### 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?

# Shared Vertex

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

Triangle $BCD$ is isosceles, as $BC=BD$. Therefore, $\angle BCD = \angle BDC = 65^\circ$.

Angles in a triangle add up to $180^\circ$, so applying this in $BCD$ gives:
$\angle CBD = 180^\circ - \angle BCD - \angle BDC = 180^\circ - 65^\circ - 65^\circ = 50^\circ$.

$\angle CBD$ and $\angle ABE$ are opposite angles at $B$, so are equal. Therefore $\angle ABE = 50^\circ$.

Since the angles in triangle $ABE$ add up to $180^\circ$, $x = 180^\circ - \angle BEA - \angle ABE = 180^\circ - 90^\circ - 50^\circ = 40^\circ$.

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