Find the area of a pentagon having the vertic | vedclass

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

Question

(A) To find the area of a polygon with vertices $(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)$, we use the shoelace formula:Area $= \frac{1}{2} |(x_1y_2

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) To find the area of a polygon with vertices $(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)$, we use the shoelace formula:Area $= \frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_5 + x_5y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_5 + y_5x_1)|$Given vertices: $(1, 5), (-2, 4), (-3, -1), (2, -3), (5, 1)$.Sum $1 = (1 	imes 4) + (-2 	imes -1) + (-3 	imes -3) + (2 	imes 1) + (5 	imes 5) = 4 + 2 + 9 + 2 + 25 = 42$Sum $2 = (5 	imes -2) + (4 	imes -3) + (-1 	imes 2) + (-3 	imes 5) + (1 	imes 1) = -10 - 12 - 2 - 15 + 1 = -38$Area $= \frac{1}{2} |42 - (-38)| = \frac{1}{2} |42 + 38| = \frac{1}{2} |80| = 40$ square units.

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

Similar Questions

Show that,for every $a \in R$,the area of the triangle having the vertices $A(5, a)$,$B(2, 5)$,and $C(2, 3)$ is $3$ square units.

Prove that $(2a, 4a)$, $(2a, 6a)$, and $(2a + \sqrt{3}a, 5a)$ are the vertices of an equilateral triangle.

For points $A(1, 7)$,$B(2, 4)$,and $C(k, 5)$,if $\angle A = 90^\circ$,then find the value of $k$.

If the distance between the points $(a, 1)$ and $(8, a)$ is $5,$ then $a=$

Find the area of the triangle whose vertices are $(-8, 4), (-6, 6)$ and $(-3, 9).$

Vedclass Test Series

Exam Paper Generator

Online Exam Module