KVPY 2019 SB SX Exam Previous Papers

Question : 10 of 120

Marks: +1, -0

Among all the parallelograms whose diagonals are 10 and 4, the one having maximum area has its perimeter lying in the interval

Solution:

Area

= 1 / 2d_1d_2 \sin θ

is maximum when θ = 90°
⇒ Parallelogram is a rhombus
⇒ perimeter =

4√{(d_1/2)^2+(d_2/2)^2

= 4√29∊ (21, 22]

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