Video: AQA GCSE Mathematics Higher Tier Pack 5 β€’ Paper 2 β€’ Question 24

A) The converse to Pythagoras’s theorem states that a triangle with lengths π‘Ž, 𝑏, and 𝑐, where 𝑐 is the hypotenuse, is a right-angled triangle if π‘ŽΒ² + 𝑏² = 𝑐². For example, a triangle with lengths 3, 4, and 5 must be right-angled because 3Β² + 4Β² = 5Β². Prove that the triangle with lengths 50.1, 120, and 130.1 is not right-angled. B) A Pythagorean triple is a set of three positive integers that form a right-angled triangle. For example, 3, 4, 5 is a Pythagorean triple. Michael notices that 3π‘˜, 4π‘˜, 5π‘˜ for any positive integer π‘˜ is also a Pythagorean triple. Joe says that the same is true for 3 + π‘˜, 4 + π‘˜, and 5 + π‘˜. Find a counterexample to show that Joe is wrong.

02:48

Video Transcript

The converse to Pythagoras’s theorem states that a triangle with lengths π‘Ž, 𝑏, and 𝑐, where 𝑐 is the hypotenuse, is a right-angled triangle if π‘Ž squared plus 𝑏 squared equals 𝑐 squared. For example, a triangle with lengths three, four, and five must be right-angled because three squared plus four squared equals five squared. Prove that the triangle with lengths 50.1, 120, and 130.1 is not right-angled.

Part b) A Pythagorean triple is a set of three positive integers that form a right-angled triangle. For example, three, four, five is a Pythagorean triple. Michael notices that three π‘˜, four π‘˜, five π‘˜ for any positive integer π‘˜ is also a Pythagorean triple. Joe says that the same is true for three plus π‘˜, four plus π‘˜, and five plus π‘˜. Find a counterexample to show that Joe is wrong.

Part a) If the triangle is right-angled, it will satisfy Pythagoras’s theorem, as the example given β€” three, four, five β€” does. We need to show then if we substitute these values into the equation for Pythagoras’s theorem, the values do not satisfy the equation.

Remember, the hypotenuse is the longest side. So 𝑐 is 130.1 in this case. Let’s begin by squaring the values of π‘Ž and 𝑏. They are 50.1 and 120. 50.1 squared plus 120 squared is equal to 16910.01. Next, we’re going to square the value of 𝑐, which we said was 130.1. That’s 16926.01. The value for π‘Ž squared plus 𝑏 squared is not equal to the value for 𝑐 squared. So this triangle cannot be right-angled.

For part b), the clue here is in the word β€œcounterexample.” A counterexample is any single example that proves the statement is incorrect. Remember, to prove something is true, it’s important to show that it’s true for every single value. But to disprove a statement, we just need to find one example where it’s not true.

A sensible starting choice here is to choose a low value of π‘˜. Let’s try one. Then three plus π‘˜ is equal to four, four plus π‘˜ is equal to five, and five plus π‘˜ is equal to six. Once again, the longest side is the hypotenuse. So six is equal to 𝑐.

We’re going to work out the values of π‘Ž squared plus 𝑏 squared. Here that’s four squared plus five squared, which is equal to 41. 𝑐 squared is six squared, and that’s 36. We can say that 41 is not equal to 36. And we’ve shown that Joe is wrong. When π‘˜ is equal to one, three plus π‘˜, four plus π‘˜, and five plus π‘˜ do not form a Pythagorean triple.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.