#### You may also like ### Some(?) of the Parts

A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle ### Triangle Midpoints

You are only given the three midpoints of the sides of a triangle. How can you construct the original triangle? ### Fermat's Poser

Find the point whose sum of distances from the vertices (corners) of a given triangle is a minimum.

# Slide Height

##### Age 14 to 16 Short Challenge Level:
Using area
On this diagram, the height we are looking for is labelled $x$ m. There are two ways to view area of the triangle. One is that it has a base of 12 and perpendicular height of 5, and so its area is $\frac{1}{2}\times5\times12 =30$m$^2$.
The other way we can view it is with a base of 13 and perpendicular height of $x$ and so it also has area $\frac{1}{2}\times13x$. We can equate these two areas to get that:
$\frac{13}{2}x = 30 \Rightarrow x=\frac{30\times2}{13}=4.62$m to the nearest cm

Using similar triangles
The height must make a right angle with the ground, which is shown in the diagram below. Also on the diagram, the red angle, the blue angle and a right angle must add up to 180$^{\text{o}}$, because the angles in a triangle add up to 180$^{\text{o}}$, and they are the angles in the triangle made by the slide and the ground. The triangle outlined in gold in the diagram below is also a right-angled triangle. It contains the blue angle and a right angle, so, to add up to 180$^{\text{o}}$, the third angle must be the same as the red angle. This means that the triangle made by the slide and the ground must be similar to the golden triangle.

The hypotenuse of the triangle made by the slide and the ground is 13 m, and the hypotenuse of the golden triangle is 12 m. So the sale factor between the two triangles is $\frac{12}{13}$.

The height, $x$ m, corresponds to the 5 m side of the larger triangle (the ladder). So the height is $\dfrac{12}{13}$ of 5, which is 4.61538..., which is 4.62 m to the nearest cm.
You can find more short problems, arranged by curriculum topic, in our short problems collection.