solve for 10 points and no it is not homework?

Question

John has been given a field in the shape of a triangle. Two sides of the triangle are exactly 10 metres long. What is the largest possible area, in square metres, of John's triangular field? Give your answer as the number of square metres. There is no need to include units.

GOOD LUCK.........................

Answer 1

50?

Answer 2

It is not an equilateral triangle. An isosceles right triangle (base 10, altitude 10) has an area of 10x10x(1/2) = 50. This is the maximum. It can be drawing the isosceles triangle with the unknown side as the (horizontal) base. Construct the altitude. That divides the triangle into 2 congruent right triangles, each with hypotenuse 10. The area of each is then 1/2 x 10sin (t) x 10cos (t), where t is 1/2 the angle between the 2 equal sides. The total area is then A = 100 x sin t x cos t. This has a maximum value at t = 45 degrees, making the triangle a right triangle. (Do the derivative of A with respect to t, set = 0).

Answer 3

1

Answer 4

Maximum area occurs when the two sides are at right angles.
Then area = half that of a square measuring 10 X 10.
therefore ans = 50 m^2

Answer 5

Scenario 1:
- The triangle is equilateral as described all above

Scenario 2:
- The sides are "exactly 10m long" so the Area is 50m^2

Scenario 3:
- The sides are 10m to the nearest m, so the largest possible length is just under 10.5m. Thus the area is 55.125m^2

Answer 6

I think it is when the angle between the two sides is 120 degrees.

why ?

because that looks good, i am pretty sure of it.

I see now,

take a circle between the sides you have a triangle , outside the sides you have two triangles. If the outside area of the 2 triangels is the same as the inner area then you are at a maximum. So 120 degrees, because then the height of the inner triangle is 1/2 and also of the outer triangles

Answer 7

maximumum area = 1/2 x 10 x 10 m² = 50 m²
This occurs when angle between the given sides is 90°

Answer 8

50

Answer 9

50

The shape with the largest area is a right triangle, as it turns out.

The easy way to "prove" this is to imagine a rhombus of sides 10, and deform it from one flat shape to the next. The maximum area, by reason of symmetry, would be in the middle of the deformation, or a square.

Answer 10

Those that said an equilateral triangle are wrong and those who have said an isosceles triangle with sides of 10m are correct. The area would indeed be 50 sq m.

Answer 11

Let the height of the triangle be h and the base be b.
(b/2)^2 = 10^2 - h^2
(b^2)/4 = 100 - h^2
b^2 = 4(100 - h^2)
b = 2*[SQRT(100 - h^2)]

A = (b*h)/2 = 2*[SQRT(100 - h^2)]*h/2 = SQRT(100h^2 - h^4)

For maximum A, (100h^2 - h^4) must be maximum.

Let x = 100h^2 - h^4
dx/dh = 200h - 4h^3

When dx/dh = 0
200h - 4h^3 = 0
4h(50 - h^2) = 0
h = 0 or h^2 = 50
h = 0 (NA) or h = SQRT50