1. ## Area of Archway

I have this really hard question which is attached at the bottom. I am not supposed to use a calculator to solve it but I've been trying a few methods to work it out, and nothing has come nearer than calculus. I used the equation of a circle graph with radius 2 and found the integral from R to P, and multiplied it by 2.

However, I haven't actually technically 'learnt' calculus so I had to solve it with my calculator and got this:

Area from P to R = $\int_{-2}^{-1}\sqrt{4-x^2}dx=\frac{4sin^{-1}(\frac{x}{2})}{90}+\frac{x\sqrt{4-x^2}}{2}$

Some devilish answer... and then there's all the tricky math involved in evaluating it. Anyways, after running it through the calculator and multiplying by 2 I get (B).

Is there any simple (non-calculus?) way to solve this? It was a question on a very time-limited test. Thanks.

2. Originally Posted by DivideBy0
I have this really hard question which is attached at the bottom. I am not supposed to use a calculator to solve it but I've been trying a few methods to work it out, and nothing has come nearer than calculus. I used the equation of a circle graph with radius 2 and found the integral from R to P, and multiplied it by 2.

However, I haven't actually technically 'learnt' calculus so I had to solve it with my calculator and got this:

Area from P to R = $\int_{-2}^{-1}\sqrt{4-x^2}dx=\frac{4sin^{-1}(\frac{x}{2})}{90}+\frac{x\sqrt{4-x^2}}{2}$

Some devilish answer... and then there's all the tricky math involved in evaluating it. Anyways, after running it through the calculator and multiplying by 2 I get (B).

Is there any simple (non-calculus?) way to solve this? It was a question on a very time-limited test. Thanks.

the area will be twice the area of sector PQR minus the area of the equilateral triagle formed by their radii. (see diagram below)

so A = 2*area of sector in radians - 2*(1/2)bh

so A = 2( (1/2)r^2(pi/3)) - 2((1/2)(1)(sqrt(3))
=> A = 2(2pi/3) - sqrt(3)
=> A = 4pi/3 - sqrt(3)

any questions?