A rectangular trampoline measures 15 meters by 18 meters. A pad of constant width is placed around the trampoline so that the total area is 340 square meters. What is the width of the pad? Enter your answer in the box.
Answers
Given is a rectangular trampoline with dimensions 15 meters by 18 meters.
A pad of uniform width is fixed about the boundary of trampoline such that total area becomes 340 square meters.
It says to find width of the pad that is placed around trampoline.
Let's assume the width of pad = 'x' meters.
Now Length of trampoline with pad would be, L = 2x + 15 meters.
And Width of trampoline with pad would be, W = 2x + 18 meters.
Given is the Total area = 340 square meters.
We know the formula for area of rectangle is given as follows :-
Area of Rectangle = Length x Width
⇒ 340 = (2x+15)·(2x+18)
⇒ 340 = 4x² + 66x + 270
⇒ 4x² + 66x + 270 - 340 = 0
⇒ 4x² + 66x - 70 = 0
⇒ 2x² + 33x - 35 = 0
⇒ 2x² + 35x - 2x - 35 = 0
⇒ x·(2x + 35) - 1·(2x + 35) = 0
⇒ (x - 1)·(2x + 35) = 0
⇒ (x - 1) = 0 or (2x + 35) = 0
x can't be negative number, so x = 1.
Hence, x = 1 meter is the width of the pad.
Answer:
The width of the pad is [tex]1\ m[/tex]
Step-by-step explanation:
Let
x----> the width of the pad
we know that
[tex](15+2x)(18+2x)=340[/tex]
Solve for x
[tex]270+30x+36x+4x^{2} =340\\ \\4x^{2}+66x-70=0[/tex]
Solve the quadratic equation by graphing
The solution is [tex]x=1\ m[/tex]
see the attached figure
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2- getting the y-intercept:
To get the y-intercept, we will choose a point from the graph, substitute with this point in the equation and solve for c.
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