Question

Find the area of the triangle with the given vertices.
A(4,2), B(2,-2), C(-2,5)

Answer 1

Answer:

15 units²

Step-by-step explanation:

ΔABC is a right triangle with:

• AB = shortest leg
• AC = longest leg
• BC = hypotenuse
• ∠BAC = 90°

Area of a triangle = 1/2 x base x height

⇒ area = 1/2 x AB x AC

Use the distance formula to find the lengths:

$\sf distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

(where $\sf (x_1,y_1) \ and \ (x_2,y_2)$ are the endpoints)

$\sf AB=\sqrt{(2-4)^2+(-2-2)^2}=2\sqrt{5}$

$\sf AC=\sqrt{(-2-4)^2+(5-2)^2}=3 \sqrt{5}$

$\sf \implies area=\dfrac12 \times 2\sqrt{5} \times 3\sqrt{5} =15 \ units^2$