# Thread: Help with Geometry math challange

1. ## Help with Geometry math challange

Hi Guys,

I'm working with my 10 year old son for his 11+ entrance exams and a bit lost.

Q: The area of a right angled triangle is 32cm squared.

The length, in cm of the two shorter sides are both different square numbers bigger than 1.

what are the lengths of the two shorter sides?

2. ## Re: Help with Geometry math challange

4 and 16

The two shorter sides are the perpendicular legs of the right triangle, call them base and height

$\dfrac{1}{2}bh = 32$

$bh = 64$

Note that $64=2^6 =2^2 \cdot 2^4 = 2^2 \cdot 4^2$

3. ## Re: Help with Geometry math challange

Let's call the two sides A and B. Given that the area is 32 cm^2, then (1/2)AB = 32. Rearrange: B = 64/A.

Now it's a matter of trial and error. Since you know A is a square number, it must be 1, or 4, or 9, or 16 etc. Try substituting these various values for A into the above equation, and see which one(s) yield a value for B that is a square number. You only need to try a few.

Post back with your results and we'll check it for you.