# A right triangle has one leg that is 5 cm longer than the other leg and the hypotenuse is 25 cm long. How do you find the length of each leg?

Feb 26, 2016

No such triangle exists.

#### Explanation:

If we denote the shorter leg as $x$ then
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + {\left(x + 5\right)}^{2} = {25}^{2}$

$\rightarrow \textcolor{w h i t e}{\text{XXXX}} 2 {x}^{2} + 10 x = 0$

$\rightarrow \textcolor{w h i t e}{\text{XXXX}} x \left(x + 10\right) = 0$

$\rightarrow \textcolor{w h i t e}{\text{XXXX}} x = 0 \mathmr{and} x = - 10$

Since neither of these are possible for any real triangle,
no triangle exists that meets the given conditions.

Feb 26, 2016

15cm and 20cm.

#### Explanation:

Since it is a right triangle use$\textcolor{b l u e}{\text{ Pythagoras's Theorem}}$

which states 'in a right triangle the square on the hypotenuse is equal to the sum of the squares on the other two sides'

This can be written as an equation. If c is the hypotenuse and a and b are the other 2 sides ( the legs ) then

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

Here , let one of the legs be x , so the other is x + 5

then ${25}^{2} = {x}^{2} + {\left(x + 5\right)}^{2} \text{ using theorem }$

now distribute brackets and collect like terms. Also going to reverse the equation so x terms are on the left side.

hence : ${x}^{2} + {x}^{2} + 10 x + 25 = 625$

so $2 {x}^{2} + 10 x + 25 = 625$

this is a quadratic equation , hence equate to zero to solve.
$2 {x}^{2} + 10 x - 600 = 0$
common factor of 2 :$2 \left({x}^{2} + 5 x - 300\right) = 0$

Require factors of -300 which sum to 5 (coefficient of x term). These are +20 and - 15 . If unsure use quadratic formula to obtain them.

$\Rightarrow 2 \left(x + 20\right) \left(x - 15\right) = 0$

the solution is x = - 20 or x = 15 : x > 0 thus x = 15

The 2 legs are x and x+5 hence 15 and 20.