Find the area of a quadrilateral having the v | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the vertices of the quadrilateral be $A(-1, -1), B(-4, -2), C(-5, -4),$ and $D(-2, -3)$.To find the area,we divide the quadrilateral into two

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Let the vertices of the quadrilateral be $A(-1, -1), B(-4, -2), C(-5, -4),$ and $D(-2, -3)$.To find the area,we divide the quadrilateral into two triangles,$	riangle ABC$ and $	riangle ADC$.The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Area of $	riangle ABC = \frac{1}{2} |(-1)(-2 - (-4)) + (-4)(-4 - (-1)) + (-5)(-1 - (-2))|$$= \frac{1}{2} |(-1)(2) + (-4)(-3) + (-5)(1)| = \frac{1}{2} |-2 + 12 - 5| = \frac{1}{2} |5| = 2.5$.Area of $	riangle ADC = \frac{1}{2} |(-1)(-4 - (-3)) + (-5)(-3 - (-1)) + (-2)(-1 - (-4))|$$= \frac{1}{2} |(-1)(-1) + (-5)(-2) + (-2)(3)| = \frac{1}{2} |1 + 10 - 6| = \frac{1}{2} |5| = 2.5$.Total Area = Area of $	riangle ABC +$ Area of $	riangle ADC = 2.5 + 2.5 = 5$ square units.

Find the area of a quadrilateral having the vertices $(-1, -1), (-4, -2), (-5, -4),$ and $(-2, -3)$.

Similar Questions

$\overline{CD}$ is parallel to the $Y$-axis and $C(4, -5)$,then the coordinates of $D$ are $\ldots \ldots \ldots$

Using the distance formula,show that the points $A(7,3)$,$B(3,0)$,$C(0,-4)$,and $D(4,-1)$ are the vertices of a rhombus.

If the vertices of $\Delta PQR$ are $P(3, 10)$,$Q(6, 5)$,and $R(1, 2)$,then $\Delta PQR$ is a $\dots$ triangle.

Show that the points $A(-2, 1)$,$B(2, -2)$,and $C(5, 2)$ are the vertices of a right-angled triangle.

Vedclass Test Series

Exam Paper Generator

Online Exam Module