Triangle PQR is a right-angled triangle with PQ = 10 cm, QR = 6 cm and ∠PRQ = 90°. M is a point on PR such that MR = 5 PM. (Refer to the diagram.) Find, giving your answer as a fraction in its simplest form, the value of

1. Find the length of $PR$ using Pythagoras' theorem in right-angled triangle $PQR$.

Since $\angle PRQ = 90^\circ$, $PQ$ is the hypotenuse.

$PR^2 + QR^2 = PQ^2$

$PR^2 + 6^2 = 10^2$

$PR^2 = 100 - 36 = 64$

$PR = 8\text{ cm}$

2. Use $MR = 5PM$ to find $MR$, then apply the tangent ratio in triangle $MQR$.

Given $MR = 5PM$, we have

$PM : MR = 1 : 5$

So $PR$ is split into $1 + 5 = 6$ equal parts.

$MR = \frac{5}{6} \times PR = \frac{5}{6} \times 8 = \frac{20}{3}\text{ cm}$

In right-angled triangle $MQR$,

$\tan \angle MQR = \frac{\text{opposite}}{\text{adjacent}} = \frac{MR}{QR}$

$\tan \angle MQR = \frac{\frac{20}{3}}{6} = \frac{20}{18} = \frac{10}{9}$

1. Find $MQ$ in right-angled triangle $MQR$.

From part (a), $PR=8$ cm and $MR=\dfrac{20}{3}$ cm.

In right-angled triangle $MQR$,
$MQ^2=MR^2+QR^2$
$MQ^2=\left(\dfrac{20}{3}\right)^2+6^2=\dfrac{400}{9}+36=\dfrac{724}{9}$
$MQ=\dfrac{2\sqrt{181}}{3}$

2. Use the supplementary-angle relationship to find $\cos \angle PMQ$.

Since $P, M, R$ are collinear, $MP$ and $MR$ are opposite rays.

Hence
$\angle PMQ=180^\circ-\angle RMQ.$

Therefore
$\cos \angle PMQ=-\cos \angle RMQ.$

In right-angled triangle $MQR$,
$\cos \angle RMQ=\dfrac{MR}{MQ}=\dfrac{\frac{20}{3}}{\frac{2\sqrt{181}}{3}}=\dfrac{10}{\sqrt{181}}.$

So
$\cos \angle PMQ=-\dfrac{10}{\sqrt{181}}.$

1. Compare the bases $PM$ and $PR$ since the two triangles have the same height from $Q$ to line $PR$.

Given $MR=5PM$,

$PR=PM+MR=PM+5PM=6PM$.

Since $\triangle PQM$ and $\triangle PQR$ have bases $PM$ and $PR$ on the same straight line $PR$, and share the same perpendicular height from $Q$,

$\frac{\text{area of }\triangle PQM}{\text{area of }\triangle PQR}=\frac{PM}{PR}=\frac{PM}{6PM}=\frac{1}{6}.$