Asked by sarah
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.
Answers
Answered by
Damon
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.
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.
Answered by
Damon
r = 2 is positive root from
http://www.mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php
so w = 4
http://www.mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php
so w = 4
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