Question

Find the area of the quadrilateral whose vertices are $(-4,5),(0,7),(5,-5)$ and $(-4,-2).$

Answer 1

Given, coordinate of vertices:

$A(x_1,y_1)\equiv (-4,5)$, $B(x_2,y_2)\equiv (0,7)$,

$C(x_3,y_3)\equiv (5,-5)$, $D(x_4,y_4)\equiv (-4,-2)$

Area of quadrilateral $ABCD=$ Area of $\Delta ABC+$ Area of $\Delta ADC$

Area of $\Delta ABC=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

$=\frac{1}{2}|(-4)(7+5)+0(-5-5)+5(5-7)|=\frac{1}{2}|-48+0-10|=29$

Area of $\Delta ACD=\frac{1}{2}|x_1(y_3-y_4)+x_3(y_4-y_1)+x_4(y_1-y_3)|$

$=\frac{1}{2}|(-4)(-5+2)+5(-2-5)-4(5+5)|=\frac{1}{2}|12-35-40|=31.5$

Area of quadrilateral $ABCD=29+31.5=60.5sq.units$