To draw the circumcircle of a triangle, we must know the point equidistant from

sides of the triangle

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any two vertices of the triangle

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all the vertices of the triangle

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any two sides of the triangle

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Solution

The correct option is B all the vertices of the triangle
The point that is equidistant from the each vertex is the circumcentre that is the centre of circumcircle.
So to draw circumcircle we must know the centre.
So option $C$ is correct.

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Similar questions

Q. Construct a triangle ABC such that: BC = 6 cm,

∠ B = 60^{\circ}

and

∠ C = 45^{\circ}

In the same figure, find a point which is equidistant from the vertices of the triangle.Name this point. Draw the Circumcircle of the triangle.

Circumcircle of a triangle passes through only two vertices of a triangle.

(a) For constructing a triangle, we must know two sides and included angle. Which triangle can be uniquely constructed by knowing two sides and any other angle?
(b) Which of the following congruency conditions is preferred to draw a triangle exactly congruent to a right-angled triangle? [2 MARKS]

For constructing a triangle, we must know two sides and included triangle. Which of the following types of triangles can be uniquely constructed by knowing two sides and an excluded angle?

Q. Assertion :Two vertices of a triangle are

(- 2, 3)

and

(- 4, - 7)

. If the orthocentre of the triangle is the origin, the coordinates of the third vertex are

(5, - 1)

. Reason: Orthocentre of a triangle is equidistant from the sides of the triangle.

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To draw the circumcircle of a triangle, we must know the point equidistant from

The correct option is B all the vertices of the triangleThe point that is equidistant from the each vertex is the circumcentre that is the centre of circumcircle.So to draw circumcircle we must know the centre.So option C is correct.

The correct option is B all the vertices of the triangle
The point that is equidistant from the each vertex is the circumcentre that is the centre of circumcircle.
So to draw circumcircle we must know the centre.
So option $C$ is correct.