what the maximum area of an triangle isosceles triangle with congruent sides of 20 cm?

Question

Is there a way to do this with the cosine law?

Answer 1

Yes, there is a way with the cosine law as you asked...

Let x be the third side of the triangle, and θ the angle opposite x.

x^2 = 20^2 + 20^2 - 2*20*20cosθ
x^2 = 800 - 800cosθ = 800(1-cosθ)

Drop a perpendicular h to side x.
Area = xh/2
and tan(θ/2) = x/(2h)
so h = x/(2tan(θ/2)) = x cot(θ/2)/2
and then:
Area = x^2 cot(θ/2)/4
and since x^2 = 800(1-cosθ),
Area = 200(1-cosθ)cot(θ/2)
= 200(1-cosθ)(1+cosθ) /sinθ
= 200(1-cos^2θ) / sinθ
= 200sin^2θ / sinθ
= 200sinθ

Now you can take the derivative with respect to θ, and set equal to zero to find maximum.

d Area/dθ = 200cosθ
200cosθ = 0
θ = π/2
Area = 200sinθ = 200*1 = 200

An even easier interpretation can be seen using the vector cross-product.

The area of the resultant parallelogram formed by two vectors of length 20 and angle θ in between is:
20*20*sin(θ)

since the area of an isosceles triangle with those two sides is half the area of the parallelogram, its area is:
200sinθ

http://mathworld.wolfram.com/CrossProduct.html

Answer 2

The area of a triangle is half the base times the height, so you want the biggest possible product for b x h. Sketch an isoceles triangle and drop a perpendicular to the unequal side, which will be the base. That gives you two half-triangles, both are right triangles. Label the two 20cm sides.

The biggest possible product for b x h will occur when the base and the height of each half-triangle are equal. That's because each half-triangle is half of a rectangle, and the biggest rectangle for a given perimeter is a square. If the base and height of each half-triangle are equal, then each half-triangle is itself an isoceles triangle with 45-degree angles in the corners and the other is 90 degrees.

In such a triangle, the ratio of the sides is 1:1:sq.rt. of 2. Divide 20 by sq.rt. of 2 (1.4142) and you get 14.142 cm. so the area of the biggest triangle will be 14.142 x 14.142 = 200 sq.cm. Calculus is not needed in my analysis.

Answer 3

Dra a perpendicular to the base. This is the altitude of the triangle and and also a median. Let x = 1/2 the base.
Then the altitude = sqrt(20^2-x^2). Hence the area A of the triangle is xsqrt(20^2-x^2)
dA/dx = sqrt(20^2-x^2) + x(1/2)*1/sqrt(20-x^2) * -2x
dA/dx = sqrt(20-x^2) - x^2/sqrt(20^2-x^2)
Set equal to 0 and solve for x
20^2-x^2 -x^2 = 0
2x^2 =400
x^2 = 200
x = 10sqrt(2)
A = 10sqrt(2) * sqrt(400-200)
A= 10sqrt(2)*10sqrt(2) = 200 cm^2

Answer 4

I'm going to take the calculus approach and use optimization

Objective: A = 1/2 * b * h
Constraint: (b/2)^2 + h^2 = 20^2

Find b:
(b^2) / 4 + 4 (h^2) / 4 = 400
[(b^2) + 4(h^2)] = 1600
b = sqrt(1600 - 4(h^2))

After making substitution for b:
A = 1/2 * h * sqrt(1600 - 4(h^2))

Take the derivative:
A' =h/2 * (-4h / sqrt(1600 - 4(h^2)) + [1/2 * sqrt(1600 - 4(h^2))]
= [-2(h^2) / sqrt(1600 - 4(h^2)] + [1/2 * sqrt(1600 - 4(h^2))]
Set the derivative equal to 0 in order to maximize:
A' = 0 = [-2(h^2) / sqrt(1600 - 4(h^2)] + [1/2 * sqrt(1600 - 4(h^2))]

Simplify:
[2(h^2)] / sqrt(1600 - 4(h^2) = [sqrt(1600 - 4(h^2))] / 2
1600 - 4(h^2) = 4(h^2)
1600 = 8(h^2)
h^2 = 200
h = sqrt(200)

Find b (your third side):
b = sqrt(1600 - 4((sqrt(200))^2))
= sqrt(1600 - 4*200)
= sqrt(800)

The maximum area:
A = 1/2 * b * h
= 1/2 * sqrt(800) * sqrt(200)
= 1/2 * sqrt(160000)
= 1/2 * 400
= 200

The maximum area for the triangle is 200 cm^2

Answer 5

in accordance on your thoughts, the fringe is the sum of length of all aspects. So, the fringe is Perimeter=(2x + 9) + (2x + 9) + (3x) because of fact the fringe is everyday to be 60 inches, 60 = (2x + 9) + (2x + 9) + (3x) or, 60 = 2(2x + 9) + 3x = 4x + 18 + 3x = 7x + 18 60 - 18 = 7x 40 two = 7x 40 two/7=x x=6 consequently, x = 6