How do you find all lengths of a triangle just by knowing the area?

Question

My problem is

The area of a right triangle is $24$ square feet. A second right triangle has a base that is $2$ times as long as the first triangle's base and a height that is $3$ times as long as the first triangle's height. What is the area of the second triangle?"

I drew it out to help but then I realized I need to know the lengths of the legs of the first triangle before I figure out the area of the second triangle.

Answer 1

Use algebra: $b = \text{base of first triangle}; h = \text{height of first triangle}$.

$\dfrac12\times b\times h = 24$

$B = \text{Second triangle base} = 2b. \\ H = \text{Second triangle height} = 3h.$

So area of second triangle is

$\dfrac12\times B\times H = \ldots\text{what?}$

Answer 2

Let $h$ and $b$ be the height and base length of the original triangle.

We know that $\frac{hb}{2} = 24$, and we know the area of the 2nd triangle is $\frac{3h2b}{2}$.

$\frac{hb}{2} = 24 \implies hb = 48$. If you replace $hb$ with $48$ in $\frac{3h2b}{2}$ then you'll find that the area is $144\ units^{2}$.

Answer 3

Actually, to answer you question, how do you figure out the legs from the area, the answer is, you can't.

$A = \frac12\times b\times h$ so $b = 2\times\frac Ah$ will yield an infinite number of possibilities.

$\frac12\times b\times h = 24$ for example can be solved by any $b = \frac{48}h$. Examples: $h=1, b=48; h=6, b=8; h = 5,000,000, b = .0000096$, etc.

To solve your problem you don't need to know what $a$ and $b$ are separately.