You may also like

problem icon

Consecutive Numbers

An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.

problem icon

Golden Thoughts

Rectangle PQRS has X and Y on the edges. Triangles PQY, YRX and XSP have equal areas. Prove X and Y divide the sides of PQRS in the golden ratio.

problem icon

Ladder and Cube

A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?

Folding in Half

Stage: 3 and 4 Short Challenge Level: Challenge Level:1

Since the original triangle is isosceles and right-angled, folding it produces a smaller triangle, also isosceles and right-angled.

By Pythagoras' Theorem, the hypotenuse of the original triangle is $\sqrt{200}=10\sqrt{2}$ cm.
Hence the difference between the perimeters of the two triangles is $(10+10+10\sqrt{2})-(5\sqrt{2}+5\sqrt{2} +10)=10$ cm.
Alternatively: let the length of the shorter sides of the new triangle be x cm, shown below. Then the perimeter of the original triangle is $(20+2x)$ cm and the perimeter of the new triangle is $(10+2x)$cm. Hence the difference between the perimeters of the two triangles is $10$cm.

This problem is taken from the UKMT Mathematical Challenges.
View the archive of all weekly problems grouped by curriculum topic

View the previous week's solution
View the current weekly problem