Question

2. A piece of property is in the shape of the right triangle. The longer leg is 10 cm shorter than twice the length of theshorter leg. The hypotenuse is 20 cm longer than the longer leg. Find the lengths of the sides of the property.

Answer 1

Let's call the longer leg "a", the shorter leg "b" and the hypotenuse "c".

So, if "a" is 10 cm shorter than twice the length of "b", we have:

$\begin{gathered} a=2b-10 \\ 2b=a+10 \\ b=\frac{a}{2}+5 \end{gathered}$

And if "c" is 20 cm longer than "a", than:

$c=a+20$

We also know the Pythagorean Theorem to be:

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

We can substitute "b" and "c" into it to find out "a":

$\begin{gathered} (a+20)^2=a^2+(\frac{a}{2}+5)^2 \\ a^2+40a+400=a^2+\frac{a^2}{4}+5a+25 \\ a^2-a^2-\frac{a^2}{4}+40a-5a+400-25=0 \\ -\frac{a^2}{4}+35a+375=0 \end{gathered}$

Now we have to use Bhaskara's Formula to find a, so:

$\begin{gathered} a=\frac{-35\pm\sqrt[]{35^2-4\cdot(-\frac{1}{4})\cdot375}}{2\cdot(-\frac{1}{4})}=\frac{-35\pm\sqrt[]{1225+375}}{-\frac{1}{2}}=-2(-35\pm\sqrt[]{1600})=-2(-35\pm40) \\ a_1=-2(5)=-10 \\ a_2=-2(-75)=150 \end{gathered}$

Since we can't have a negative lentgh, we have a = 150 cm.

Now we use the equations to find "b" and "c":

$b=\frac{a}{2}+5=\frac{150}{2}+5=75+5=80$$c=a+20=150+20=170$

Finally, we have a = 150 cm, b = 80 cm and c = 170 cm.