SOLUTION: PROBLEM 1: A rectangle of area 48 sq. ft. is inscribed in a circle of an area 78 4/7sq. ft. Find the dimensions of the rectangle. ( USE 22/7 FOR PI)
PROBLEM 2: The area of the r
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-> SOLUTION: PROBLEM 1: A rectangle of area 48 sq. ft. is inscribed in a circle of an area 78 4/7sq. ft. Find the dimensions of the rectangle. ( USE 22/7 FOR PI)
PROBLEM 2: The area of the r
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Question 350810: PROBLEM 1: A rectangle of area 48 sq. ft. is inscribed in a circle of an area 78 4/7sq. ft. Find the dimensions of the rectangle. ( USE 22/7 FOR PI)
PROBLEM 2: The area of the rectangle is 350 sq. cm. If the length exceeds 3 times its width by 5, find the dimensions of the rectangle.
You can put this solution on YOUR website! Area of circle = 78 4/7
pi*r^2 =550/7
r^2=560/7 *7/22
r^2=25
r=5 radius
..
l^2+(48/L)^2=100
L^4+2304-100L^2=0
L^4-64L^2+36L^2+2304 =0
L^2(L^2-64)+36(L^2-64)=0
(L^2-64)(L^2+36)=0
(L+8)(L-8)(L^2+36)=0
ignoring negative values L=8
width = 6
...
let width be x
Length = 3x+5
Area = 350 cm^2
x(3x+5)=350
3x^2+5x=350
3x^2+5x-350=0
..
use quadratic formula
a=3,b=5,c=-350
x1= (-5+sqrt(25+4200)) / 6
x1=(-5+sqrt(4225))/6
x1=(-5+65)/6
x1=60/6
x1= 10.
x2 will be negative which can be ignored
width = 10 cm
length = 3x+5
=30+5
=35 cm
C H E C K
Area = 10*35 = 350cm^2
