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Difficulty:
35%
(medium)
Question Stats:
74% (02:01) correct
26%
(02:27)
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based on 35
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A wedding stage is a square of side 10 metres each. At a corner, a triangular section of the stage is cordoned off using a rope in a straight line. If the length of the rope is 6 metres, what is the maximum area (in square metres) which can be cordoned off in the triangle?
A. 9
B. 16
C. 25
D. 36
E. 49
Source : Wizako
A. 9
B. 16
C. 25
D. 36
E. 49
Source : Wizako
Suraj0184
Triangular section cordoned off is a right triangle because square has all angles as 90. And we are asked to find the maximum area and area of a right triangle is maximum when it is an isosceles right triangle
So the hypotenuse of the 45-45-90 right triangle is 6m so the 2 legs will be equal to 6/Root(2)
So maximum area = 1/2 * 6/Root(2) * 6/Root(2) = 36/2*2 = 9
Answer - A
rajtare
Joined: 05 Nov 2014
Last visit: 16 Jul 2026
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Location: India
Concentration: Operations, Leadership
Schools: Sloan '26 Columbia CBS'26 Goizueta '26 Kellogg '26
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Schools: Sloan '26 Columbia CBS'26 Goizueta '26 Kellogg '26
Posts: 183
Suraj0184
Area of triangle is given as = 1/2 (l)*(b)
now, the area will have a maximum value if product of l, and b is maximum. Also from triangle inequality we know that sum of any two sides of triangle is always greater than the other.
so let two sides of triangle be x, and y so x+y>6
we also know that sqrt(x^2+ y^2) = 36
so that means no side of the triangle can be greater than 6. i.e x+y<11 so the possible value of x+y can be 7,8,9,10. Also, we know that product of two numbers is maximum when they are equal provided there sum is constant.
i.e. x = y
using this we have sqrt(2)x = 6
or x =3sqrt(2)
so Area = 1/2(3sqrt(2)*(3sqrt(2))
= 9
hence A
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