Question: The area of this isosceles triangle is $108$ cm$^2$.
What is the length of $PQ$?
Give your answer in centimetres (cm) and give any decimal answers to 1 d.p.
$R$ $Q$
$18$ cm
Isosceles triangle with top vertex $P$ and base $RQ$; a dashed vertical height drops from $P$ to the midpoint of $RQ$, with a right-angle marker at the foot. The base $RQ$ is labeled $18$ cm and the triangle’s area is $108$ cm$^2$.
1. **State the problem:**
We have an isosceles triangle $PQR$ with base $RQ = 18$ cm and area $108$ cm$^2$. We need to find the length of side $PQ$.
2. **Recall the formula for the area of a triangle:**
$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$
3. **Find the height:**
Let the height from $P$ to $RQ$ be $h$. Using the area formula:
$$108 = \frac{1}{2} \times 18 \times h$$
Simplify:
$$108 = 9h$$
Divide both sides by 9:
$$h = \frac{108}{9}$$
$$h = 12$$
4. **Use the height to find $PQ$:**
Since the triangle is isosceles, the height drops to the midpoint of $RQ$, splitting $RQ$ into two segments each of length:
$$\frac{18}{2} = 9$$
Now, consider the right triangle formed by $P$, the midpoint of $RQ$, and $Q$. The legs are $h = 12$ cm and half the base $9$ cm, and the hypotenuse is $PQ$.
Use the Pythagorean theorem:
$$PQ = \sqrt{h^2 + 9^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225}$$
$$PQ = 15$$
5. **Final answer:**
The length of $PQ$ is $15.0$ cm (to 1 decimal place).