Question

find the area of a quadrilateral ABCD whose vertices are A(1,0) B(5,3) C(2,7) and D(-2,4)

Answer 1

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Answer 2

Answer: The answer is 25 sq. units.

Step-by-step explanation:  We are given to find the area of a quadrilateral with vertices A(1, 0), B(5, 3), C(2, 7) and D(-2, 4).

The quadrilateral is drawn on the co-ordinate plane as shown in the attached file.

First, we will find the lengths of all the sides using distance formula, which states that

If (a, b) and (c, d) be any two points on a plane, then the distance between them is given by

$D=\sqrt{(c-a)^2+(d-b)^2}.$

So, we have

$AB=\sqrt{(5-1)^2+(3-0)^2}=\sqrt{16+9}=\sqrt{25}=5,\\\\BC=\sqrt{(2-5)^2+(7-3)^2}=\sqrt{9+16}=\sqrt{25}=5,\\\\CD=\sqrt{(-2-2)^2+(4-7)^2}=\sqrt{16+9}=\sqrt{25}=5,\\\\DA=\sqrt{(1+2)^2+(0-4)^2}=\sqrt{9+16}=\sqrt{25}=5.$

Also, the slopes of AB(m), BC(n), CD(o) and DA(p) are given by

$m=\dfrac{3-0}{5-1}=\dfrac{3}{4},\\\\n=\dfrac{7-3}{2-5}=-\dfrac{4}{3},\\\\o=\dfrac{4-7}{-2-2}=\dfrac{3}{4},\\\\p=\dfrac{0-4}{1+2}=-\dfrac{4}{3}.$

Since m × n = n × o = o × p = p × m =-1, m = o, n = p and AB = BC = CD = DA, so we have

all the sides are equal, opposite sides are parallel and adjacent sides are perpendicular, therefore the quadrilateral ABCD is a square.

Hence, the area of quadrilateral ABCD is given by

$Area=5\times 5=25~\textup{sq. units.}$

Thus, the answer is 25 sq. units.