An equilateral triangle of side 9 cm is inscribed in a circle. Find th

1.	An equilateral triangle of side 9 cm is inscribed in a circle. Find the radius of the circle.
Answer» Let ABC be an equilateral triangle of side 9 cm Let, AD be one of its medians and G be the centroids of the triangle ABC Then, AG : GD = 2: 1 We know that, In an equilateral triangle centroid coincides with the circumcentre Therefore, G is the centre of the circumference with circum radius GA Also, G is the centre and GD is perpendicular to BC Therefore, In right triangle ADB, we have AB² = AD² + DB² 9² = AD² + DB² AD = \(\frac{9\sqrt 3}{2}\) cm Therefore, Radius = AG = \(\frac{2}{3}\)AD = 3\(\sqrt{3}\) cm

An equilateral triangle of side 9 cm is inscribed in a circle. Find the radius of the circle.

Answer»

Let ABC be an equilateral triangle of side 9 cm

Let, AD be one of its medians and G be the centroids of the triangle ABC

Then,

AG : GD = 2: 1

We know that,

In an equilateral triangle centroid coincides with the circumcentre

Therefore,

G is the centre of the circumference with circum radius GA

Also,

G is the centre and GD is perpendicular to BC

Therefore,

In right triangle ADB, we have

AB² = AD² + DB²

9² = AD² + DB²

AD = \(\frac{9\sqrt 3}{2}\) cm

Therefore,

Radius = AG = \(\frac{2}{3}\)AD

= 3\(\sqrt{3}\) cm

Discussion

No Comment Found

Related InterviewSolutions

The circle described on the line joining the points `(0,1), (a, b)` as diameter cuts the x-axis in points whose abscissae are roots of the equationA. `x^(2)+ax+b=0`B. `x^(2)-ax+b=0`C. `x^(2)+ax-b=0`D. `x^(2)-ax-b=0`
The equation of the circle whose one diameter is PQ, where the ordinates of P, Q are the roots of the equation `x^(2)+2x-3=0` and the abscissae are the roots of the equation `y^(2)+4y-12=0` isA. `x^(2)+y^(2)+2x+4y-15=0`B. `x^(2)+y^(2)-4x-2y-15=0`C. `x^(2)+y^(2)+4x+2y-15=0`D. none of these
Statement-1: The equation of a circle through the origin and belonging to the coaxial system, of which limiting points are (1, 1) and (3, 3) is `2x^(2)+2y^(2)-3x-3y=0` Statement-2: The equation of a circle passing through the points (1, 1) and (3, 3) is `2x^(2)+y^(2)-2x-6y+6=0`.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is not a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True.
A circle touches y-axis at (0, 2) and has an intercept of 4 units on the positive side of x-axis. The equation of the circle, isA. `x^(2)+y6(2)-4(sqrt(2)x+y)+4=0`B. `x^(2)+y^(2)-4(x+sqrt(2)y)+4=0`C. `x^(2)+y^(2)-2(sqrt(2)x+y)+4=0`D. none of these
If (1, 2) is a limiting point of a coaxial system of circles containing the circle `x^(2)+y^(2)+x-5y+9=0`, then the equation of the radical axis, isA. `x+2y+9=0`B. `3x-y+4=0`C. `x+9y-4=0`D. `3x-y-1=0`
The equation of the circle passing through (0, 0) and belonging to the system of circles of which (3, 1) and (-1, 5) are limiting points, isA. `x^(2)+y^(2)-x+3y=0`B. `x^(2)+y^(2)-11x+3y=0`C. `x^(2)+y^(2)=1`D. none of these
One of the limit point of the coaxial system of circles containing `x^(2)+y^(2)-6x-6y+4=0, x^(2)+y^(2)-2x-4y+3=0`, isA. (-1, 1)B. (-1, 2)C. (-2, 1)D. (-2, 2)
The limiting points of the system of circles represented by the equation `2(x^(2)+y^(2))+lambda x+(9)/(2)=0`, areA. `(pm(3)/(2), 0)`B. `(0, 0), ((9)/(2), 0)`C. `(pm (9)/(2), 0)`D. `(pm 3, 0)`
A ray of light incident at the point (-2, -1) gets reflected from the tangent at (0, -1) to the circle `x^2 +y^2=1`. The reflected ray touches the circle. The equation of the line along which the incident ray moved isA. `4x-3y+11=0`B. `4x+3y+11=0`C. `3x+4y+11=0`D. none of these
Diameter of a circle is 26 cm and length of a chord of the circle is 24 cm. Find the distance of the chord from the centre.Given: In a circle with centre O, PO is radius and PQ is its chord, seg OR ⊥ chord PQ, P-R-Q, PQ = 24 cm, diameter (d) = 26 cm To Find: Distance of the chord from the centre (OR)