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# Help. Thanks.

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A triangle has side lengths of $10,$ $24,$ and $26.$ Let $a$ be the area of the circumcircle. Let $b$ be the area of the incircle. Compute $a - b.$

May 9, 2020

#1
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Right scalene triangle.
Sides: a = 10 b = 24 c = 26

Area: T = 120
Perimeter: p = 60
Semiperimeter: s = 30

Angle ∠ A = α = 22.62° = 22°37'11″ = 0.395 rad
Angle ∠ B = β = 67.38° = 67°22'49″ = 1.176 rad
Angle ∠ C = γ = 90° = 1.571 rad

Height: ha = 24
Height: hb = 10
Height: hc = 9.231

Median: ma = 24.515
Median: mb = 15.62
Median: mc = 13

Area of circumcircle =13^2 x pi
Area of incircle =4^2 x pi
[13^2 x pi] - [4^2 x pi] =153pi

May 9, 2020
#2
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Guest May 10, 2020
#3
+109519
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And that is not how you show any credit for a person's effort.

You can at least say thanks for the effort and explain why you believe it to be wrong.

In fact, you could go through it line by line and explain which line/s you believe to be incorrect.

Melody  May 10, 2020
#5
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It is. 153 pi is. No explanation for why it is wrong was given, yet great explanations for how it is correct are given.

Guest May 10, 2020
#4
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Yes, that is the correct answer.

My explaination is

we know that Triangle ABC is a right triangle because we know the Pythagorean Theorem's converse is true.

The hypotenuse of a right triangle is a diameter of the triangle's circumcircle.

Therefore, the area of the circumcircle shoud be 13^2*pi = 169pi

To find the incircle's area, we see $[ABC] = (10)(24)/2 = 120$

and $s = (10+24+26)/2 = 30$, so $r = [ABC]/s = 120/30 = 4$.

When $[ABC] = rs$, where $r$ is the inradius and $s$ is the semiperimeter of the triangle.

Therefore, the area of the incircle is $4^2\pi = 16\pi$.Finally, the area of the circumcircle is $169\pi - 16\pi = \boxed{153\pi}$

So whoever discredited the guest who got the correct answer with a great explanation, is wrong, and

Melody is absolutly correct although that guest may not be the same guest who asked the problem. :p

So yes, the answer should be 153pi.

May 10, 2020
edited by Guest  May 10, 2020