Question

The hypotenuse of a triangle is one foot more than twice the length of the shorter leg. The longer leg is seven feet longer than the shorter leg. Find the dimensions of the triangle

Answer 1

Start by calling the shortest side $s$. We then know that the longer leg (call it $l$) is 7 feet longer i.e. $l=s+7$. We also know that the hypotenuse, $h$ is 1 foot more than twice the short leg i.e. $h=2s+1$. We now have 3 unknowns but only 2 equations to solve them. Luckily we know this is a right triangle so we can use the Pythagorean Theorem as our third equation:
$a^{2}+b^{2}=c^{2}$
Substituting known values:
$s^{2}+(s+7)^{2}=(2s+1)^{2}$
This is now an equation in one variable and can be solved algebraically for $s$. Back-substitution can then be used for the other sides. 

Answer 2

The length of the sides of a triangle is required.

The shortest side is 8 units, the longer side is 15 units and hypotenuse is 17 units.

a = Shorter side

b = Longer side = a+7

c = Hypotenuse = 2a+1

From the Pythagoras theorem

$a^2+b^2=c^2\\\Rightarrow a^2+(a+7)^2=(2a+1)^2\\\Rightarrow a^2+a^2+49+14a=4a^2+1+4a\\\Rightarrow 2a^2-10a-48=0\\\Rightarrow a^2-5a-24=0\\\Rightarrow a=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\times 1\times\left(-24\right)}}{2\times 1}\\\Rightarrow a=8,-3$

$b=a+7=8+7\\\Rightarrow b=15$

$c=2a+1=2\times 8+1\\\Rightarrow c=17$

The shortest side is 8 units, the longer side is 15 units and hypotenuse is 17 units.

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