Final answer:
To prove that the area of an isosceles triangle is A = (1/2)x²sin(θ), we use the definition of sine in a right triangle formed by dropping a perpendicular from the vertex angle to the base, which gives us the height. Multiplying this height by the base length and 1/2 gives the area of the isosceles triangle.
Step-by-step explanation:
To show that the area of an isosceles triangle with equal sides of length x is A = (1/2)x²sin(θ), where θ is the angle formed by the two equal sides, let's consider the isosceles triangle ABC where AB = AC = x and θ is the angle at vertex A.
When we drop a perpendicular BD from vertex B to base AC, we create two right triangles ABD and CBD inside the original isosceles triangle. In right triangle ABD, we can use the definition of sin(θ) which is opposite/hypotenuse, hence BD/x. Multiplying both sides by x/2, we get the area of one of the right triangles as (1/2) * BD * x.
The area of the isosceles triangle is twice the area of one of the right triangles, hence A = 2 * (1/2) * BD * x = x * BD. Since BD = x * sin(θ), it follows that A = x * (x * sin(θ)) = (1/2)x²sin(θ), proving the initial statement.