What is the area of the triangle whose vertices are (0,0,0) (3, 4, 0) and (3, 4, 6) ?

Question

What is the area of the triangle whose vertices are (0,0,0) (3, 4, 0) and (3, 4, 6) ? (A) 12 square units (B) 15 square units (C) 30 square units (D) 36 square units

Tardigrade Admin · Accepted Answer

Let A (0, 0, 0), B (3, 4, 0) and C (3, 4, 6)
Area,

(Δ) = (Δ x^{2} + Δ y^{2} + Δ^{2})

Now,

Δ x = \frac{1}{2} ∣ ∣ y_{1} y_{2} y_{3} z_{1} z_{2} z_{3} 111 ∣ ∣ = \frac{1}{2} ∣ ∣ 044006111 ∣ ∣

= \frac{1}{2} [1 (24)] = 12

Similarly,

Δ y = \frac{1}{2} ∣ ∣ 006033111 ∣ ∣ = \frac{1}{2} [1 (- 8)] = - 9

and

Δ z = \frac{1}{2} ∣ ∣ 033044111 ∣ ∣ = \frac{1}{2} [12 - 12] = 0

Area of

Δ = 1 2^{2} + (- 9)^{2} + (0)^{2}

= 144 + 81 = 225 = 15

square units

Q. What is the area of the triangle whose vertices are (0,0,0) (3, 4, 0) and (3, 4, 6) ?

12 square units

15 square units

30 square units

36 square units