Question

# 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 $$ABCD$$.

Solution

## Let the vertices of the quadrilateral be $$A(-5,7),B(-4,-5),C(-1,-6),D(4,5)$$.Joining $$AC$$ there are two triangles $$ABC,ADC$$Therefore area of quadrilateral $$ABCD=$$Area of $$\Delta ABC+$$Area of $$\Delta ADC$$Area of $$\Delta ABC=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$$Here $$x_1=-5,y_1=7,x_2=-4,y_2=-5,x_3=-1,y_3=-6$$Area of $$\Delta ABC=\dfrac{1}{2}[-5(-5+6)-4(-6-7)-1(7+5)]=\dfrac{1}{2}(35)$$ square units.Area of $$\Delta ADC=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$$Here $$x_1=-5,y_1=7,x_2=4,y_2=5,x_3=-1,y_3=-6$$Area of $$\Delta ADC=\dfrac{1}{2}[-5(5+6)+4(-6-7)-1(7-5)]=\dfrac{1}{2}(-109)=\dfrac{1}{2}(109)$$ square units. (since area cannot be negative)Therefore area of quadrilateral $$ABCD=$$Area of $$\Delta ABC+$$Area of $$\Delta ADC$$                                                                   $$=\dfrac{1}{2}(35)+\dfrac{1}{2}109$$                                                                   $$=\dfrac{1}{2}(144)$$                                                                   $$=72$$Therefore area of quadrilateral $$ABCD=72$$ square unitsMathematics

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