208. If the extremities of the base of an isosceles triangle are the points (2a, 0) and (0, a) and the equation of one of the sides is x=2a,
then the area of the triangle is ..
Answers
Answer:
Area of the triangle is 5a²/2.
*a graph is added for a = 1.
Step-by-step explanation:
Let the third point be C, as this point lies of x = 2a, it must be satisfied by this equation and its x-coordinate must be 2a.
So, let the required point be (2a, y).
Since this is isosceles triangle, length of two sides must be equal. (sides other than base) [using distance formula]
⇒ AC = CB
⇒ √(2a - 0)² + (y - a)² = √(2a - 2a)² + (y - 0)²
⇒ (2a)² + (y - a)²= (0)² + (y)²
⇒ 4a² + y² + a² - 2ay = y²
⇒ 5a² = 2ay
⇒ (5a/2) = y
Hence point C ≡ (2a , 5a/2)
Using distance formula:
AC = CB = √(2a - 2a)² + (y - 0)²
= y
= (5a/2)
BC = √(2a - 0)² + (0 - a)²
= a√5
Area of isosceles triangle =
= 1/2 * b * √(y² - b²/4)
= 1/2 * a√5 * [ (5a/2)² - (a√5²/4]
= 1/2 * a√5 * (a√5)
= (a²/2) * √5 * √5
= 5a²/2
Step-by-step explanation:
5a²/2 this is your answer ok it will help you