A right triangle with integer leg lengths is called cool if the number of square units in its area is equal to five times the number of units in the sum of the lengths of its legs. What is the sum of all the different possible areas of cool right triangles?

learnmgcat Jun 12, 2024

#1**0 **

We an represent this problem as \(\frac{1}{2}ab = 5(a+b)\), where a and b are the two legs. We can now simplify this equation further.

\(\frac{1}{2}ab = 5a+5b\)

\(ab=10a+10b\)

\(ab-10a-10b=0\)

We can now use Simon's Favourite Factoring Trick to simplify this.

\(ab-10a-10b+100=100\)

\((a-10)(b-10)=100\)

The factor pairs of 100 are (1, 100), (2,50), (4, 25), (5, 20), and (10, 10). We now choose a and b to make thepart in the parentheses equal to one of these pairs. Therefore, all possibilities for the legs (a and b) are (11, 110), (12, 60), (14, 35), (15, 30), and (20, 20). We can calculate the area for each of these.

\(\frac{1}{2}(11)(110)=605\)

\(\frac{1}{2}(12)(60)=360\)

\(\frac{1}{2}(14)(35)=245\)

\(\frac{1}{2}(15)(30)=225\)

\(\frac{1}{2}(20)(20)=200\)

Therefore, the sum of all areas is 605 + 360 + 245 + 225 + 200 = **1635**

Maxematics Jun 12, 2024