KVPY SA Exam 05-Nov-2017 Question Paper

Question : 62 of 80

Marks: +1, -0

In an isosceles trapezium, the length of one of the parallel sides, and the lengths of the non-parallel sides are all equal to 30. In order to maximize the area of the trapezium, the smallest angle should be

$π/6$
$π/4$
$π/3$
$π/2$

Validate

Solution:

In ∆ABF, y = 30 cosθ
Area is

A=1/2[60+ 60\cosθ ] (30 \sinθ)

30 (1+\cosθ) (30\sinθ)

900 (\sinθ+\sinθ\cosθ)

For maximum or minimum

{dA}/{dθ}=900[\cosθ +(- \sin ^2θ+\cos θ)] =0

\cosθ=1 + \cos^2θ + \cos^2θ= 0

2\cos^2θ+\cosθ-1 = 0

(2\cosθ-1) (\cosθ+1) = 0

\cosθ=1/2

\cosθ=-1

(not possible)
θ = 60°

Go to Question:

Prev Question

Next Question