Charice is painting the lines for her own basketball court. The free throw section will be a rectangle with a semi-circle on top. The length of the rectangle will be 2.25 metres greater than the width. Using 3.14 for mc004-1.jpg, the area of the court is 31.28 m2. Determine the width, w, of the free throw section.
width = 2 r
length of rectangle = 2 r + 2.25
area of semicircle = (1/2) pi r^2
area of rectangle = 2r(2r+2.25)
total area of free throw section
= (pi/2) r^2 + 4 r^2 + 4.5 r
Pi/2 + 4 = 5.57
5.57 r^2 + 4.5 r = 31.28
5.57 r^2 + 4.5 r - 31.28 = 0
use quadratic equation to solve or r or go to same site I sent you for the other tank problem.
when you find r, multiply it by 2 to get the width.
r = 2 is positive root from
http://www.mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php
so w = 4
Let's denote the width of the rectangle as 'w'.
According to the problem statement, the length of the rectangle is 2.25 meters greater than the width. Therefore, the length of the rectangle can be expressed as 'w + 2.25'.
The area of the court is the sum of the area of the rectangle and the area of the semi-circle on top.
The area of the rectangle is given by the formula: length * width.
So, the area of the rectangle is: (w + 2.25) * w.
The area of a semi-circle is given by the formula: (1/2) * pi * radius^2.
In this case, the radius of the semi-circle is equal to half of the width of the rectangle. So, the radius is: w / 2.
The area of the semi-circle is: (1/2) * 3.14 * (w / 2)^2.
The total area of the court is given to be 31.28 square meters. Therefore, we have the equation:
(w + 2.25) * w + (1/2) * 3.14 * (w / 2)^2 = 31.28.
Now, we can solve this equation to find the value of 'w'.
To determine the width of the free throw section, we can set up the equation for the area of the court.
First, let's assign variables to the dimensions. Let w represent the width of the free throw section, and since the length is 2.25 meters greater, the length would be w + 2.25.
The formula for the area of a rectangle is A = length × width.
The area of the rectangle section is the area of the court minus the area of the semi-circle. Thus, we have:
A = (w + 2.25) × w - (1/2) × 3.14 × (w/2)²
Given that the area of the court is 31.28 m², we can substitute this value into the equation:
31.28 = (w + 2.25) × w - (1/2) × 3.14 × (w/2)²
Simplifying the equation further, we have:
31.28 = w² + 2.25w - (1/2)(3.14)(w/2)²
Now, let's solve the equation to find the value of w.