A tennis court has a length of \( x \mathrm{~cm} \) and the breath of \( y \mathrm{~m} \). It has an area of \( 40 \mathrm{~m}^{2} \) and a perimeter of 28 cm . Determine the values of \( x \) and \( y \).
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To solve for the values of \( x \) and \( y \) given that the area of the tennis court is \( 40 \, \text{m}^2 \) and the perimeter is \( 28 \, \text{cm} \), we can start by using the formulas for area and perimeter. 1. **Convert cm to m**: Since we want to work in the same units for the calculations, we convert \( x \) from cm to meters: \( x \, \text{cm} = \frac{x}{100} \, \text{m} \). 2. **Area Formula**: The area of the rectangle is given by \( \text{Area} = \text{length} \times \text{width} \). \[ 40 = \left( \frac{x}{100} \right) \times y \implies xy = 4000 \quad \text{(1)} \] 3. **Perimeter Formula**: The perimeter of the rectangle is calculated by \( \text{Perimeter} = 2(\text{length} + \text{width}) \). \[ 28 = 2\left( \frac{x}{100} + y \right) \implies \frac{x}{100} + y = 14 \quad \text{(2)} \] Now we solve these two equations. From equation (2): \[ \frac{x}{100} + y = 14 \implies y = 14 - \frac{x}{100} \] Substituting \( y \) in equation (1): \[ x\left(14 - \frac{x}{100}\right) = 4000 \] \[ 14x - \frac{x^2}{100} = 4000 \] Multiplying the entire equation by 100 to eliminate the fraction: \[ 1400x - x^2 = 400000 \] Rearranging gives us: \[ x^2 - 1400x + 400000 = 0 \] Now we can solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -1400, c = 400000 \): \[ b^2 - 4ac = (-1400)^2 - 4(1)(400000) = 1960000 - 1600000 = 360000 \] \[ x = \frac{1400 \pm \sqrt{360000}}{2} \] \[ \sqrt{360000} = 600, \text{ so } x = \frac{1400 \pm 600}{2} \] This gives two possible values for \( x \): \[ x = \frac{2000}{2} = 1000 \quad \text{or} \quad x = \frac{800}{2} = 400 \] Using \( x = 1000 \: y = 14 - \frac{1000}{100} \rightarrow y = 14 - 10 = 4 \) Using \( x = 400 \): \[ y = 14 - \frac{400}{100} \rightarrow y = 14 - 4 = 10 \] Thus, the two potential solutions are: 1. \( (x, y) = (1000 \, \text{cm}, 4 \, \text{m}) \) 2. \( (x, y) = (400 \, \text{cm}, 10 \, \text{m}) \) Now converting these back to the length in cm and m: - 1000 cm = 10 m - 4 m = 400 cm - or vice versa. Hence, the dimensions can be \( x = 400 \, \text{cm} \) and \( y = 10 \, \text{m} \).
