Advertisements
Advertisements
Question
ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.
Advertisements
Solution
Circumcentre of a triangle is always equidistant from all the vertices of that triangle. Circumcentre is the point where perpendicular bisectors of all the sides of the triangle meet together.
In ΔABC, we can find the circumcentre by drawing the perpendicular bisectors of sides AB, BC, and CA of this triangle. O is the point where these bisectors are meeting together. Therefore, O is the point which is equidistant from all the vertices of ΔABC.
RELATED QUESTIONS
Show that in a right angled triangle, the hypotenuse is the longest side.
In the following figure, ∠BAC = 60o and ∠ABC = 65o.
Prove that:
(i) CF > AF
(ii) DC > DF
In the following figure ; AC = CD; ∠BAD = 110o and ∠ACB = 74o.
Prove that: BC > CD.
Name the smallest angle in each of these triangles:
In ΔXYZ, XY = 6.2cm, XY = 6.8cm and YZ = 5cm
Prove that the hypotenuse is the longest side in a right-angled triangle.
For any quadrilateral, prove that its perimeter is greater than the sum of its diagonals.
In ABC, P, Q and R are points on AB, BC and AC respectively. Prove that AB + BC + AC > PQ + QR + PR.
In the given figure, ∠QPR = 50° and ∠PQR = 60°. Show that : PN < RN
In ΔPQR, PS ⊥ QR ; prove that: PQ > QS and PQ > PS
In ΔPQR, PS ⊥ QR ; prove that: PQ > QS and PR > PS
