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Answer:
16 squares
Step-by-step explanation:
The area of an isosceles is
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Area of pool edge = Area of big rectangle - Area of the pool
Area of pool edge = Area of the pool = 40 × 60 = 2400 ft²
2400 = Area of big rectangle - 2400
Area of big rectangle = 2400 + 2400
Area of big rectangle = 4800
Length × width = 4800
From the diagram, we need the length to be [x + x] more than the length of the pool, where x is the distance from the pool edge to the patio edge.
We also need the width of the big rectangle to be [[x + x] more than the width of the pool.
Length = 60 + 2x
Width = 40 + 2x
Length × Width = [60+2x] × [40+2x]
4800 = 2400 + 120x + 80x + 4x²
0 = 4x² + 200x - 2400
0 = 4[x² + 50x - 600]
0 = x² + 50x - 600
0 = [x - 60] [x + 10]
x - 60 = 0 OR x + 10 = 0
x = 60 OR x = -10
We can only use the positive value of x since the context is length
Hence, x = 60
Area of pool edge = Area of the pool = 40 × 60 = 2400 ft²
2400 = Area of big rectangle - 2400
Area of big rectangle = 2400 + 2400
Area of big rectangle = 4800
Length × width = 4800
From the diagram, we need the length to be [x + x] more than the length of the pool, where x is the distance from the pool edge to the patio edge.
We also need the width of the big rectangle to be [[x + x] more than the width of the pool.
Length = 60 + 2x
Width = 40 + 2x
Length × Width = [60+2x] × [40+2x]
4800 = 2400 + 120x + 80x + 4x²
0 = 4x² + 200x - 2400
0 = 4[x² + 50x - 600]
0 = x² + 50x - 600
0 = [x - 60] [x + 10]
x - 60 = 0 OR x + 10 = 0
x = 60 OR x = -10
We can only use the positive value of x since the context is length
Hence, x = 60
1 unit to the left
Check the table I attached
Check the table I attached