Question

ffind the area of the square with vertices (0,3), (6,0), (3,9), and (9,6)

Answer 1

Since it is a square:

⇒ All length are equal

Find one length:

$\text {Length = }\sqrt{(X_2-X_1)^2 + (Y_2 - Y_1)^2}$

$\text {Length = }\sqrt{(3-0)^2 + (0-6)^2}$

$\text {Length = }\sqrt{3^2 + 6^2}$

$\text {Length = }\sqrt{45}$

$\text {Length = }3\sqrt{5}$

Find the area:

$\text {Area = Length}^2$

$\text {Area = } (3\sqrt{5})^2$

$\text {Area = } 45 \text { units}^2$

Answer: The square is 45 units²