Can you answer the following math problem? (geometry/algebra)?

Question

The are 3 fields, one square, one circular, one an equilateral triangle. The area of the square one is 50% large than the area of the circular one. The area of the square one is 75% larger than the area of triangular one. The perimeter of all 3 added together is 4000. What is the area of each field?
this isnt a homework problem, I came across it playing a silly little online game. Now I used to be pretty good at math, but I have not done anything but add and subtract for decades now :p Guess Im rusty. Can anyone help me out here? Thanks

Answer 1

Let s denote the length of a side of the equilateral triangle, a the length of a side of the square, and r the radius of the circle.

We are trying to find

A=\sqrt{3}/4 s^2+a^2+\pi r^2 (*)

subject to

a^2=7/4 \sqrt{3}/4 s^2 (1)
a^2=3/2 \pi r^2 (2)
3s+4a+\2 pi r=4000 (3)

(1) implies that s=4a/(3^(1/4) \sqrt{7})
(2) implies that r=a \sqrt{2}/\sqrt{3 \pi}

Subsituting these values into (3) and simplifying gives us

2/21 a (42+6*3^(3/4) \sqrt{7}+7 \sqrt{6 \pi})=4000

so

a=42000/(42+6*3^(3/4) \sqrt{7}+7 \sqrt{6 \pi})

Now subsituting all of these values into (*) gives us

A=\sqrt{3}/4 s^2+a^2+\pi r^2
=47a^2/21
=3948000000/(42+6*3^(3/4) \sqrt{7}+7 \sqrt{6 \pi})^2
\approx 334888 m^2

Be sure I used the area formulas correctly! :)

Answer 2

T, C, and S will mean triangle, circle, and square.

S = 1.75T = 1.5C
T = (4/7)S; C = (2/3)S

T = bh/2 = (b/2)[b sqrt(3)/2] = (b^2) sqrt(3)/4
base b = 2 sqrt[T/sqrt(3)]

C = pi r^2 ==> r = sqrt(C/pi)
Circumference c = 2 pi r = 2 sqrt(pi C)

4 sqrt S + 6 sqrt[T/sqrt(3)] + 2 sqrt(pi C) = 4000
4 sqrt S + 6 sqrt[4S/(7 sqrt(3))] + 2 sqrt(2 pi S/3) = 4000
{4 + 12 / sqrt[7 sqrt(3)] + 2 sqrt(2 pi/3)} sqrt S = 4000
10.34 sqrt S = 4000
S = 149,630.6 sq m.
T = (4/7)S = 85,503.2 sq m.
C = (2/3)S = 99,753.7 sq m.

Total area = S(1 + 4/7 + 2/3) = S(21+12+14)/21 = (47/21)S = 334,887.57 sq m.

Check: Perimeter of square = 1547.3 m
Perimeter of triangle = 1333.1 m.
Circumference of circle = 1119.6 m.

Total fencing = 4000 m

Answer 3

S(square)=a^2=p(square)^2/16
S(circular)=pi*r^2=p(circular)^2/4pi
S(triangle)=a^2*sqrt(3)/4=p(triangle)^2sqrt(3)/36
let's mark
p(square)^2 as x
p(circular)^2 as y
p(triangle)^2 as z
we have the followin 3 equations
x/4=1,5y/pi from it y=x*pi/6
x/4=1,75z*sqrt(3)/9 from it z=x*3sqrt(3)/7
x+y+z=4000 putting above equations we have
x(1+pi/6+3sqrt(3)/7)=4000
x=4000/(1+pi/6+3sqrt(3)/7)
the rest i'm sure u can do urself

P.S. and sry for my poor english;-)

Answer 4

nicely the situation-unfastened answer to the question is as follows. You draw strains from each and every verticee to the middle of the alternative side. the place they intersect is the circumcentre. to try this you will desire to discover the midpoints and discover the equations of those sort of strains, then set 2 of them equivalent to locate the intersecting element.

Answer 5

like this question was Answered in this link:
http://answers.yahoo.com/question/index;_ylt=AoGIDqb_rXuT2UggRdo9lt_zy6IX?qid=20060811015428AA0ebGw

Answer 6

Easy.

TFTP