Question

the area of Pentagon whose vertices are (4,1) (3,6) (-5,1) (-3,-3)and (-3,0) is​

Answer 1

Answer:

$30$ $sq.units$

Step-by-step explanation:

We know that for a pentagon of sides

$(x_1,y_1)$

$(x_2,y_2)$

$(x_3,y_3)$

$(x_4,y_4)$

$(x_5,y_5)$

Its Area = $\frac{1}{2}|(x_1y_2+x_2y_3 +x_3y_4+x_4y_5+x_5y_1)-(y_1x_2+y_2x_3+y_3x_4+y_4x_5+y_5x_1)|$

⇒ In our question

$(x_1,y_1) = (4,1)$

$(x_2,y_2)= (3,6)$

$(x_3,y_3) = (-5,1)$

$(x_4,y_4) = (-3,-3)$

$(x_5,y_5) = (-3,0)$

⇒ Its Area = $\frac{1}{2} |[(4)(6)+(3)(1)+(-5)(-3)+(-3)(0)+(-3)(1)]-[(1)(3)+(6)(-5)+(1)(-3)+(-3)(-3)+(0)(4)]$

⇒ $\frac{1}{2} |(24+3+15+0-3)-(3-30-3+9+0)|$

⇒ $\frac{1}{2} |39 + 21|$

⇒ $\frac{1}{2} |60|$

⇒ $\frac{1}{2} *60 = 30sq.units$