You may also like

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

Building Tetrahedra

Can you make a tetrahedron whose faces all have the same perimeter?

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: 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