Question

The area of a rectangular pool is at most 1260 ft². What could be the possible dimensions of the pool, if one side of the pool is 48 ft more than three times the other side? ​

Answer 1

Let :

• Larger side of pool = x
• Smaller side of pool = y

Given :

• Area of rectangular pool = 1260 ft²
• Larger side is 48 ft more than three times the snaller side

To find :

Sides of pool

Formula used :

1) Area of rectangle = length× breadth

{ Here Area of rectangular pool = larger side × smaller side }

$\sf 2) \: a {y}^{2} + bx + c = 0 \\ \\ \sf \implies \: x \: = \: \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac } }{2a}$

Solution :

We , know that Larger side is 48 ft more than three times the snaller side

➝ x = 48 + (3y) .... equation 1

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Also , given that area of rectangular pool = 1260 ft²

➝ xy = 1260 .... equation 2

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Putting value of x from equation 1 into equation 2, we get;

➝ (48 +3y) y = 1260

➝ 3y² + 48y = 1260

Dividing both side by 3 , we get;

➝ y² + 16y = 420

➝ y² + 16y - 420 = 0..... equation 3

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Comparing equation 3 , with general form of quadratic equations { ay² + by + c = 0}, we get

• a = 1
• b = 16
• c = -420

Using quadratic formula , and solving for y

$\sf \implies \: y \: = \: \dfrac{ -16 \: \pm \: \sqrt{ {16}^{2} - 4 \times 1 \times ( - 420) } }{2 \times 1} \\ \\ \\ \sf \implies \: y \: = \: \dfrac{ -16 \: \pm \: \sqrt{ 256 + 1680} }{2 } \\ \\ \\ \sf \implies \: y \: = \: \dfrac{ -16 \: \pm \: \sqrt{ 1936} }{2 } \\ \\ \sf \implies \: y \: = \: \dfrac{ -16 \: \pm \: 44}{2 } \\ \\ \sf \implies \: y \: = \: \dfrac{ -16 }{2} \: \pm \: \dfrac{44}{2} \\ \\ \sf \implies \: y \: = \: - 8 \: \pm \: 22$

This give ,

either , y = -8 + 22 ......or .....y = -8 -22

either , y = 14 ........or........ y = -30

{ as side cannot be negative , y ≠ -30 }

Therefore, y = 14ft

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Putting value of y in equation 1 we get,

➝ x = 48 + 3(14)

➝ x = 48 + 42

➝ x = 90 ft

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ANSWER :

• Larger side = 90 ft
• Smaller side = 14 ft