Question

Which of the following is the area of a quadrilateral with vertices at (-4,2),(1,2),(1,-3) and (-4,-3)

Answer 1

Answer:

The correct option is B.

Step-by-step explanation:

The given vertices are (-4,2),(1,2),(1,-3) and (-4,-3).

Plot these point on a coordinate plate. From the graph it is noticed that the given quadrilateral is a square.

Distance formula:

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

Use distance formula to find the side length.

$AB=\sqrt{(1-(-4))^2+(2-2)^2}=\sqrt{5^2}=5$

$BC=\sqrt{(1-1)^2+(-3-2)^2}=\sqrt{(25}=5$

Since both consecutive sides are equal therefore it is a square.

Area of a square is

$A=a^2$

Where, a is side length.

The side length of the square is 5. So, area of ABCD is

$A=(5)^2$

$A=25$

Therefore the area of quadrilateral is 25 units square. Option B is correct.