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14 Aug 2006, 04:29
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A farmer has three fields. One field is an equilateral triangle, one field is a circle, and one field is a square. The square field is 75% larger in area than the triangular field, and 50% larger in area than the circular field. In order to completely fence all three of the fields, exactly 4000 metres of fencing is required.
What is the total area of all three fields, in square metres?
What is the total area of all three fields, in square metres?
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14 Aug 2006, 13:27
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This is not a difficult questions, but it is time consuming.
Given,
Area of square = 75% more than that of triangle.
Is L is length, L^2= (1/2*Height of triangle*Base)1.75
L^2= (1/2*root(5)/2*t^2)*1.75
(how we get to this depends on a property of equilateral triangle)
Also area of square = 50% more than that of the circle
L^2=1/5(pi*r^2)
L = root(1.5pi)*r
4000 is total for covering perimeter of all three shapes
4000 = 4L+2pir+3t
Replace values of r and t with derived values of L and we can solve for length of square.
Similarly, derive values for t (side of triangle) and r(radius of circle) and we get total area.
This seems too complicated.. I'm wondering if there is a simpler method.
Given,
Area of square = 75% more than that of triangle.
Is L is length, L^2= (1/2*Height of triangle*Base)1.75
L^2= (1/2*root(5)/2*t^2)*1.75
(how we get to this depends on a property of equilateral triangle)
Also area of square = 50% more than that of the circle
L^2=1/5(pi*r^2)
L = root(1.5pi)*r
4000 is total for covering perimeter of all three shapes
4000 = 4L+2pir+3t
Replace values of r and t with derived values of L and we can solve for length of square.
Similarly, derive values for t (side of triangle) and r(radius of circle) and we get total area.
This seems too complicated.. I'm wondering if there is a simpler method.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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