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प्रश्न
ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.
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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 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)
From the following figure, prove that: AB > CD.
In the following figure, ∠BAC = 60o and ∠ABC = 65o.
Prove that:
(i) CF > AF
(ii) DC > DF
Name the greatest and the smallest sides in the following triangles:
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Name the smallest angle in each of these triangles:
In ΔPQR, PQ = 8.3cm, QR = 5.4cm and PR = 7.2cm
In ΔABC, the exterior ∠PBC > exterior ∠QCB. Prove that AB > AC.
ABCD is a trapezium. Prove that:
CD + DA + AB + BC > 2AC.
In the given figure, ∠QPR = 50° and ∠PQR = 60°. Show that: SN < SR
In ΔPQR, PS ⊥ QR ; prove that: PQ > QS and PR > PS
Prove that in an isosceles triangle any of its equal sides is greater than the straight line joining the vertex to any point on the base of the triangle.
