# Expressing the Area of a Triangle in Surd Form

1. Jul 27, 2012

### odolwa99

1. The problem statement, all variables and given/known data

Having a bit of trouble with this one, specifically the answer of (ii) (Q. diagram included as attachment). Can anyone help?

Many thanks.

Q. In the given diagram, the angles prq & pqs are both right angles. The $\angle$rpq = X, $\angle$qps = Y, |pr| = a & |ps| = b. (i) By considering the triangles prq & psq, show that b = $\frac{a}{cosXcosY}$, (ii) If X = 45o & Y = 30o, express the area of $\triangle$pqs in terms of a. Give your answer in surd form

3. The attempt at a solution

Attempt: (i): cos = $\frac{adjacent}{hypotenuse}$
Thus, cosX = $\frac{a}{|pq|}$ & cosY = $\frac{|pq|}{b}$
Therefore, cosX$\cdot$cosY = $\frac{a}{|pq|}\cdot\frac{|pq|}{b}$ = $\frac{a}{b}$
And so, cosX$\cdot$cosY = $\frac{a}{b}$ => b = $\frac{a}{cosX\cdot cosY}$

(ii) Area of triangle: $\frac{1}{2}b\cdot|pq|\cdot sin30^o$
1st, note that: cos45o = $\frac{1}{\sqrt{2}}$ & cos30o = $\frac{\sqrt{3}}{2}$
Thus, b = $\frac{a}{cosX\cdot cosY}$ => $\frac{a} {1/\sqrt{2}\cdot\sqrt{3}/2}$ => $\frac{a}{\sqrt{3}/2\sqrt{2}}$ => $\frac{2\sqrt{2}a}{\sqrt{3}}$
2nd, note that: cosX = $\frac{a}{|pq|}$ => |pq| = $\frac{a}{cos45^o}$ => |pq| = $\frac{a}{1/\sqrt{2}}$ => |pq| = $\sqrt{2}a$
Now, $\frac{1}{2}b\cdot|pq|\cdot sin30^o$ = $\frac{1}{2}\cdot\frac{2\sqrt{2}a}{\sqrt{3}}\cdot \sqrt{2}a\cdot\frac{1}{2}$ => $\frac{(2\sqrt{2}a)(\sqrt{2}a)}{4\sqrt{3}}$ => $\frac{(\sqrt{2}a)(\sqrt{2}a)}{2\sqrt{3}}$ => $\frac{2a^2}{2\sqrt{3}}$ => $\frac{a^2}{\sqrt{3}}$

Ans.: (From text book): $\frac{a^2\sqrt{3}}{3}$

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2. Jul 27, 2012

### LCKurtz

I didn't check your steps, but look what happens if you rationalize the denominator in your answer.

3. Jul 27, 2012

### odolwa99

Great. Thank you very much.

4. Jul 27, 2012

### Staff: Mentor

Minor point: You are using => incorrectly in some (not all) places. For example, in the line that starts "Thus, b..." All of the => implications in that line should be =. Likewise in the last line.

Use = between two expressions that have the same value. Use => between two statements for which the first statement implies the second, such as 2x + 1 = 3 => x = 1.

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