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Problem:

$PQRS$ is a trapezium in which base $PQ=2$ units and base $RS = 3$ units. Draw a line from point $P$ to $R$. Triangle $PQR$ is formed. Shade in that triangle.

Now, what fraction of the trapezium is shaded?

I am having some trouble solving this problem as the height of the trapezium is not mentioned. I believe that knowing that height is unnecessary, but I am not too sure.

My Attempt:

Lemma: Given a trapezium with bases $a$ and $b$, and an altitude (height) $h$, the area is $\dfrac{h(a+b)}{2}$.

In this example, $a,b=PQ,RS$ respectively, so the area is $$\dfrac{h(2+3)}{2} = \dfrac{5h}{2}.$$

Now I know that if it were a square, then the two bases would be equal to each other, and the line $PR$ would be splitting it into equal halves.

So, $\Delta PQR = \dfrac 12$.

But, since on of the bases is larger, and the smaller base is a side of $\Delta PQR$, then my assumption is that the area of $\Delta PQR$ is less than $\dfrac 12$.

Nevertheless, the height of $PQRS$ is not stated so I do not know how to solve this. Intuitively, I believe the answer is $\dfrac 13$ but how can it be proven?

Thank you in advance.

Edit:

Doing some research, I found a very similar post over here.

General Rule (edited just now):

After looking at the answer, I came up with a general rule:

You are given a quadrilateral $PQRS$ (i.e. a square; trapezium; etc) in which base $PQ = a$ units; base $RS = b$ units; the two bases are parallel; and height $h\perp PQ\land RS$. By drawing a diagonal line $PR = c$ units to form two congruent triangles $PQR$ and $PSR$ that make up $PQRS$, then by denoting $S_{ABCD}$ the area of $ABCD$ and $\Delta{UVW}$ a triangle $UVW$, $$\frac{S_{\Delta PQR}}{S_{PQRS}} = \frac{a}{a+b}\,\text{ and }\,\frac{S_{\Delta PSR}}{S_{PQRS}} = \frac{b}{b+a},$$ regardless of the value of $h$.

Just clarifying: is this true? If so, does this theorem have a name?

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Mr Pie
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  • Think why the area of $,\triangle PQR,$ is $,\dfrac{h \cdot a}{2},$. – dxiv May 10 '18 at 04:32
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    @dxiv ahhh I never thought about finding the area of the triangle :)) – Mr Pie May 10 '18 at 04:36
  • "I am having some trouble solving this problem as the height of the trapezium is not mentioned." Doesn't matter as both the triangle and trapezium of will have the same height so both the area of the triangle and the trapezium will have areas proportional to the height. – fleablood May 10 '18 at 05:08
  • "I never thought about finding the area of the triangle" Out of curiosity, how could you be asked to find what proportion the area of a triangle is without thinking about finding the area of a triangle? What else is there to think about? – fleablood May 10 '18 at 05:10
  • @fleablood sorry, I may not be that advanced as you, I suppose. – Mr Pie May 10 '18 at 11:39
  • Your "General Rule" is correct. The ratios cancel the $\frac12$s and $h$s in the various area formulas $$|\triangle PQR| = \frac12 h a \qquad |\triangle PSR| = \frac12 h b \qquad |\square PQRS| = \frac12 h (a+b)$$ – Blue May 14 '18 at 01:22

1 Answers1

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$$\text{Required fraction} = \frac{S_{\triangle PQR}}{S_{PQRS}} = \frac{\frac12 \cdot 2h}{\frac12 \cdot 5h} = \frac25$$

GNUSupporter 8964民主女神 地下教會
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    I see, because it is a ratio that also include the area of the triangle and not just the trapezium. Thank you very much for that. If it was that easy, you might have given me a hint, but it doesn't matter now as you get $$\color{green}{\checkmark}$$ I have to wait $17$ hours before I can upvote, but congratulations nonetheless! – Mr Pie May 10 '18 at 04:37
  • @user477343 You may think of "half of the trapezoid is shaded" (with a picture in your mind) in our natural language, so that the desired fraction is $$\frac{S_{\rm shaded}}{S_{\rm trapezoid}} = \frac12.$$ You may try other shapes to develop your sense towards sth like "$p/q$ of $A$ is $B$". – GNUSupporter 8964民主女神 地下教會 May 10 '18 at 04:46
  • Thank you for that :) By the way, when you write $S_n$, what does the $S$ stand for? I know from your answer that you mean area, and I use the same notation but with $A$ instead to stand for Area. Does $S$ stand for Shape or something like that? Or does it stand for nothing at all, pretty much? – Mr Pie May 10 '18 at 04:50
  • @user477343 I often saw authors of [tag:contest-math] books using $S_{\rm sth}$ to denote the area of $\rm sth$ when I was a high school student. – GNUSupporter 8964民主女神 地下教會 May 10 '18 at 04:58
  • Yes, this problem is an old contest-math problem that I found around my desk. It was from 2007 or 2008 from what I could remember (I think the former) and I found it difficult to solve. I am currently now a high school student so I still have much to learn :) – Mr Pie May 10 '18 at 05:09
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    @user477343 I recommend this question to you since its answers are suitable to students like you. – GNUSupporter 8964民主女神 地下教會 May 10 '18 at 05:11