Review question

# Which of these area and perimeter statements is true? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8330

## Solution

The vertices of an equilateral triangle are labelled $X$, $Y$ and $Z$. The points $X$, $Y$ and $Z$ lie on a circle of circumference $10$ units. Let $P$ and $A$ be the numerical values of the triangle’s perimeter and area, respectively. Which of the following is true?

1. $\dfrac{A}{P}=\dfrac{5}{4\pi}$;

2. $P<A$;

3. $\dfrac{P}{A}=\dfrac{10}{3\pi}$;

4. $P^2$ is rational.

Since the circumference $C=2\pi r=10$, we have $r=\dfrac{5}{\pi}$.

We know that the area of a triangle $ABC$ is $\dfrac{1}{2}ab\sin C$. Using this,the area of the smaller triangle shown in the diagram is $\frac{1}{2}r^2\sin\left( \frac{2\pi}{3}\right)=\left(\dfrac{5}{\pi}\right)^2\frac{\sqrt3}{4} =\dfrac{25\sqrt{3}}{4\pi^2},$ since $\sin\left( \dfrac{2\pi}{3}\right)=\dfrac{\sqrt{3}}{2}$.

Thus the area of the whole equilateral triangle is $A=\dfrac{75\sqrt{3}}{4\pi^2}.$

To find $P$, we note that $\dfrac{1}{2}XZ = r\sin \left(\frac{\pi}{3}\right)=\dfrac{5}{\pi}\frac{\sqrt{3}}{2},$ and hence $P=\dfrac{15\sqrt{3}}{\pi}.$ Therefore $\frac{A}{P}=\frac{75\sqrt{3}}{4\pi^2}\frac{\pi}{15\sqrt{3}}=\frac{5}{4\pi}$ so the answer is (a).

To check, it is easy to see that $P^2$ is irrational, since it involves $\pi^2$, so (d) is not correct.

Since $\dfrac{A}{P} = \dfrac{5}{4\pi}, P > A$, so (b) is wrong.

From our work, (c) is clearly wrong.

Alternatively, the cosine rule applied to the smaller triangle could be used to find $P$: $P=3\times\sqrt{2\times r^2-2\times r^2\cos \left(\frac{2\pi}{3}\right)},$ which again gives $P=\dfrac{15\sqrt3}{\pi}$.