If the adjacent sides of a rectangle are -x-2 - x - 2- units and -x-2 - x -2- units long- find the area of the rectangle-x-4 - 5x-3 - x-2- square units-x-4 - x-3 - 4x-2 - 4- nbsp-square units-x-4 - x-3 -x-2 -4x- nbsp-square units-x-4 -x-2 - 4x -4- nbsp-square units

The correct option is D (x^4 x^2 + 4x 4) nbsp;square unitsWe know that area of a rectangle = length x breadth Hence, area =(x^2 x + 2) × (x^2 + x 2) =x^ ...

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Byju's Answer

Standard VIII

Mathematics

Multiplication of a Binomial by a Trinomial

If the adjace...

Question

If the adjacent sides of a rectangle are $(x^{2} - x + 2)$ units and $(x^{2} + x - 2)$ units long, find the area of the rectangle.

(x^{4} - 5 x^{3} - x^{2})

square units

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(x^{4} - x^{3} - 4 x^{2} + 4)

square units

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(x^{4} - x^{3} - x^{2} - 4 x)

square units

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(x^{4} - x^{2} + 4 x - 4)

square units

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Solution

The correct option is D $(x^{4} - x^{2} + 4 x - 4)$ square units
We know that
area of a rectangle = length x breadth
Hence, area $= (x^{2} - x + 2) \times (x^{2} + x - 2)$
$= x^{4} + x^{3} - 2 x^{2} - x^{3} - x^{2} + 2 x + 2 x^{2} + 2 x - 4$
$= x^{4} - x^{2} + 4 x - 4$

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If the adjacent sides of a rectangle are (x2−x+2) units and (x2+x−2) units long, find the area of the rectangle.

The correct option is D (x4−x2+4x−4) square unitsWe know that area of a rectangle = length x breadth Hence, area =(x2−x+2)× (x2+x−2) =x4+x3−2x2−x3−x2+2x+2x2+2x−4 =x4−x2+4x−4

If the adjacent sides of a rectangle are $(x^{2} - x + 2)$ units and $(x^{2} + x - 2)$ units long, find the area of the rectangle.

The correct option is D $(x^{4} - x^{2} + 4 x - 4)$ square units
We know that
area of a rectangle = length x breadth
Hence, area $= (x^{2} - x + 2) \times (x^{2} + x - 2)$
$= x^{4} + x^{3} - 2 x^{2} - x^{3} - x^{2} + 2 x + 2 x^{2} + 2 x - 4$
$= x^{4} - x^{2} + 4 x - 4$