Finding the area of shaded region HELP

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What would I need to do to answer this question?

What I have attempted to do is to find the total area of the rhombus first that is in the centre. I know that this makes up 4 congruent or identical triangles. So to find the area I can work out the area of a single triangle and then multiply that by 4. So in doing so I got this:

With the triangle's area it has to be 'base * height' divided by 2. But I don't know the base's length of let's say OB (shown in the diagram), however that can be found using Pythagoras theorem so...

OB^2 + 2^2 = 4^2

This turned out to be 2 sqrt (3) for the length of OB.

Then the rhombus's area is 4(1/2 * 2 sqrt (3) * 2) which gives 8 sqrt (3)

I'm not too sure what to do now

If anyone can help what to next now that would be really great. Thanks!

(edited 5 years ago)

Reply 1

mqb2766

Original post by Yatayyat

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Calculate the area of the two segments (both are the same).
The circle has radius 4. What is the angle DAB?
This would enable you to calculate the area of the sector and segment.

Reply 2

Yatayyat

Original post by mqb2766

Calculate the area of the two segments (both are the same).
The circle has radius 4. What is the angle DAB?
This would enable you to calculate the area of the sector and segment.

Wait how would you find the segment's area?

I tried working out the sector's area as I can now see now the the diagram is just showing the intersection of two circles.
So the area of a sector is '1/2 * r^2 * theta' when working in radians, where I know r to be 4.
I don't know what angle theta is but I'm assuming that theta must be angle DAB in this case.
Therefore by using trig I know that cos (angle BAO) = 1/2, giving angle BAO to be pi/3 when doing cos^-1(1/2)
Multiplying angle BAO by 2 will then give angle DAB which is 'pi/3 * 2 = 2pi/3'

Substituting all of that into the area of a sector formula gives:

A = 1/2 * (4)^2 * (2pi/3) = 16pi/3

I think to find segment area, it has to be the area of the sector minus the area of the top triangle DAB
So area of top triangle DAB is 1/2 * b * h, where b = 4 sqrt(3) and h = 2
Substitution all those values into the area of a triangle gave a value of 4 sqrt(3)
So I would say the segment area to be 16pi/3 - 4 sqrt(3)

Area of the two segments combined is 2(16pi/3 - 4 sqrt(3)) = 32pi/3 - 8 sqrt(3)

So I can say that shaded region area is equal to 32pi/3 - 8 sqrt(3) - 8 sqrt(3) (from taking away the total area of the rhombus)

This then simplified gave 32pi/3 - 16 sqrt (3), which then becomes 2/3 (16pi -24 sqrt(3)) when taking out factor of 2/3 in the end.

Does the working out seem fine here?

(edited 5 years ago)

Reply 3

begbie68

Original post by Yatayyat

...

This then simplified gave 32pi/3 - 16 sqrt (3), which then becomes 2/3 (16pi -24 sqrt(3)) when taking out factor of 2/3 in the end.

Does the working out seem fine here?

Yep.

But why they didn't have 16/3 (2pi - sqrt(3)) is a bit weird. Where's the question from?

Reply 4

Yatayyat

Original post by begbie68

Yep.

But why they didn't have 16/3 (2pi - sqrt(3)) is a bit weird. Where's the question from?

What do you exactly mean when you say '16/3 (2pi - sqrt(3))' you had to prove the area was 2/3(16pi - 24 sqrt(3)) right?

The question was an end of unit topic question on radians that my teacher had sent me to complete for h/w. Not exactly sure if the question is from a past paper though.

Reply 5

begbie68

Original post by Yatayyat

What do you exactly mean when you say '16/3 (2pi - sqrt(3))' you had to prove the area was 2/3(16pi - 24 sqrt(3)) right?

yes. I know what you had to prove. That wasn't my question.
When you factorise an expression, the convention is to 'take out' denominator, and HCF, so that the bracketed part that you leave is 'clean'.
Most decent question makers would do this.
This wasn't done in this question. Hence a bit odd. That's all.

Reply 6

Yatayyat

Original post by begbie68

Ahh, yeah sorry I see what you mean now.

But isn't 16/3(2pi - sqrt(3)) not equal to 32pi/3 -16 sqrt(3)

You get 32pi/3 -16/3 sqrt(3) instead.

(you only get the first term correct when you expand the brackets out)

Is it not meant to be 16(2pi/3 - sqrt(3)) as 16 here is HCF in the expression?