Can someone help me with this maths question - A-level

audreygal

This is probably very easy , but my mind is blank! Can someone help me with part A.

Original post by audreygal

What book is this question from?
Edexcel AS Core mathematics ...

Reply 3

the bear

the sector area is 1/2r²{2Θ}

the edge of the square is 2{rsinΘ} ....

edited ^^^

(edited 7 years ago)

Reply 4

audreygal

Original post by Rashina

What book is this question from?
Edexcel AS Core mathematics ...

Yeah I think so

Reply 5

RogerOxon

Original post by the bear

the sector area is 1.2r²{2Θ}

the edge of the square is 2{rsinΘ} ....

Perhaps I'm not reading that correctly, but the sector area doesn't look right. It can be calculated as a proportion of the circle, i.e.

\frac{2\theta}{2\pi}\pi r^2 = \theta r^2

I agree that each side of the square is

2rsin\theta

.

Calculate the area of the square and set it equal to the area of the sector.

Reply 6

audreygal

Original post by RogerOxon

Perhaps I'm not reading that correctly, but the sector area doesn't look right. It can be calculated as a proportion of the circle, i.e.

\frac{2\theta}{2\pi}\pi r^2 = \theta r^2

I agree that each side of the square is

2rsin\theta

.

Calculate the area of the square and set it equal to the area of the sector.

Thank you, yes that is right as I have just proved it.

Reply 7

the bear

Original post by RogerOxon

Perhaps I'm not reading that correctly, but the sector area doesn't look right. It can be calculated as a proportion of the circle, i.e.

\frac{2\theta}{2\pi}\pi r^2 = \theta r^2

I agree that each side of the square is

2rsin\theta

.

Calculate the area of the square and set it equal to the area of the sector.

sorry... it was supposed to say 1/2, not 1.2

:ashamed:

Reply 8

BrasenoseAdm

Official Rep

Original post by audreygal

Use the cosine rule first to work out CD and then the double-angle formula for cos(2-theta).

You should end up with 2r^2 - 2r^2 (1 - 2sin^theta). This is the area of ABCD and the question says this area is equal to the area of the sector, which is 1/2 r^2 2-theta

From here a bit of factorising and manipulation should give the identity you need.

.

(edited 7 years ago)

Reply 9

audreygal

Original post by BrasenoseAdm

Use the cosine rule first to work out CD and then the double-angle formula for cos(2-theta).

You should end up with 2r^2 - 2r^2 = (1 - 2sin^theta). This is the area of ABCD and the question says this area is equal to the area of the sector, which is 1/2 r^2 2-theta

From here a bit of factorising and manipulation should give the identity you need.

.

Thanks! I don't know why I didn't see this!