The side of a square is $x$ cm. There is a rectangle with length $3$ more than the side of the square and the breadth is $2$ less than the side of the square. Find the sum of the areas of square and rectangle.

x^{2} - x + 6

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2 x^{2} + x - 6

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2 x^{2} - x + 18

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None of the above

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Solution

The correct option is B $2 x^{2} + x - 6$
We know the area of a square is $s i d e \times s i d e$
$\Rightarrow A = x \times x = x^{2}$
Lenght and Breadth of Rectangle are $(x + 3)$ and $(x - 2)$ respectively.
Area of rectangle $= l e n g t h \times b r e a d t h = (x + 3) \times (x - 2)$
$\Rightarrow x^{2} + x - 6$
Sum of areas is given by $x^{2} + x^{2} + x - 6 = 2 x^{2} + x - 6$

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The side of a square is xcm. There is a rectangle with length 3 more than the side of the square and the breadth is 2 less than the side of the square. Find the sum of the areas of square and rectangle.

The correct option is B 2x2+x−6We know the area of a square is side×side⇒A=x×x=x2Lenght and Breadth of Rectangle are (x+3) and (x−2) respectively.Area of rectangle =length×breadth=(x+3)×(x−2)⇒x2+x−6Sum of areas is given by x2+x2+x−6=2x2+x−6

The side of a square is $x$ cm. There is a rectangle with length $3$ more than the side of the square and the breadth is $2$ less than the side of the square. Find the sum of the areas of square and rectangle.