If the area of an equilateral triangle inscribed in the circle, x2 + y2 + 10x + 12y + c = 0 is 27 √3 sq. units then c is equal to :

Question

If the area of an equilateral triangle inscribed in the circle, x2 + y2 + 10x + 12y + c = 0 is 27 √3 sq. units then c is equal to : (A) 20 (B) 25 (C) 13 (D) -25

Tardigrade Admin · Accepted Answer

3 ((1/2) r2 . sin120°) =27 √3 (r2/2) (√3/2) = (27√3/3) r2 = (108/3) =36 Radius = √25+36 -C= √36 C = 25

Q. If the area of an equilateral triangle inscribed in the circle, $x^{2} + y^{2} + 10 x + 12 y + c = 0$ is $273 s q$ . units then c is equal to :

20

25

13

-25

$3 (\frac{1}{2} r^{2} . sin 12 0^{\circ}) = 273$
$\frac{r ^{2}}{2} \frac{3}{2} = \frac{27 3}{3}$
$r^{2} = \frac{108}{3} = 36$
Radius $= 25 + 36 - C = 36$
$C = 25$

Q. If the area of an equilateral triangle inscribed in the circle, x2+y2+10x+12y+c=0 is 273​sq. units then c is equal to :

20

25

13

-25

3(21​r2.sin120∘)=273​ 2r2​23​​=3273​​ r2=3108​=36 Radius =25+36−C​=36​ C=25

Q. If the area of an equilateral triangle inscribed in the circle, $x^{2} + y^{2} + 10 x + 12 y + c = 0$ is $273 s q$ . units then c is equal to :

$3 (\frac{1}{2} r^{2} . sin 12 0^{\circ}) = 273$
$\frac{r ^{2}}{2} \frac{3}{2} = \frac{27 3}{3}$
$r^{2} = \frac{108}{3} = 36$
Radius $= 25 + 36 - C = 36$
$C = 25$