Question
Download Solution PDFIf 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.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
Semi-perimeter (s) = 12 cm
Area (A) = 28 cm2
Formula used:
Semi-perimeter of rectangle: s = (x + y)
Area of rectangle: A = x × y
Calculation:
x + y = 12
xy = 28
We need to find x4 + x2y2 + y4.
Using the identity (x + y)2 = x2 + y2 + 2xy:
(x + y)2 = 122 = 144
⇒ x2 + y2 + 2xy = 144
⇒ x2 + y2 + 2 × 28 = 144
⇒ x2 + y2 + 56 = 144
⇒ x2 + y2 = 88
Now, (x2 + y2)2 = x4 + y4 + 2x2y2
(88)2 = x4 + y4 + 2 × (28)2
7744 = x4 + y4 + 2 × 784
7744 = x4 + y4 + 1568
x4 + y4 = 7744 - 1568
x4 + y4 = 6176
Finally, x4 + x2y2 + y4 = x4 + y4 + x2y2
= 6176 + 784
= 6960
∴ The value of x4 + x2y2 + y4 is 6960.
Last updated on Jun 3, 2025
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