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

What type of a quadrilateral do the points $A(2, -2)$,$B(7, 3)$,$C(11, -1)$,and $D(6, -6)$ taken in that order,form?

$A (h, k), B (1, 1)$ and $C (2, 1)$ are the vertices of $\Delta ABC$. If the area of $\Delta ABC$ is $1$,then find the possible values of $k$.

If $A (-5, 7)$,$B (-4, -5)$,$C (-1, -6)$,and $D (4, 5)$ are the vertices of a quadrilateral,find the area of the quadrilateral $ABCD$ (in square units).

State whether the following statements are true or false. Justify your answer.
The points $(0,5), (0,-9)$ and $(3,6)$ are collinear.

Vedclass Test Series

Exam Paper Generator

Online Exam Module