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Naya [18.7K]
3 years ago
7

The dimensions of a rectangular garden were 4m by 5m. Each dimension was increased by the same amount the garden then had an are

a of 56M^2 find the dimension of the new garden
Mathematics
1 answer:
pogonyaev3 years ago
5 0

The dimension of new garden is 7 meter by 8 meter

<em><u>Solution:</u></em>

The dimensions of a rectangular garden were 4m by 5m

Length = 4 m

Width = 5 m

Each dimension was increased by the same amount the garden then had an area of 56 square meter

Let "x" be the amount of increase

Then,

Length = 4 + x

Width = 5 + x

Also, given that,

New area = 56 square meter

<em><u>The area of rectangle is given as:</u></em>

Area = length \times width

<em><u>Substituting the values we get,</u></em>

56 = (4+x)(5+x)\\\\56 = 20 + 4x + 5x + x^2\\\\x^2 + 9x + 20 - 56 = 0\\\\x^2 + 9x -36=0

<em><u>Let us solve the above equation by quadratic formula</u></em>

\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

<em><u>Using the above formula,</u></em>

\text{For } x^2+9x-36 \text{ we get , }

a = 1 and b = 9 and c = -36

<em><u>Substituting the values of a = 1 ; b = 9 ; c = -36 in above quadratic formula we get,</u></em>

x = \frac{-9\pm \sqrt{9^2-4\cdot \:1\left(-36\right)}}{2\cdot \:1}\\\\x = \frac{-9\pm \sqrt{81+144}}{2}\\\\x = \frac{-9 \pm 15}{2}

<em><u>We get two solutions,</u></em>

x = \frac{-9+15}{2} \text{ or } x = \frac{-9-15}{2}\\\\x = 3 \text{ or } x = -12

Since dimension cannot be negative, ignore x = -12

Thus solution is x = 3

Length = 4 + x = 4 + 3 = 7

Width = 5 + x = 5 + 3 = 8

Thus dimension of new garden is 7 meter by 8 meter

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