Yeah, It's Right

denote pts A , B , C , E , F as shown in the above diagram by the power of a point AF = AB and FE = DE . Since AB = 5 -r and DE = 12 - r . This implies AB + FE = 17 - 2r . But AB + FE = AE = 13 . Hence, 17 - 2r = 13 implying r = 2 .

Therefore , area of the inscribed circle = (2)^2 pi = 4pi

$\frac{b\times h}{2}=sr$

$\Rightarrow \frac{5\times12}{2}=\frac{12+5+13}{2} \times r \therefore r=2$ and the circle's area is $2^2\pi=\boxed{4\pi}$

5² + 12² = 13² so this is a right angled triangle.

a is the base, b is the height and c the hypothenuse.

ab / (a+b+c) = r

5 x 12 / (5+12+13) = 2

π r² = area

=4π

The inradius for any triangle with sides $a$ , $b$ , $c$ with area $[ABC]$ is given by $r=\dfrac{2[ABC]}{a+b+c}$ .

But we can check that this is a right triangle because $a=5$ , $b=12$ and $c=13$ make the Pythagorean theorem true: $a^2+b^2=c^2$ .

So we can simplify the formula: the area of a right triangle with legs $a$ and $b$ is $\dfrac{ab}{2}$ , so:

$r=\dfrac{ab}{a+b+c}$

Now, multiply the numerator and the denominator by $a+b-c$ :

$r=\dfrac{ab(a+b-c)}{(a+b+c)(a+b-c)}=\dfrac{ab(a+b-c)}{a^2+2ab+b^2-c^2}$

Again, by the Pythagorean theorem that simplifies to:

$r=\dfrac{ab(a+b-c)}{2ab}=\dfrac{a+b-c}{2}$

So, $r=\dfrac{12+5-13}{2}=\boxed{2}$ .

what about this?

(Semi perimeter)*(inradius)= area of triangle. So the inradius comes out to be 2.

i just accidentaly press b

We can apply the formula $[ABC]=rs$ , where $r$ is the inradius and $s$ is the semiperimeter of the triangle. The area of triangle ABC is $\frac{5\cdot 12}{2}=30$ . The semiperimeter of triangle ABC is $\frac{5+12+13}{2}=15$ . Plugging this into the formula, we get: $30=r\cdot 15\rightarrow r=2$ So, the area of the circle is $\pi r^{2}=\boxed{4\pi}.$

We have, $\displaystyle R_{incentre} = \frac{\Delta}{s} \;\; \text{where }\Delta = \text{Area and } s = \text{Semi perimeter}$

$\displaystyle\Rightarrow R = \frac{\frac{12\times 5}{2}}{\frac{5+12+13}{2}} = \frac{60}{30} = 2$

$\displaystyle \begin{aligned} ∴ \text{Area of incircle} &= \pi \times R^2\\ &=\boxed{4\pi}\end{aligned}$

r=(12+5-13)/2=2 ok thanks

In- radius=area/semi perimeter

delta = sr [Where 'Delta' => area of triangle, S => Semi-perimeter, R => in-radious ] r = [1/2(12 X 5)] / [1/2 (12+13+5)] = 2

Area = (Pi x r^2) = 4pi

Just use the formula that inradius of a triangle is (p+b-h)/2

To be honest i didnt actually solve it using math. The base is 5 and the diameter can not be longer than the base so going to the answer to actually answer the question. All answers have Pi so that information is irrelevant so you are looking for the radius squared. If you are given the answers you must remember that you can use them to find your answer.

Process of elimination of the given options. Radius can't be over half of 5, and clearly it is not 1. Only one option even makes sense.

² + 12² = 13² so this is a right angled triangle. a is the base, b is the height and c the hypothenuse. ab / (a+b+c) = r 5 x 12 / (5+12+13) = 2 π r² = area =4π

I drew it on autocad.is that allowed

In a triangle rectangle is valid this formula: A (5)+B (12)-2r=c (17)