
Madhu P. answered 12/17/15
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Maths tutor believing Mathematics is Fun and Fundamental
Hi Aisha,
The patio is 15ft by 10ft. So its a rectangle with width and length given. So area of the patio is 150 sq ft...........(1)
Now John wants to put a flower bed AROUND the patio. So imagine another rectangle bigger than the first such that there is a constant width between the 2 rectangles (I am unable to draw here but u can draw 2 rectangles one inside the other such that the gap between the 2 will represent the flower bed area. )
Now if the width of the flower bed is say 'w' ft then as it goes around the patio, the new length and width of the patio PLUS the flower bed will be (15+2w)ft and (10+2w) ft
( we add 2w to the length and width of patio as the flower bed goes around it)
So area of the flowerbed with patio inside is (15+2w)*(10+2w) sq ft .............(2)
Now given that area of flower bed is 100 ft. So the area of flower bed is the difference of equations (1) and (2)
(15+2w)*(10+2w) - 150 = 100
So need to solve this quadratic equation to get the answer.
Simplifying the equation we get
150+30w+20w+4w2 -150 = 100
Rearranging we get
4w2 + 50w -100 = 0
Dividing the equation by 2 through out we simplify as
2w2 +25w -50 =0
Solving the equation we get w =3.187 and -15.687
Since the width cannot be negative we take the required width of flower bed = 3.187