# The hypotenuse of a right triangle is 17 cm long. Another side of the triangle is 7 cm longer than the third side. How do you find the unknown side lengths?

Jun 2, 2018

8 cm and 15 cm

#### Explanation:

Using the Pythagorean theorem we know that any right triangle with sides a, b and c the hypotenuse:

${a}^{2} + {b}^{2} = {c}^{2}$

$c = 17$

$a = x$

$b = x + 7$

${a}^{2} + {b}^{2} = {c}^{2}$

${x}^{2} + {\left(x + 7\right)}^{2} = {17}^{2}$

${x}^{2} + {x}^{2} + 14 x + 49 = 289$

$2 {x}^{2} + 14 x = 240$

${x}^{2} + 7 x - 120 = 0$

$\left(x + 15\right) \left(x - 8\right) = 0$

$x = - 15$

$x = 8$

obviously the length of a side cannot be negative so the unknown sides are:

$8$

and

$8 + 7 = 15$

Jun 2, 2018

$8 \text{ and } 15$

#### Explanation:

$\text{let the third side } = x$

$\text{then the other side "=x+7larrcolor(blue)"7 cm longer}$

$\text{using "color(blue)"Pythagoras' theorem}$

$\text{square on the hypotenuse "=" sum of squares of other sides}$

${\left(x + 7\right)}^{2} + {x}^{2} = {17}^{2}$

${x}^{2} + 14 x + 49 + {x}^{2} = 289$

$2 {x}^{2} + 14 x - 240 = 0 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\text{divide through by 2}$

${x}^{2} + 7 x - 120 = 0$

$\text{the factors of - 120 which sum to + 7 are + 15 and - 8}$

$\left(x + 15\right) \left(x - 8\right) = 0$

$\text{equate each factor to zero and solve for x}$

$x + 15 = 0 \Rightarrow x = - 15$

$x - 8 = 0 \Rightarrow x = 8$

$x > 0 \Rightarrow x = 8$

$\text{lengths of unknown sides are}$

$x = 8 \text{ and } x + 7 = 8 + 7 = 15$