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-A- x4-5 x3-x2 square unitsB- x4-x3-4 x2-4 square unitsC- x4-x3-x2-4 x square unitsD- x4-x2-4 x-4 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 Binomials

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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Similar questions

If the length and breadth of a rectangle are $(x^{2} - x + 2)$ cm and $(x^{2} + x - 2)$ cm respectively, find the area of the rectangle.

Q. If the sides of a rectangle are

(x^{2} - x - 2)

and

(x^{2} + x - 2)

, find the area of the rectangle.

Answer:

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

If each side of a square is of length

(x + 5) units, then find its area?

Q. The total area bounded by the closed region of

y^{2} + x^{4} = x^{2}

in square units is

Q. Five real numbers

x_{1}, x_{2}, x_{3}, x_{4}, x_{5}

are such that:

\sqrt{x_{1} - 1} + 2 \sqrt{x_{2} - 4} + 3 \sqrt{x_{3} - 9} + 4 \sqrt{x_{4} - 16} + 5 \sqrt{x_{5} - 25} = \frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5}}{2}

The value of

\frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5}}{2}

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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 $= (x^{2} - x + 2) \times (x^{2} + x - 2)$
$= x^{4} + x^{3} - 2 x^{2} - x^{3} - x^{2} + 2 x + 2 x^{2} + 2 x - 4$
$= x^{4} - x^{2} + 4 x - 4$