If the semi-perimeter and area of a rectangular field whose length and breadth are 'x' and 'y' is 12 cm and 28 cm2, respectively, then find the value of x4 + x2y2 + y4.
Show answer & explanation
Correct answer: B
Explanation
The correct answer is Option 2 i.e. 6960.
Given, (x + y) = 12
And xy = 28
⇒ x2 + 2xy + y2 = 144
⇒ x2 + y2 = 144 - 56
⇒ x2 + y2 = 88
Squarying both sides,
⇒ (x2 + y2)2 = 7744
⇒ x4 + 2x2y2 + y4 = 7744
⇒ x4 + y4 = 7744 - 1568
⇒ x4 + y4 = 6176
Required answer = x4 + x2y2 + y4
⇒ 6176 + 784 = 6960