is there a formula for this?

[englsinger_J]

A problem states that the hypotenuse of a right triangle is 12 inches and the area is 24 square inches. find the dimensions of the triangle, correct to one decimal point.

is their a formula I could use to solve this? I tried solving it by using the pythagorean therom and finding the dimensions of the base and height by also using 1/2 bh formula. with this method I got that b is about 4.3 and h is about 11.3.

is there a formula to solve this easier? If someone knows could they tell me without actually telling me the answer.
Thanks

 

[stapel]

Without seeing your work, it is difficult to say whether there might be an easier way. I'm not familiar with any "formula" for this. I would just set up a system of (non-linear) equations, using the area formula and the Pythagorean Formula:

. . . . .A = (1/2)bh = 24

. . . . .b² + h² = (12)²

Then solve for the base "b" and the height "h".

Eliz.

 

[englsinger_J]

thanks

 

[Unco]

A problem states that the hypotenuse of a right triangle is 12 inches and the area is 24 square inches. find the dimensions of the triangle, correct to one decimal point.

is their a formula I could use to solve this? I tried solving it by using the pythagorean therom and finding the dimensions of the base and height by also using 1/2 bh formula. with this method I got that b is about 4.3 and h is about 11.3.

is there a formula to solve this easier? If someone knows could they tell me without actually telling me the answer.
Thanks

Your dimensions are correct (just watch the rounding with 11.21..).

I wonder from the references to a formula as to whether you used technology to solve it.

Substituting b in terms of h from Eliz's first equation into her second gives

\(\displaystyle \mbox{ \frac{2304}{h^2} + h^2 = 144}\)

Multiply both sides by h², let u = h² and you have a quadratic to solve for u. Solve for h from those.