Question

Find the area of quadrilateral formed by joining the vertices (-5,2) (0,3) (10,-2) (0,5). On a graph

Answer 1

Step-by-step explanation:

you can do it ans is right

Answer 2

Given :  vertices (-5,2) (0,3) (10,-2) (0,5).

To find : area of the  quadrilateral

Explanation:

using formula

area of triangle

$\frac{1}{2}(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))$

here ,

in triangle ABC

$(x_1,y_1)= (-5,2)\\\\(x_2,y_2)= (0,3)\\\\(x_3,y_3)= (10,-2)$

then area of triangle ABC

$\frac{1}{2}(-5(3+2)+0(-2-2)+10(2-3))\\\\\frac{1}{2}(-5(5)+10(-1))\\\\\frac{1}{2}(-25-10)\\\\\frac{1}{2}(-35)\\\\=\frac{-35}{2}$

similarly

in triangle ADC

$(x_1,y_1)= (-5,2)\\\\(x_2,y_2)= (10,-2)\\\\(x_3,y_3)= (0,5)$

area of triangle ADC

$\frac{1}{2}(-5(-2-5)+10(5-2)+0)\\\\\frac{1}{2}(-5(-7)+10(3))\\\\\frac{1}{2}(35+30)\\\\\frac{1}{2}(65)\\\\=\frac{65}{2}$

now , area  of  quadrilateral ABCD = area of triangle ABC + area of triangle ADC

$\frac{65}{2}-\frac{35}{2}\\\\=\frac{30}{2}$

= 15 units

hence , area  of  quadrilateral ABCD is 15 units

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