Ratio of Areas of Similar Triangles

Question

First step, I can't find the height. How do you find the height?

Answer 1

If you assume one of the answers must be the correct one, here's a way to see that it can only be (E):

$\triangle SMN$ is similar to $\triangle SQR$ and of half its dimensions, therefore a quarter of its area. $\triangle SQR$ is strictly smaller than $\triangle NQR$, which, because $N$ is the midpoint of $PQ$, is half the area of $\triangle PQR$. The area of $\triangle SMN$ is therefore strictly smaller than one-eight the area of $\triangle PQR$.

The only ratio amongst the proffered answers that meets this condition is the ratio $1{:}12$ of (E).

Answer 2

HINT : Do you see why the followings hold?

$$\triangle PQR:\triangle MQR=2:1$$ $$\triangle MQR:\triangle SQR=3:2$$ $$\triangle SQR:\triangle MNS=2^2:1^2$$

P.S. For exmaple, since $MQ:SQ=3:2$, we have $\triangle MQR:\triangle SQR=3:2$. Do you see why?

Answer 3

Hint 1: $\triangle MNS$ is similar to $\triangle QRS$ (why?)

Note that $MN$ is parallel to $QR$ and compare the angles of the two triangles.

Hint 2: You know the ratio of $MN$ to $QR$. Can you then find the ratio of the areas of $\triangle MNS$ and $\triangle QRS$

See Oleg567's comment: if the lengths of the sides are in ratio $a:b$, the the ratio of the areas is $a^2:b^2$.

Hint 3: $$\frac{\triangle MNS}{\triangle PQR} = \frac{\triangle MNS}{\triangle PNM} \frac{\triangle PNM}{\triangle PQR}$$

Answer 4

This is quite an interesting question. Could you share with us which level or type of test or exam this is from?

• Draw a line from P to S and extend it such that it meets QR at L.
• Let K be the point of intersection of PL and MN.
• Note that MN is parallel to QR but half the length (as M and N are midpoints). i.e. $MN=\frac 12 QR$.
• As such, $PK=\frac 12 PL$.
• Also, $PS=\frac 23 PL$.
• Hence, $KS=(\frac 23 - \frac 12)PL =\frac 16 PL$.
• Note that KS and PS are proportional to the height of triangles MNS and PQR respectively (for actual heights, both KS and PS should be multiplied by $\sin\theta$ where $\theta$ is the angle between PS (or KS) and QR (or MN)).
• Hence $$\begin{align} \dfrac {\triangle MNS}{\triangle PQR}&=\dfrac {\;\;\frac 12 \cdot MN\cdot KS \sin\theta } {\frac 12\cdot QR\cdot PL\sin\theta}\\ &= \dfrac {MN\cdot KS} {QR\cdot PL}\\ &= \dfrac {\frac 12QR\cdot \frac 16 PL} {QR\cdot PL}\\ &=\frac 1{12} \end{align}$$