Answer
The length of the other two sides of the triangle are 12 and 16.
Work Step by Step
We have to equations.
1. The sum of the lengths of the two sides is 28 cm.
$$a + b = 28$$
2. The hypotenuse of this right triangle has length 20 cm. As we know, the equation for two sides: a and b, and a hypotenuse is:
$$a^2 + b^2 = h^2$$ $$a^2 + b^2 = (20)^2 = 400$$
- Solve for a in the first equation, and substitute its value into the second equation.
$$a = 28 - b$$
$$(28 - b)^2 + b^2 = 400$$ $$28^2 - 56b + b^2 + b^2 = 400$$ $$784 - 400 - 56b + 2b^2 = 0$$ $$384 - 56b + 2b^2 = 0$$ $$b_1 = \frac{-(-56) + \sqrt{(-56)^2 - 4 \times 2 \times 384}}{2(2)}$$ $$b_1 = \frac{56 + \sqrt{3136 - 3072}}{4} = \frac{56 + \sqrt {64}}{4} = \frac{64}{4} = 16$$
$$b_2 = \frac{56 - \sqrt {64}}{4} = \frac{48}{4} = 12$$
- Calculate $a$:
$a_1 = 28 - b = 28 - 16 = 12$
$a_2 = 28 - b = 28 - 12 = 16$
Therefore, the length of the other two sides of the triangle are 12 and 16.
