IMC 2016 Q18
18. The sum of the areas of the squares on the sides of a right-angled isosceles triangle is 72 cm2.
What is the area of the triangle?
- A 6 cm2
- B 8 cm2
- C 9 cm2
- D 12 cm2
- E 18 cm2
Intermediate Mathematical Challenge 2016, Question 18. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: C 9 cm2
Let the length of each of the shorter sides of the triangle be \(x\) cm and the length of its hypotenuse be \(y\) cm. Then, by Pythagoras' Theorem: \(x^2 + x^2 = y^2\). So \(y^2 = 2x^2\). Also, \(x^2 + x^2 + y^2 = 72\), so \(4x^2 = 72\), that is \(x^2 = 18\). Now the area of the triangle, in cm2, is \(\frac{1}{2} \times x \times x = \frac{1}{2}x^2 = 9\).
Solution © UK Mathematics Trust, from the IMC 2016 solutions. Longer versions with investigations: Solutions and investigations (PDF).