Question

Which set of numbers could represent the lengths of the sides of a right triangle?

Answer 1

Answer:

The first set: 8, 15, and 17.

Step-by-step explanation:

Pair: 8, 15, 17

By the pythagorean theorem, a triangle is a right triangle if and only if

$\text{longest side}^2 = \text{first shorter side}^2 + \text{second shorter side}^2$.

In this case,

$\text{longest side}^2 = 17^2 = 289$.

$\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= 8^2 + 15^2\\ &=64 + 225 = 289 \end{aligned}$.

In other words, indeed $\text{hypotenuse}^2 = \text{first leg}^2 + \text{second leg}^2$. Hence, 8, 15, 17 does form a right triangle.

Similarly, check the other pairs. Keep in mind that the square of the longest side should be equal to the sum of the square of the two

Pair: 10, 15, 20

Factor out the common factor $2$ to simplify the calculations.

$\text{longest side}^2 = 20^2 = 400$

$\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= 10^2 + 15^2\\ &=100 + 225 = 325 \end{aligned}$.

$\text{longest side}^2 \ne \text{first shorter side}^2 + \text{second shorter side}^2$.

Hence, by the pythagorean theorem, these three sides don't form a right triangle.

Pair: 12, 18, 22

$\text{longest side}^2 = (2\times 11)^2 = 2^2 \times 121$.

$\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= (2 \times 6)^2 + (2 \times 9)^2\\ &=2^2 \times(36 + 81) = 2^2 \times 117 \end{aligned}$.

$\text{longest side}^2 \ne \text{first shorter side}^2 + \text{second shorter side}^2$.

Hence, by the pythagorean theorem, these three sides don't form a right triangle.

Pair: 7, 9, 11

$\text{longest side}^2 = 11^2 = 121$.

$\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= 7^2 + 9^2\\ &=49+ 81 = 130 \end{aligned}$.

$\text{longest side}^2 \ne \text{first shorter side}^2 + \text{second shorter side}^2$.

Hence, by the pythagorean theorem, these three sides don't form a right triangle.