Alex's inscribed circle

r = (area of triangle)/(semiperimeter) = 9. Hence are of circle =81*(p1) , so a=81

Let $X$ , $Y$ , and $Z$ be the points of tangency of the circle to the triangle sides $AB$ , $BC$ , and $CA$ , respectively. By the ice-cream cone theorem (a special case of a Power of a Point), $AX=AZ$ , $BX=BY$ , and $CY=CX$ . Thus we have three unknowns for the lengths of the perimeter of the triangle, and three equations, namely $AX+BX=51$ , $BY+CY=45$ , and $AZ+CZ=24$ .

After solving, we are especially interested in $CY=CZ=9$ . Since we know $\angle IYC=\angle IZC=\angle YCZ=90^\circ$ and $CY=CZ$ , $IYCZ$ is a square. Thus the radius of the circle is $IY=YC=9$ . The area is $81\pi$ , so our answer is $81$ .

We know the radius of a circle inscribed inside a right triangle can be found from

Radius = (45+24-51)/2 =9

Therefore the area of the circle is (pi)(9^2)

Therefore a=81

We know that, the radius $r$ of the circle inscribed in a right triangle can be defined as,

$r=\frac{a+b-c}{2}$

where, $c$ is the hypotenuse and $a$ and $b$ are the other two sides.

Putting in the values of $AB$ , $AC$ and $CB$ in the identity, we get,

$r=\frac{45+24-51}{2}=\frac{18}{2}=9$

So, the area of the circle is,

$\pi r^{2}=\pi \times 9^{2}=81\pi$

Hence, the value of $a$ is $81$ .

inradius=area/semi perimeter=(1/2 * 45 *24)/(45+24+51)/2=540/60=9,so area of circle=pi * 9^2=81pi

radius= triangular area/k = √k(k-a)(k-b)(k-c) / k where k=1/2(a+b+c) where a, b and c are the side lengths of the triangle a= 51 b=24 c=45

k= 1/2(51+24+45) k= 1/2(120) k=60

radius= √60(60-51)(60-24)(60-45) / 60 radius=√60(9)(36)(15) / 60 radius=√291600 / 60 radius=540/60 radius=9

area=r^2 π area=9^2 π area=81 π therefore a=81

En español:

Siendo un circulo inscrito en un triángulo el radio del circulo sera igual al inradio. Teniendo la formula del área de un triangulo

$A=sr$
Donde: A es el area, s el semiperimetro y r el inradio.

Despejando \­(r ) \­( r=A/s \­)

El área es igual a $A=(Base*Altura)/2 = (24*45)/2=540$ . El semiperimetro es igual a $s=(AB+AC+CB)/2=(51+24+45)/2=60$ .

Reemplazando $r=A/s=540/60=9$ Como el área del circulo es igual a $r^2* \pi$ el resultado es $9^2=81$

Using the relation, in-radius, $r = \frac {\Delta}{s}$ where $\Delta$ is the are of the triangle and $s$ the semi-perimeter. Hence $r = 9$ . So, the answer is $81$ .

Let point of contact be D,E and F on AC,CB and BA respectively. Let CD=CE=x, since tangent drawn from a point to a circle are of equal length then BE=BF=45-x, therefore again using above theorem we have, AF=AD=6+x , since AF+FB=AB=51, this implies (6+x)+x=24 (i.e. AD+DC=AC) so, x=9, which is the radius of circle area=9 9 pi so a=81

r = area of triangle/semi-perimeter As area of in-circle = aπ, r = √a = (0.5 24 45)/{0.5(24+45+51)} = (24*45)/120 = 9 or a = 81

Let us draw perpendiculars from point I on sides AB, BC, CA. Let's call the feet of these perpendiculars as points D,E,F respectively. Now let the radius of the circle be r. therefore, ID = IE = IF = FC = CE = r Also BE=(45-r) and AF=(24-r) By properties of congruence of two triangles, we can easily show that the triangles IFA and IDA are congruent, so are the triangles IEB and IDB. Hence, EB = DB = (45-r) and AF = AD = (24-r) but AB = AD + DB so 51 = (45-r) + (24-r) Solving this we get r = 9 and area of circle = 81 pi

You can find the radius of the inscribed circle of a triangle by formula r=L/s. L=(24x45)/2=540 and s=(24+45+51)/2=60 and we get r=540/60=9. The area of the circle can be ound by formula L=r^2xpi and a=r^2=81

Note 24^2 + 45^ 2= 51^2, therefore ABC is a right triangle, and therefore its circumradius is equal to half the hypotenuse, or 51/ 2.

Using equation r = a b c/ 4 s R, where r is the triangle's in radius, a,b,c its sides, and s its semi perimeter, so r=9, therefore the inscribed circle's area is equal to 81π, and a=81.

We see that the incircle formula gives inradius=(45+24-51)/2=9

Let the intersection of circle centred at I with $\bigtriangleup ABC,$ 's sides AB, BC , CA be D,E,F Since , $AD = AF, BD = BE, CF = CE .$ & radius of circle $r = CF$ ( since, $CB \bot AC$ ) , s = semi perimeter of triangle .

Since , $s = (AB + BC + AC)/2 = AD + BE + CF .= AB + r ( AD+ BE = AD+BD )$

We have $(51+24+45)/2 = 51 + r .$ ,

$r = 9 .$

So , area of circle $= \pi* r^2 = 81*r$

a*pi is the area of triangle

a=r^2

x=51; y=24; z=45; s=(x+y+z)/2 R=(s (s-x) (s-y)*(s-z))^(1/2)/s

a=(s (s-x) (s-y)*(s-z))/s^2

a=60 9 36 15/60 60=81

Let r=radius;

r=(BC+CA-AB)/2 r=(45+24-51)/2 r=9

The area of the circle is 81π. Therefore a=81.

First, we must find the radius r, from the equation of the area of the triangle ABC: (24 r)/2 + (45 r)/2 + (51 r)/2 = (24)(45)/2. We get r = 9. Then the Area is aπ = 81 π. So a = 81.

Note $24^2 + 45^2 = 51^2$ , therefore $ABC$ is a right triangle, and therefore its circumradius is equal to half the hypotenuse, or $\frac{51}{2}$ . Through the equation $r = \frac{abc}{4sR}$ , where $r$ is the triangle's inradius, $a, b, c$ its sides, and $s$ its semiperimeter, we obtain $r = 9$ , therefore the inscribed circle's area is equal to $81\pi$ , and $a = 81$ .

we know that tangent lines in a certain intersection point have the same length. and by using ratios, we can get the length of the three pair of tangent lines: 9, 15,36..

the area of a circle is $\pi$ $r^{2}$ the radius of the circle is the tangent line with a right angle, in this case, 9 so the area is $\pi$ $9^{2}$ = 81 $\pi$ and a = 81

The semiperimeter of the triangle is 60, and as the area is $24 \cdot 45$ , the radius of the inscribed triangle is $\frac{24 \cdot 45}{60}=9$ . As the radius is 9, the area is $81\pi$ .