A triangle has vertices A, B, and C. Vertex A has an angle of #pi/2 #, vertex B has an angle of #( pi)/6 #, and the triangle's area is #98 #. 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)/6  #, and the triangle's area is #98 #.  What is the area of the triangle's incircle?

Sadie Adams · Accepted Answer

#color(chocolate)("Area of incircle " = pi r^2 = pi * (3.89)^2 = 47.59#

#hat A = pi/2, hat B pi/6, hat C = pi/3, A_t = 98#

#A_t = (1/2) ab sin C = (1/2(bc sin A = (1/2) ac sin B#

#ab =( 2A_t) /sin C = (2 * 98) / sin (pi/3) = 226.32#

#bc = (2 * 98) / sin (pi/2) = 196#

#ca = (2 * 98) / sin (pi/6) = 392#
4169.97
#a = sqrt(abc)^2 / (bc) =sqrt (226.32 * 196 * 392) / 196 =21.28#

#b = sqrt(abc)^2 / (ac) =sqrt (226.32 * 196 * 392) / 392 =10.64#

#c = sqrt(abc)^2 / (ab) =sqrt (226.32 * 196 * 392) / 226.32 =18.43#

#"Semiperimeter " = s = (a = b + c) / 2 = (21.28 + 10.64 + 18.43) / 2 = 25.18#

#color(chocolate)("Incircle radius " = r = A_t / s = 98 / 25.18 = 3.89#

#color(chocolate)("Area of incircle " = pi r^2 = pi * (3.89)^2 = 47.59#

Parker Allen · Answer

undefined To find the area of the triangle's incircle, follow these steps:

1. Use the formula for the area of a triangle: $ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $.
2. Identify the base and height of the triangle.
3. Use the formula for the radius of the incircle of a triangle: $ r = \frac{\text{Area of the triangle}}{\text{Semi-perimeter of the triangle}} $.
4. Calculate the semi-perimeter of the triangle: $ s = \frac{a + b + c}{2} $, where $ a $, $ b $, and $ c $ are the lengths of the triangle's sides.
5. Calculate the radius $ r $ of the incircle using the formula from step 3.
6. Use the formula for the area of the incircle of a triangle: $ \text{Area of incircle} = \pi \times r^2 $.

Given that the triangle has a right angle at vertex A, we can use trigonometric relationships to find the lengths of the triangle's sides.

Let's denote the sides as follows:
- $ AB $ is the side opposite the angle $ \frac{\pi}{6} $.
- $ AC $ is the side opposite the right angle $ \frac{\pi}{2} $.
- $ BC $ is the hypotenuse.

From the given information, we know that the area of the triangle is 98. We also know that the base of the triangle is $ AB $ and the height is $ AC $.

Using trigonometric relationships in a $30^\circ$-$60^\circ$-$90^\circ$ triangle, we find that $ AB = AC\sqrt{3} $. Thus, the base and height are equal.

Using the area formula, we have:
$$ 98 = \frac{1}{2} \times AB \times AC $$

Since $ AB = AC\sqrt{3} $, we can rewrite the equation as:
$$ 98 = \frac{1}{2} \times (AC\sqrt{3}) \times AC $$

Solving for $ AC $, we get:
$$ AC^2 = \frac{98 \times 2}{\sqrt{3}} $$

Then, calculate the semi-perimeter $ s $ using the lengths of the sides $ AB $ and $ AC $.

Next, compute the radius $ r $ of the incircle using the formula mentioned above.

Finally, use the formula for the area of the incircle to find the answer.