Pythagorean Galore

Geometry Level 2

A right triangle has two sides of length $6$ and $8$ . The sum of all possible areas of the right triangle can be expressed as $a+b\sqrt{c}$ , where $a$ , $b$ , and $c$ are integers with $c$ being square-free. Find $100a+10b+c$ .

The answer is 2467.

2 solutions

Utsav Playz
Feb 3, 2021

Notice that these two right triangles gives us all the possible areas. Thus, the sum of the areas of those 2 triangles is

$= \dfrac{1}{2} (6) (8)$ $+$ $\dfrac{1}{2} (6) (2\sqrt7)$ $=$ $24 + 6\sqrt7$

Therefore, $100a + 10b + c$ $=$ $2400 + 60 + 7 = \boxed{2467}$

Ephram Chun
Jan 27, 2021

Either $6$ and $8$ are the legs in which case the hypotenuse is $10$ so the area is $24$ or $8$ is the hypotenuse and $6$ is the leg in which case the other leg is $\sqrt{28}=2\sqrt{7}$ thus the area of the triangle is $6\sqrt{7}$ the sum is $24+6\sqrt{7}$ so our answer is $2400+60+7=\boxed{2467}$

Since you didn't require $c$ to be square-free, $24+3\sqrt 28$ is also a correct answer.

Fletcher Mattox - 4 months, 2 weeks ago

Yes, you have to mention $c$ is square-free. You need to mention $a$ , $b$ , and $c$ are integers, if not there are infinite possibility. I have amended the problem for you.

Chew-Seong Cheong - 4 months, 2 weeks ago

Pythagorean Galore

The answer is 2467.

2 solutions

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