# trapezoids similarity

I found this geometry problem in an IGCSE text. Can't seem to find the missing lengths that they are asking for. I got an estimate for f (f=2.6666... cm) by drawing a line parallel to the left side of the trapezium that intersects the top right-hand vertex and forms two similar triangles.

is it just possible that some data is missing in this problem?

Screenshot of the problem with accompanying diagram: • Apologies for not mentioning this before. The info in that screen shot was all that was provided. – Indula Dec 21 '18 at 9:18
• First thoughts: Looking at the angle that the slanted edge makes with the vertical edge, surely it must be true that $$\frac{4}{3} = \frac{f}{2}$$ Do you agree? – Matti P. Dec 21 '18 at 9:23
• @Indula Similar quadrilaterals work in much the same way as similar triangles – Shubham Johri Dec 21 '18 at 9:28
• Yes, you could call it the '$AAAA$' criterion for quadrilaterals. Anywho, you could have a solution using triangles. – Shubham Johri Dec 21 '18 at 9:50
• The two trapezoids are NOT similar. Having the same angles does not mean they are merely scaled up versions of one another. For example rectangles are not all similar - they are only similar if they have the same aspect ratio. However, in this problem, if you extend the sloped sides of the trapezoid till they meet, you can use three triangles, and triangles are similar if they have the same angles. – Jaap Scherphuis Dec 21 '18 at 10:38

You wrote that you drew a line parallel to the left side from the top-right corner down to the base. That line will also be split into lengths of exactly $$3$$ and $$2$$, just like the left side. This is because you have parallelograms on the left. This mean that the ratios on the triangles on the right is $$3:(2+3)$$, and this gives you that the right hand side of the trapezoid is $$4*5/3=20/3$$, making $$f=20/3 - 4 = 8/3$$.

To find $$e$$ you will have to make some assumption. From the right hand end of the horizontal line of length $$6$$, you can draw a line down to the base, parallel to the left side of the trapezoid. To the left of that line you have a parallelogram, so that line is of length $$2$$, and the part of the base to the left of it is of length $$6$$. So to find $$e$$, you need the part of the base to the right of it. That unknown part is the base of a triangle with sides $$2$$ and $$f=8/3$$. Unfortunately, it is impossible to find that third length unless we know one of the angles of that triangle.

To illustrate this, I have drawn the diagram in two ways: You can clearly see the length of the base $$e$$ differs, even though the given lengths and the length of $$f$$ all remain unchanged.

Presumably the left side is supposed to be perpendicular to the base. In that case you can use Pythagoras to find out the exact length.

• Thanks so much! There was a typo in the very first post I made which indicated f=1.6666... which should have been f=8/3=2.666... Thanks for verifying my method for finding the length f. I was thinking about assuming that the trapezium is right-angled (it looks like that after all) and using the Pythagorean theorem to find the length e but was hesitant about explicitly adopting that assumption seeing as they haven't indicated that explicitly. – Indula Dec 21 '18 at 15:15
• But on another note, if it hadn't been right-angled, and instead had been something akin to what you've sketched in your first diagram, is it possible to find the length, even if we knew the degrees of the angle? – Indula Dec 21 '18 at 15:17
• Yes, in that case you will have to use the law of cosines, which can be considered an extension of the Pythagorean theorem that works for an arbitrary angle, not just a right angle. – Jaap Scherphuis Dec 21 '18 at 15:19
• Thanks a lot! That was really helpful! – Indula Dec 21 '18 at 15:25

Since the trapezium is right, base length of smaller $$\triangle$$ by Pythagoras theorem is $$\sqrt{4^2-3^2}=\sqrt{7}$$.

So remnant of the base of smaller trapezium is $$6-\sqrt{7}$$.

Clearly $$f=\frac83$$ by similarity of triangles.

Now finding the base of bigger $$\triangle$$ (say $$E$$) by similarity of triangles: $$\frac{E}{\sqrt{7}}=\frac{4+\frac83}{4}$$ or $$E=\frac{5\sqrt7}{3}$$

So that, $$e=6-\sqrt{7}+\frac{5\sqrt7}{3}=6+\frac{2\sqrt7}{3}$$.