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:
2 Answers
$\begingroup$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.
$\endgroup$ 4 $\begingroup$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, $e=6-\sqrt{7}+\frac{5\sqrt7}{3}=6+\frac{2\sqrt7}{3}$.
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