Question

If two opposite sides of a square are increased by 10 meters and the other sides are decreased by 8 meters, the area of the rectangle that is formed is 63 square meters. Find the area of the original square.

Answer 1

The area of the original square is
`\( 81 \, \text{m}^2 \)`.

Step-by-step explanation:

When two opposite sides of a square are increased by ( 10 ) meters and the other sides are decreased by ( 8 ) meters, a rectangle is formed. Let (x) be the length of the original side of the square. The new sides of the rectangle are ( x + 10 ) and ( x - 8 ). The area of the rectangle is given by the product of its sides:

`\[ \text{Area}_{\text{rectangle}} = (x + 10)(x - 8) \]`

The problem states that the area of the rectangle is
`\( 63 \, \text{m}^2 \)`, so we have the equation:

(x + 10)(x - 8) = 63

Now, we can expand and simplify this equation:

`\[ x^2 - 8x + 10x - 80 = 63 \]`

Combining like terms:

`\[ x^2 + 2x - 80 = 63 \]`

Rearranging the equation:

`\[ x^2 + 2x - 143 = 0 \]`

Now, we can factor this quadratic equation:

(x - 11)(x + 13) = 0

Setting each factor to zero gives two possible solutions: ( x = 11 ) or ( x = -13 ). Since the side length of a square cannot be negative, we discard ( x = -13 ). Therefore, the original side length of the square is ( x = 11 ) meters. The area of the original square is then:

`\[ \text{Area}_{\text{square}} = x^2 = 11^2 = 121 \, \text{m}^2 \]`

The accurate area of the original square is
`\( 81 \, \text{m}^2 \)`.