Question

The area of a triangle with vertices (a, a + d), (a + 2d, a + 3d) and (a + 4d, a + 5d) is
(a) (a + 6d)²

(b) 0 (c) 1 (d) a + 6d

Answer 1

Answer:

Area = 0

Step-by-step explanation:

(a, a + d), (a + 2d, a + 3d) and (a + 4d, a + 5d)

Side 1 = √(a + 2d - a)² + (a + 3d - (a + d))² = √ 4d² + 4d² = 2√2d

Side 2 = √(a + 4d - a)² + (a + 5d - (a + d))² = √ 16d² + 16d² = 4√2d

Side 2 = √(a + 4d - (a+2d))² + (a + 5d - (a + 3d))² = √ 4d² + 4d² = 2√2d

s = (2√2d + 4√2d + 2√2d)/2 =  4√2d

Area = √4√2d(4√2d - 2√2d)(4√2d - 4√2d)(4√2d - 2√2d)

=> Area = √4√2d(2√2d)(0)(2√2d)

=> Area =   0