Question

The hypotenuse of a right triangle is 1 meter longer than twice the length of one leg. If the other leg is measures 3 meters find the lengths of each unknown side to the nearest tenth of a meter.

Answer 1

Answer:

$AC= h = 3.2\ m$  

$AB= b = 1.1\ m$

$BC= p = 3\ m$

Step-by-step explanation:

Let x be the length of the one side of a right angle triangle.

Given:

The hypotenuse of a right triangle is 1 meter longer than twice the length of one leg, and the length of the other leg measure 3 meters.

Length of one leg b = x

So length of the hypotenuse h = 2x + 1

Length of other leg p = 3 m

We need to find the length of each unknown side.

Solution.

From the figure h is the length of the hypotenuse AC and b, p is length of the legs AB and BC

Using Pythagoras theorem.

$(AC)^{2} = (AB)^{2}+(BC)^{2}$

$h^{2} = (b)^{2}+(p)^{2}$

Substitute all given value in above equation and then simplify.

$(2x+1)^{2} = x^{2}+(3)^{2}$

$4x^{2}+1+4x= x^{2}+9$

$4x^{2}-x^{2}-9+1+4x= 0$

$3x^{2}+4x-8 = 0$

Now, we first find the root of the above equation.

Use quadratic formula with $a=3, b=4, c=-8$.

$x=\frac{-b\pm \sqrt{(b)^{2}-4ac}}{2a}$

Put a, b and c value in above equation.

$x=\frac{-4\pm \sqrt{(4)^{2}-4(3)(-8)}}{2(3)}$

$x=\frac{-4\pm \sqrt{16+96}}{6}$

$x=\frac{-4\pm \sqrt{112}}{6}$

$x=\frac{-4\pm 4\sqrt{7}}{6}$

For positive sign

x = 1.1 m

So the length of the hypotenuse $h = 2x+1$  

$h = 2\times 1.1+1$  

$h = 2.2+1$

$h = 3.2\ m$

Therefore, the length of the each side of the right angle triangle is given below.

$AC= h = 3.2\ m$  

$AB= b = 1.1\ m$

$BC= p = 3\ m$