Advertisements
Advertisements
प्रश्न
ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.
Advertisements
उत्तर
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.
संबंधित प्रश्न
In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see the given figure). Show that ∠A > ∠C and ∠B > ∠D.

In a huge park people are concentrated at three points (see the given figure):

A: where there are different slides and swings for children,
B: near which a man-made lake is situated,
C: which is near to a large parking and exit.
Where should an ice-cream parlour be set up so that maximum number of persons can approach it?
(Hint: The parlor should be equidistant from A, B and C)
If two sides of a triangle are 8 cm and 13 cm, then the length of the third side is between a cm and b cm. Find the values of a and b such that a is less than b.
D is a point in side BC of triangle ABC. If AD > AC, show that AB > AC.
Prove that the perimeter of a triangle is greater than the sum of its three medians.
In the given figure, ∠QPR = 50° and ∠PQR = 60°. Show that: SN < SR
In ΔABC, BC produced to D, such that, AC = CD; ∠BAD = 125° and ∠ACD = 105°. Show that BC > CD.
In the given figure, T is a point on the side PR of an equilateral triangle PQR. Show that PT < QT
In ΔABC, D is a point in the interior of the triangle. Prove that DB + DC < AB + AC.
