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Find the image for the question here

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

user642796
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4 Answers4

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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?

mathlove
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  • I don't see the second and third followings. –  Aug 19 '14 at 20:28
  • @Ben: For the second: Since $\triangle PNM$ and $\triangle PQR$ are similar, $MN:RQ=1:2$. Since $\triangle MNS$ and $\triangle QRS$ are similar, $MS:QS=MN:QR=1:2\Rightarrow MQ:SQ=3:2$. When you see $MQ$ and $SQ$ as the base of $\triangle MQR$ and $\triangle SQR$ respectively, we have $\triangle MQR:\triangle SQR=MQ:SQ=3:2$ because their height are the same. – mathlove Aug 20 '14 at 11:30
  • @Ben: For the third: Since $\triangle SQR$ and $\triangle MNS$ are similar and the ratio of their edges is $2:1$, the ratio of their areas is $2^2:1^2$. see for example here : http://www.mathwarehouse.com/geometry/similar/triangles/area-and-perimeter-of-similar-triangles.php – mathlove Aug 20 '14 at 11:36
  • I don't understand the conclusion that $MQ:SQ=3:2$ "Since △MNS and △QRS are similar, $MS:QS=MN:QR=1:2⇒$". How can you talk about sides $MQ and SQ$, which are part of triangles $NMS and QSR$, if they are unrelated to $△MNS and △QRS$? –  Aug 20 '14 at 12:59
  • @Ben: If you understand that $MS:SQ=1:2\iff 2MS=SQ\iff MS=(1/2)SQ$, then I'm sure that you can understand that $MQ=MS+SQ=(1/2)SQ+SQ=(3/2)SQ$, which can be simplified as $2MQ=3SQ\iff MQ:SQ=3:2$. – mathlove Aug 21 '14 at 08:48
  • My problem is that I did not think of splitting $MQ$ into $2$ parts, $MS and QS$. –  Aug 21 '14 at 14:13
  • If I had realized that, I would have gotten "$2MQ=3SQ⟺MQ:SQ=3:2$ and solved the problem. –  Aug 21 '14 at 14:28
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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: $$\frac{\triangle MNS}{\triangle PQR} = \frac{\triangle MNS}{\triangle PNM} \frac{\triangle PNM}{\triangle PQR}$$

angryavian
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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 $$\begin{align} \dfrac {\triangle MNS}{\triangle PQR}&=\dfrac {\;\;\frac 12 \cdot MN\cdot KS \sin\theta } {\frac 12\cdot QR\cdot PL\sin\theta}\\ &= \dfrac {MN\cdot KS} {QR\cdot PL}\\ &= \dfrac {\frac 12QR\cdot \frac 16 PL} {QR\cdot PL}\\ &=\frac 1{12} \end{align}$$
Hypergeometricx
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  • scholastic apitude test –  Aug 19 '14 at 20:11
  • I don't see why PS=2/3PL. Can you explain? –  Aug 19 '14 at 20:37
  • This is a characteristic of lines from each apex to the midpoint of the opposite side, where the point of intersection of the lines is \frac 23 of the way. This can also be seen by noting that MNS and QRS are similar triangles but with MNS being half in length as MN is half QR. This means KS = \frac 12 SL, i.e. KS=\frac 13 KL. But PK=KL=\frac 12 PL. Hnece KS=(1/3)(1/2)PL=(1/6)PL. Hence – Hypergeometricx Aug 20 '14 at 17:07
  • [ignore above comment which can no longer be edited or deleted; use this one instead]

    This is a characteristic of lines from each apex to the midpoint of the opposite side, where the point of intersection of the lines is $\frac 23$ of the way.

    This can also be seen by noting that MNS and QRS are similar triangles but with MNS being half in length as MN is half QR.

    This means $KS = \frac 12 SL$, i.e. $KS=\frac 13 KL$.

    But $PK=KL=\frac 12 PL$. Hence $KS=\frac 13 \cdot \frac 12 PL=\frac 16 PL$.

     – Hypergeometricx Aug 20 '14 at 17:15
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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).

Barry Cipra
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  • I don't see that "The area of △SMN is therefore strictly smaller than one-eight the area of △PQR." –  Aug 19 '14 at 21:35
  • @Ben, $\triangle SMN=(1/4)\triangle SQR$ and $\triangle SQR\lt \triangle NQR=(1/2)\triangle PQR$. – Barry Cipra Aug 20 '14 at 01:16
  • I did not realize at first that $△NQR=(1/2)△PQR$. Second, although I knew $NM:QR=1:2$, I did not conclude that the areas would be respectively $△SQR:△MNS=4:1$ . I would have solved the problem if I had realized those two things. –  Aug 21 '14 at 14:15