### Right Time

At the time of writing the hour and minute hands of my clock are at right angles. How long will it be before they are at right angles again?

### Isosceles

Prove that a triangle with sides of length 5, 5 and 6 has the same area as a triangle with sides of length 5, 5 and 8. Find other pairs of non-congruent isosceles triangles which have equal areas.

Four rods, two of length a and two of length b, are linked to form a kite. The linkage is moveable so that the angles change. What is the maximum area of the kite?

##### Age 11 to 14 Short Challenge Level:

There are two different ways of calculating the angle $x$.

The first method is as follows:

The blue angle can be calculated since angles on a straight line add to $180^\circ$. This means it is $180^\circ - 100^\circ = 80^\circ$.

The orange angle can be calculated similarly, to be $180^\circ - 93^\circ = 87^\circ$.

The green angle can then be calculated since the angles in the top-left triangle add up to $180^\circ$. This is $180^\circ - 80^\circ - 58^\circ = 42^\circ$.

The green and red angles are opposite angles. This means that they are equal, so the red angle is also $42^\circ$.

Then, the angles in the left-hand triangle must add up to $180^\circ$, so $x^\circ = 180^\circ - 42^\circ - 87^\circ= 51^\circ$.

The other method involves looking at the farthest right angle in the diagram.

The angles in the red and purple triangle must add up to $180^\circ$, so the green angle is $180^\circ - 58^\circ - 93^\circ = 29^\circ$.

Then, the angles in the blue triangle must also add up to $180^\circ$. Therefore, $x^\circ = 180^\circ - 29^\circ - 100^\circ = 51^\circ$.

This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.