Question

Q.6) Find the area of quadrilateral whose
vertices are
A(-3, 1), B(-2,-2), C(1,4), D(3,-1)
B)D
CD
D)​

Answer 1

Answer:

Area of Quadrilateral = 25 sq. units.

Step-by-step explanation :

Given that :

• A(-3, 1)
• B(-2,-2)
• C(1,4)
• D(3,-1)

To Find :

The area of quadrilateral.

Solution :

First of all, We have to find the lengths of all the sides,

We know that :

$\bigstar\;\boxed{\sf D=\sqrt{(c-a)^2+(d-b)^2}}}$

Values in Equation,

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

We know :

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

So,

$\sf \\ \\\implies m \times n = n \times o = o \times p\\ \\\implies p \times m =-1, m = o, n = p\\ \\\implies AB = BC = CD = DA$

We also know that :

• Adjacent sides are perpendicular.
• All the sides are equal.
• Opposite sides are parallel.

$\therefore$ The quadrilateral ABCD is a square.

As we know :

$\sf \\ \\\implies Area=5\times 5=25\;sq\;units.$

$\bigstar\;\textbf{\underline{\underline{The answer is 25 sq. units.}}}$