Maximise Area of Triangle Trapped in Semi-Circle

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In summary, the homework statement is that a right angled triangle is trapped within a semi-circle of radius r. The triangle has side lengths of a, b and 2r. The triangle can be maximised in area if a and b are maximised.
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steven10137
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Homework Statement


A right angled triangle with sides of length a, b and 2r is trapped within a semi-circle of fixed radius r. Given that all three vertices are on the semi-circle, find the values of a and b that will maximise the area of the triangle

http://img100.imageshack.us/img100/1804/areaoftrainglebl3.th.jpg

Homework Equations


A(circle) = pi r ^2
A(triangle) using hero's formula (sqrt(s(s-a)(s-b)(s-c))) where s=(a+b+c)/2
OR
A(triangle)= (b x h)/2

The Attempt at a Solution


I am familiar with optimisation skills but am really stumped with this question, given that it is a scalene triangle.
If someone could give me a pointer in the right direction as to what formulas to use or attempt to derive, it would be appreciated
cheers
 
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Hint - Since the vertices are always on the semi-circle, then the triangle will always be a right-triangle -- area of triangle will be (1/2) a * b
 
  • #4
ahhhh

that is so obvious, can't believe i didn't figure that out.
I suppose I just didn't visualise the traingle in that way OR notice that it actually told me that it was right-angled ... :blushing:

cheers
i shall now attempt to optimise it
thanks fellas
 
  • #5
ok

so i have found that the area of the triangle is (ab)/2
Now i want to optimise this so that the area is maximised, so we need another equation.

I know that the area of the half-circle is (pi x r^2)
how can i now derive some other equation to maximise a and b?
 
  • #6
There is a relation - What can you say about the sides of a right triangle?
 
  • #7
pythagoras ...
a^2 + b^2 = (2r)^2

take that in terms of 'a' or 'b' and then replace than in the area equation, find the derivate and then optain max/min ... cheers

sorry not thinking very well tonight ... thankyou for you patience.
 

1. How do you determine the maximum area of a triangle trapped in a semi-circle?

To determine the maximum area of a triangle trapped in a semi-circle, you need to use the formula A = 1/2 * b * h, where A is the area of the triangle, b is the base of the triangle, and h is the height of the triangle. The base of the triangle should be the diameter of the semi-circle, and the height should be the distance from the midpoint of the base to the top of the semi-circle.

2. What is the relationship between the maximum area of the triangle and the radius of the semi-circle?

The maximum area of the triangle is directly proportional to the radius of the semi-circle. This means that as the radius increases, the maximum area of the triangle also increases. This relationship can be seen in the formula A = 1/2 * b * h, where the base of the triangle (b) is equal to the diameter of the semi-circle (2r).

3. Can the maximum area of the triangle trapped in a semi-circle be greater than the area of the semi-circle?

No, the maximum area of the triangle trapped in a semi-circle can never be greater than the area of the semi-circle itself. This is because the area of the semi-circle is already maximized and is equal to half of the area of the circle that it is inscribed in. Any triangle trapped in the semi-circle will have a smaller area than the semi-circle itself.

4. How can you prove that the maximum area of the triangle trapped in a semi-circle is half of the area of the semi-circle?

This can be proved using calculus. By taking the derivative of the area formula (A = 1/2 * b * h) with respect to the base (b), and setting it to 0, we can find the maximum value for the area. This value will be equal to half of the area of the semi-circle. This can also be visually proven by constructing a right triangle inside the semi-circle, with one leg being the diameter of the semi-circle and the other leg being the height of the triangle. The area of this triangle will be equal to half of the area of the semi-circle.

5. Are there any real-life applications of maximizing the area of a triangle trapped in a semi-circle?

Yes, this concept has real-life applications in engineering and architecture. The maximum area of a triangle trapped in a semi-circle can help determine the optimal size and shape of structures, such as bridges or arches, that are built on a semi-circular base. It can also be used in landscaping and gardening to maximize the area of triangular flower beds or garden plots that are placed in semi-circular spaces.

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