A triangle has vertices A, B, and C. Vertex A has an angle of #pi/2 #, vertex B has an angle of #( pi)/4 #, and the triangle's area is #16 #. What is the area of the triangle's incircle?

Question

A triangle has vertices A, B, and C.  Vertex A has an angle of #pi/2 #, vertex B has an angle of #( pi)/4  #, and the triangle's area is #16 #.  What is the area of the triangle's incircle?

Claire Akin · Accepted Answer

#color(crimson)("Area of inscribed circle " = A_i = pi r^2 = 8.6186#

#"Area of "Delta = A_t = (1/2) a b sin C = (1/2) b c sin A = (1/2) c a sin B#

#"Given " hat A = pi/2, hat B = pi/4, hat C = pi/4, A_t = 16#

It's an isosceles right triangle.

#a b = (2 A_t) / sin C = 32 / sin (pi/4) = 45.25#

#b c = (2 A_t) / sin A = 32 / sin (pi/2) = 32#

#c a = (2 A_t) / sin B = 32 / sin (pi/4) = 45.25#

#a = (a b c) / (b c) = sqrt(45.25 * 45.25 * 32) / 32 = 8#

#b = (a b c) / (c a) = sqrt(45.25 * 45.25 * 32) / 45.25 = 5.66#

#c = (a b c) / (a b) = sqrt(45.25 * 45.25 * 32) / 45.25 = 5.66#

#"Semi-perimeter " = s = (a + b + c) / 2 = 19.32 / 2 = 9.66#

#"Radius of inscribed circle " = r = A_t / s = 16 / 9.66 = 1.66#

#"Area of inscribed circle " = A_i = pi r^2 = pi * 1.66^2 = 8.6186#

Maverick Smith · Answer

undefined To find the area of the triangle's incircle, we can use the formula $A = rs$, where $A$ is the area of the triangle, $r$ is the radius of the incircle, and $s$ is the semi-perimeter of the triangle.

The semi-perimeter of the triangle $s$ can be calculated as $s = \frac{a + b + c}{2}$, where $a$, $b$, and $c$ are the lengths of the sides of the triangle.

Given that vertex $A$ has a right angle ($\pi/2$), we can deduce that side $c$ is the hypotenuse. Using the Pythagorean theorem, we can find the lengths of sides $a$ and $b$ based on the angles given.

Let's denote side $a$ as the side opposite the angle $\pi/4$ (vertex $B$), and side $b$ as the remaining side. Using the ratios of a $45^\circ$-$45^\circ$-$90^\circ$ triangle, we find that $a = b$.

Now, let's find the lengths of sides $a$ and $c$. Since the area of the triangle is given as 16, we can use the formula for the area of a triangle: $A = \frac{1}{2}ab$. Substituting $a = b$ and $c$ (the hypotenuse), we have $16 = \frac{1}{2}(a^2)$. Solving for $a$, we get $a = b = 4\sqrt{2}$.

Using these side lengths, we can find the semi-perimeter $s$ of the triangle.

Then, knowing the area of the triangle and its semi-perimeter, we can solve for the radius $r$ of the incircle.

Finally, we can find the area of the incircle using the formula for the area of a circle: $A_{\text{incircle}} = \pi r^2$.