In the figure, the side of the inside square is $x$ ; the side of the outside square is $x + 2$ . How much greater is the area of the outer square than the area of the inside square?

4

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4 x

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4 x + 4

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4 x + 1

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x + 4

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Solution

The correct option is C $4 x + 4$
The area of inner square is $x \times x = x^{2}$
The area of outer square is $(x + 2) (x + 2) = x^{2} + 4 x + 4$
the difference between areas is $x^{2} + 4 x + 4 - x^{2} = 4 x + 4$
The area of the outer square than the area of the inside square is $4 x + 4$ .

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Question 16

The area of a square having each side $x$ is

(a) $x \times x$

(b) $4 x$

(d) $4 + x$

(x^{2} - x + 2)

(x^{2} + x - 2)

6

A = {(x, y) : y^{2} \leq 4 x, y - 2 x ⩾ - 4}

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