14. The triangle shown has an area of 88 square units. What is the value of $y$ ?
A 17.6
B $2 \sqrt{46}$
C $6 \sqrt{10}$
D $13 \sqrt{2}$
E $8 \sqrt{5}$
Solution
$\mathbf{E}$
Method 1
We label the vertices of the triangle as shown in the figure. We let $N$ be the point where the perpendicular from $P$ to $Q R$ meets $Q R$, and we let $h$ be the length of $P N$.

The triangle has area 88. So, from the formula area $=$ $\frac{1}{2}$ (base $\times$ height) for a triangle, we have
$$
88=\frac{1}{2}(22 \times h),
$$
that is, $88=11 h$. It follows that $h=8$.
By Pythagoras' Theorem applied to the right-angled triangle $Q N P$, we have $Q N^{2}+h^{2}=10^{2}$. Therefore $Q N^{2}=10^{2}-h^{2}=10^{2}-8^{2}=100-64=36$. Since $Q N$ is a length and therefore positive, we deduce that $Q N=6$. [Alternatively, you could just note that $P Q N$ is a right angled triangle with sides in the ratio $5: 4: 3$.]
It follows that $N R=Q R-Q N=22-6=16$. Now, applying Pythagoras' Theorem to the rightangled triangle $P N R$, we have $N R^{2}+h^{2}=y^{2}$, that is, $16^{2}+8^{2}=y^{2}$. Hence $y^{2}=256+64=320$. We deduce that $y=\sqrt{320}=\sqrt{64 \times 5}=8 \sqrt{5}$.
Method 2
We use the fact that the area of a triangle is $\frac{1}{2} a b \sin \theta$, where $a$ and $b$ are the lengths of two of the sides and $\theta$ is the angle between these two sides, together with the Cosine Rule.

We label the vertices of the triangle as shown in the figure and let $\angle P Q R=\theta$. It follows that $88=\frac{1}{2}(22 \times 10) \sin \theta$. That is, $88=110 \sin \theta$. Hence
$$
\sin \theta=\frac{88}{110}=\frac{4}{5} .
$$

From the identity $\cos ^{2} \theta+\sin ^{2} \theta=1$, it follow that
$$
\cos ^{2} \theta=1-\sin ^{2} \theta=1-\left(\frac{4}{5}\right)^{2}=1-\frac{16}{25}=\frac{9}{25}=\left(\frac{3}{5}\right)^{2} \text {, }
$$
and therefore $\cos \theta= \pm \frac{3}{5}$.
From the diagram in the question, we see that $\angle P Q R$ is an acute angle, that is, $00$ and it follows that $\cos \theta=\frac{3}{5}$.
We can now deduce, using the Cosine Rule, that
$$
\begin{aligned}
y^{2} & =P Q^{2}+Q R^{2}-2 P Q . Q R \cos \theta \\
& =10^{2}+22^{2}-2 \times 10 \times 22 \cos \theta \\
& =100+484-2 \times 10 \times 22 \times \frac{3}{5} \\
& =100+484-264 \\
& =320 .
\end{aligned}
$$

Hence $y=\sqrt{320}=8 \sqrt{5}$.
Commentary
In the problems below we cover a third method for answering Question 14. This method uses Heron's Formula for the area of a triangle.
In Method 1 we used a formula for the area of a triangle in terms of the height of the triangle. In Method 2 we used a formula which involves the angle between two sides. The advantage of Heron's Formula is that it gives the area of a triangle just in terms of the lengths of the sides of the triangle. However, as you will see if you tackle the problems below, this method involves some quite complicated algebra.
Heron's Formula: The area, $A$, of a triangle whose sides have lengths $a, b$ and $c$ is given by
$$
A=\frac{1}{4} \sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)} .
$$

This formula may also be written as
$$
A=\sqrt{s(s-a)(s-b)(s-c)},
$$
where $s=\frac{1}{2}(a+b+c)$, is half of the length of the perimeter of the triangle.

For INVESTIGATION
14.1 Use Heron's Formula to find the third side, $y$, of an acute-angled triangle with sides 10 and 22 , and area 88 .
14.2 Deduce Heron's Formula from the formula $A=\frac{1}{2} a h$, where $h$ is the height of the triangle. [Hint: in the figure for Method 1, let $Q N=x$. Then use Pythagoras' Theorem, applied to triangles $P N Q$ and $P N R$, to obtain an expression for $x$ in terms of $a, b$ and $c$.]
14.3 Deduce Heron's Formula from the formula $A=\frac{1}{2} a b \sin \theta$. [Hint: use the Cosine Formula to obtain an expression for $\cos \theta$ in terms of $a, b$ and $c$.]
Note
Heron of Alexandria lived in the first century of the current era. He was a noted geometer and writer on mechanics. He invented many machines including a steam turbine. His proof of his formula for the area of a triangle is to be found in his book Metrica.