# The hypotenuse of a right triangle is 6.1 units long. The longer leg is 4.9 units longer than the shorter leg. How do you find the lengths of the sides of the triangle?

Jan 13, 2016

The sides are
color(blue)(1.1 cm  and color(green)( 6cm

#### Explanation:

The hypotenuse: $\textcolor{b l u e}{A B} = 6.1$ cm (assuming length to be in cm)

Let the shorter leg: $\textcolor{b l u e}{B C} = x$ cm

Let the longer leg: $\textcolor{b l u e}{C A} = \left(x + 4.9\right)$ cm

As per Pythagoras Theorem :

${\left(A B\right)}^{2} = {\left(B C\right)}^{2} + {\left(C A\right)}^{2}$

${\left(6.1\right)}^{2} = {\left(x\right)}^{2} + {\left(x + 4.9\right)}^{2}$

37.21 = (x)^2 +color(green)( (x+4.9)^2

Applying the below property to  color(green)( (x+4.9)^2 :
color(blue)((a+b)^2 = a^2 +2ab +b^2

$37.21 = {\left(x\right)}^{2} + \left[\textcolor{g r e e n}{{x}^{2} + 2 \times x \times 4.9 + 24.01}\right]$

$37.21 = {\left(x\right)}^{2} + \left[\textcolor{g r e e n}{{x}^{2} + 9.8 x + 24.01}\right]$

$37.21 = 2 {x}^{2} + 9.8 x + 24.01$

$13.2 = 2 {x}^{2} + 9.8 x$

$2 {x}^{2} + 9.8 x - 13.2 = 0$

Multiplying the entire equation by $10$ to remove the decimal

$20 {x}^{2} + 98 x - 132 = 0$

Dividing the entire equation by $2$ for simplicity
$10 {x}^{2} + 49 x - 66 = 0$

The equation is now of the form color(blue)(ax^2+bx+c=0 where:
$a = 10 , b = 49 , c = - 66$

The Discriminant is given by:
$\Delta = {b}^{2} - 4 \cdot a \cdot c$
$= {\left(49\right)}^{2} - \left(4 \cdot \left(10\right) \cdot \left(- 66\right)\right)$
$= 2401 + 2640 = 5041$

The solutions are found using the formula
$x = \frac{- b \pm \sqrt{\Delta}}{2 \cdot a}$

$x = \frac{\left(- 49\right) \pm \sqrt{5041}}{2 \cdot 10} = \frac{- 49 \pm \left(71\right)}{20}$

$x = = \frac{- 49 + \left(71\right)}{20} = \frac{22}{20} = 1.1$

$x = = \frac{- 49 - \left(71\right)}{20}$ (not applicable since side cannot be negative)

So, the shorter side color(blue)(x=1.1 cm

The longer side = color(blue)(x +4.9 = 6cm