Area of isoceles triangle.?

Question

I am having a few problems solving this one.

"Show that the area A of an isoceles triangle is... A=a^2sinθcosθ, where a is the length of one of the two equal sides and θ is the measure of one of the two equal angles..

,,,/..\
a/....\a
,/......\
/θ...θ \
---------
(thats the diagram they show, and the periods and commas are only to help it maintain shape in yahoo)....

I'm not completely helpless, we just started this chapter and I kind of understand what you have to do and I know all 4 angles should be 60degrees, but it doesnt solve right when I do it.

An english translation would be helpful :/... thanks!

Answer 1

A = 1/2 bh [definition of Area of Triange]
b = 2 * a sinθ [sin θ = adj/hyp; base = 2 adj; hyp = a; 2*adj = 2*a sinθ]
h = a cos θ [cos θ = opp/hyp; hyp = a; opp = a cos θ]

,,,/.|
a/..|
,/...|
/θ..|
----|

A = 1/2 *2 a sin θ * a cos θ
= a^2sinθcosθ

Answer 2

The measure of the angles don't have to be 60°. The formula works for any isosceles triangle, regardless of the measure of the base angles.

Think of an isosceles triangle as two congruent right triangles merged together, with their right angles butted up against each other in the middle. The sine of both the base angles is h/a, where h is the height of the isosceles triangle, and a is the length of one of the legs of the isosceles triangle. The cosine of both the base angles is half the length of the base of the isosceles triangle divided by a.

sin Θ = h/a ----> a sin Θ = h
cos Θ = (b/2)/a ----> a cos Θ = b/2

Now, recalling that the area formula for any triangle is A = ½ b h, all we have to do is plug the trig equivalent for b and h directly into this equation and we have our answer.

A = ½ b h
A = (a cos Θ)(a sin Θ)
A = a² cos Θ sin Θ
or
A = a² sin Θ cos Θ

Answer 3

Draw a line down the center of the triangle from the top to halfway down the base and call that h, for the height. You now have two smaller triangles, each with the sides, 4, h, and 3 (since you cut your base in half). Now you can use the Pythagorean theorum to find h: 4^2 = 3^2 + h^2 Now that you have h, you can use Area = (1/2) * b * h

Answer 4

Ok -- I'm going to assume you know that:
Area of a triangle = 1/2 * base * height
sin(theta) = opposite/hypotenuse
cos(theta) = adjacent/hypotenuse

Take your diagram and split it down the center to form 2 right triangles.
On one of these right triangles label the perpendicular height h, the base b, and the hypotenuse is a.
Since your length b is half the original base of the whole triangle, the area of the triangle is now b * h.

So now you can write:
sin(theta) = h/a
sin(theta) * a = h

And you can also write:
cos(theta) = b/a
cos(theta) * a = b

Since the Area = b * h (from above), you can now write:
Area = cos(theta) * a * sin(theta) * a
Area = a^2 * sin(theta) * cos(theta)

Ta-daaaa! Hope this helps!

Answer 5

length x width divided by two