Right traingles C and D share a leg. If the ratio of the area of traingle C to the area of traingle D is 5:9, C’s hypotenuse is of length 25, and D’s hypotenuse is of length 39, what is the area of C?
For the non-shared legs,
let the shorter be x and the longer be y
let the length of the shared leg be h
so x^2 + h^2 = 25^2 and y^2 + h^2 = 39^2
h^2 = 625-x^2 and h^2 = 1521 - y^2
625 - x^2 = 1521 - y^2
y^2 - x^2 = 896
Also, (1/2)xh /(1/2)yh = 5/9
x/y = 5/9
9x = 5y
x = 5y/9
y^2 - 25y^2/81 = 896
times 81
81y^2 - 25y^2 = 72576
56y^2 = 72576
y^2 = 1296
y = ± 36 , rejecting the negative, since we can't have a negative length of a line
y= 36, then in x=5y/9 -------->
take it from here
To find the area of triangle C, we first need to find the length of the leg it shares with triangle D.
Let's assume the shared leg of both triangles C and D is x. Since the ratio of the areas of triangle C to triangle D is 5:9, the ratio of their corresponding sides (legs) is the square root of the area ratio, which is $\sqrt{\frac{5}{9}}$.
So, the length of the leg of triangle D can be found by multiplying the length of the shared leg (x) by $\sqrt{\frac{5}{9}}$.
Therefore, the length of the leg of triangle D = x * $\sqrt{\frac{5}{9}}$.
Given that the hypotenuse of triangle C is 25, we can use the Pythagorean theorem to find the other leg of triangle C. The Pythagorean theorem states that in a right triangle:
$a^2 + b^2 = c^2$,
where a and b are the legs, and c is the hypotenuse.
Let's assume the other leg of triangle C is y. Using the Pythagorean theorem, we have:
$y^2 + x^2 = 25^2$.
Now, let's find the length of the leg of triangle D. We know that the hypotenuse of triangle D is 39 and the shared leg is x, so using the Pythagorean theorem, we have:
$x^2 + (x * \sqrt{\frac{5}{9}})^2 = 39^2$.
Simplifying this equation, we get:
$x^2 + \frac{5}{9}x^2 = 39^2$.
Combining the terms, we have:
$\frac{14}{9}x^2 = 39^2$.
Now, we can solve for x by dividing both sides by $\frac{14}{9}$:
$x^2 = \frac{39^2}{\frac{14}{9}}$.
$x^2 = 3^2 * \frac{3^4}{2^2}$.
$x^2 = 3^2 * (\frac{3^2}{2})^2$.
$x^2 = 3^2 * (\frac{9}{2})^2$.
$x^2 = 3^2 * (\frac{81}{4})$.
$x^2 = 3^2 * \frac{9 \times 9}{4}$.
$x^2 = 3^2 * 9 \times \frac{9}{4}$.
$x^2 = 3^2 * 3^2 * \frac{9}{4}$.
$x^2 = (3 \times 3 \times \frac{3}{2})^2$.
$x^2 = 9 \times (\frac{9}{2})^2$.
$x^2 = 9 \times (\frac{81}{4})$.
$x^2 = 9 \times \frac{81}{4}$.
$x^2 = \frac{9 \times 81}{4}$.
$x^2 = \frac{729}{4}$.
Taking the square root of both sides, we have:
$x = \frac{\sqrt{729}}{\sqrt{4}}$.
$x = \frac{27}{2}$.
Therefore, the length of the shared leg, x, is $\frac{27}{2}$.
Now, let's find the area of triangle C using the length of the shared leg, x, and the other leg, y:
Area of triangle C = 1/2 * x * y.
Area of triangle C = 1/2 * $\frac{27}{2}$ * y.
Area of triangle C = $\frac{27y}{4}$.
We need to find the value of y.
Using the Pythagorean theorem for triangle C, we have:
$y^2 + (\frac{27}{2})^2 = 25^2$.
$y^2 + \frac{729}{4} = 625$.
$y^2 = 625 - \frac{729}{4}$.
$y^2 = \frac{2500}{4} - \frac{729}{4}$.
$y^2 = \frac{1771}{4}$.
Taking the square root of both sides, we have:
$y = \frac{\sqrt{1771}}{\sqrt{4}}$.
$y = \frac{\sqrt{1771}}{2}$.
Now that we have the value of y, we can find the area of triangle C:
Area of triangle C = $\frac{27y}{4}$.
Substituting the value of y, we have:
Area of triangle C = $\frac{27}{4} * \frac{\sqrt{1771}}{2}$.
Simplifying this, we get:
Area of triangle C = $\frac{27\sqrt{1771}}{8}$.
Therefore, the area of triangle C is $\frac{27\sqrt{1771}}{8}$.