The perimeter of a rectangle $8 a^{2} - 10 a b + 6 b^{2}$ and one of its side $6 a^{2} - 15 a b + 2 b^{2}$ . Find its other adjacent side.

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Solution

Given the perimeter of a rectangle $8 a^{2} - 10 a b + 6 b^{2}$ and one of its side $6 a^{2} - 15 a b + 2 b^{2}$ .

Let the other adjacent side be $x$ .

Then we've,

Perimeter $= 2 (l + b)$

$2 (x + 6 a^{2} - 15 a b + 2 b^{2}) = 8 a^{2} - 10 a b + 6 b^{2}$

or, $(x + 6 a^{2} - 15 a b + 2 b^{2}) = 4 a^{2} - 5 a b + 3 b^{2}$

or, $x = - 2 a^{2} + 10 a b + b^{2}$ .

So the other side will be $b^{2} + 10 a b - 2 a^{2}$ .

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