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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)
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उत्तर
Maximum number of persons can approach the ice-cream parlour if it is equidistant from A, B and C. Now, A, B and C form a triangle. In a triangle, the circumcentre is the only point that is equidistant from its vertices. So, the ice-cream parlour should be set up at the circumcentre O of ΔABC.

In this situation, maximum number of persons can approach it. We can find circumcentre O of this triangle by drawing perpendicular bisectors of the sides of this triangle.
संबंधित प्रश्न
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 the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR >∠PSQ.

In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
In the following figure, ∠BAC = 60o and ∠ABC = 65o.

Prove that:
(i) CF > AF
(ii) DC > DF
“Caste inequalities are still prevalent in India.” Examine the statement.
Name the greatest and the smallest sides in the following triangles:
ΔDEF, ∠D = 32°, ∠E = 56° and ∠F = 92°.
ABCD is a trapezium. Prove that:
CD + DA + AB + BC > 2AC.
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 + PR > QR and PQ + QR >2PS.
