Which of these area and perimeter statements is true?

Which of these area and perimeter statements is true?

Question

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

---

1. $P<A$;

---

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

---

1. $P^2$ is rational.

Suggestion

Can we draw a diagram? What is the radius of the circle?

Solution

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 $P=\dfrac{15\sqrt3}{\pi}$.