeter:


2. Calculate the area and perimeter of the following segments:
a. A segment with a diameter of 20 cm.
b. A segment, when it is attached to a right angled triangle which has two side lengths of 12 cm.

3. Calculate the area and perimeter of one part of the yin-yang symbol (so, I guess just the yin part).

Thank you! Please reply when you can, as soon as possible if possible!

2) & 3) I will help you when you will show your work & where you got stuck

Thank you so much!

I really don't know... But here is my attempt:

2a. Asegment = Asector - Atriangle

2a. 60/360 X Pi X 8^2 - 100 (1/2 bh)
= ?

I really don't understand this - and would 3 be half of Pi?

Originally Posted by JadeKiara

Thank you so much!

I really don't know... But here is my attempt:

2a. Asegment = Asector - Atriangle

2a) that's the right way

If you want a formula see this



Put the data that you have with you, its not the real question that you supplied us

b) 2ways of doing it (try drawing diagram)

The right angle can be made at center of circle & at circumference

3) A very simple approach is that since both the colors are symmetrical, the area will be half of the circle

I actually did figure out the yin-yang before you mentioned!!

How do I find the perimeter of a segment after I have found the area?

Originally Posted by JadeKiara

I actually did figure out the yin-yang before you mentioned!!

How do I find the perimeter of a segment after I have found the area?

For perimeter, I think you will have to do

- Join the diameter from top to bottom of the figure I gave

- This will approximately form

• 2 black & white semicircle on either sides
• each semicircle of a color has the opposite color semicircle
• the smaller semicircle has radius half to that of the complete circle

So the perimeter of a single part(black or white) is

Half of Bigger circle perimeter + total perimeter of the smaller circle (as 2 semicircle on each sides)

So ultimately
it will be= perimeter of the Bigger circle

Sorry, I meant what would the perimeter of 2 be?

Originally Posted by JadeKiara

Sorry, I meant what would the perimeter of 2 be?

No problem, it seems I misread

everything you want to know

Oh, what a bugger! For you and me!

So is the formula for perimeter of a segment on this page?

Originally Posted by JadeKiara

Oh, what a bugger! For you and me!

So is the formula for perimeter of a segment on this page?

You want arc length its there , you want chord length its there
.......

Thank you ADARSH!

I must go to bed now - so shall we meet again!