Given all angles and sides

Geometry Level 3

Let $\triangle ABC$ be a right triangle with a hypotenuse of length $12$ and an angle $\theta$ such that $\tan{\theta} = \dfrac{4}{3}$ . If $A(\triangle)+P(\triangle) = \dfrac{a}{b}$ , where $A(\triangle)$ is the area of the triangle, $P(\triangle)$ is the perimeter of the triangle, and $a$ and $b$ are coprime positive integers, find $a+b$ .

The answer is 1609.

2 solutions

Edwin Gray
Feb 7, 2019

The sides of the triangle must be 12, 3x, and 4x. By the Pythagorean Theorem, 9x^2 + 16x^2 = 144 = 25x^2, so x = 12/5, The sides are then 36/5, 48/5, and 60/5. The area = (1/2) (36/5) (48/5) = 1728/50. The perimeter = (36 + 48 + 60)/5 = 1440/50. The sum = 3168/50 = 1584/25. Since (1584,25) = 1, a + b = 1609. Ed Gray

Lucas Chaves Meyles
Feb 23, 2018

Because $\triangle ABC$ has a hypotenuse of $12$ and $\tan \theta = \frac{4}{3}$ , its legs must equal $\frac{48}{5}$ and $\frac{36}{5}$ . Thus, $A(\triangle) = \big(\frac{48}{5}\big)\big(\frac{36}{5}\big)\big(\frac{1}{2}\big) = \frac{864}{25}$ and $P(\triangle) = \frac{60}{5} + \frac{48}{5} + \frac{36}{5} = \frac{144}{5}$ . Then, $A(\triangle) + P(\triangle) = \frac{864}{25} + \frac{144}{5} = \frac{1584}{25}$ . Our answer is $a + b = 1584 + 25 = \boxed{1609}$

Given all angles and sides

The answer is 1609.

2 solutions

0 pending reports