find the area of an equliteral triangle inscribed in the circle x2+y2-6x+2y-15=0
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Given equation of circle is:
x2 + y2 - 6x + 2y - 15 = 0
Centre O = (3, -1)
radius r = √{32 + (-1)2 -(-15)} = √{9 + 1 + 15 } = √25 = 5
Let ABC is an equilateral triangle inscribed in the give circle, then its altitude AD passes through the cneter of the circle
and D is the mid-point of BC
Also ∠OBD = 60
From right angle triangle OBD,
sin 60 = BD/OB
=> √3/2 = BD/r
=> BD = r*√3/2
Also, BC = 2*BD = 2*r*√3/2 = √3r
Now, area of the triangle ABC = √3/4 * a2 {a is the side of the triangle ABC}
= √3/4 * (√3r)2
= 3√3/4 * r2
= 3√3/4 * 52
= 3√3/4 * 25
= 75√3/4 square unit
x2 + y2 - 6x + 2y - 15 = 0
Centre O = (3, -1)
radius r = √{32 + (-1)2 -(-15)} = √{9 + 1 + 15 } = √25 = 5
Let ABC is an equilateral triangle inscribed in the give circle, then its altitude AD passes through the cneter of the circle
and D is the mid-point of BC
Also ∠OBD = 60
From right angle triangle OBD,
sin 60 = BD/OB
=> √3/2 = BD/r
=> BD = r*√3/2
Also, BC = 2*BD = 2*r*√3/2 = √3r
Now, area of the triangle ABC = √3/4 * a2 {a is the side of the triangle ABC}
= √3/4 * (√3r)2
= 3√3/4 * r2
= 3√3/4 * 52
= 3√3/4 * 25
= 75√3/4 square unit
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