Question

# If $$A(-7, 5), B(-6, -7), C(-3, -8)$$ and $$D(2, 3)$$ are the vertices of a quadrilateral ABCD then find the area of the quadrilateral.

Solution

## Given: ABCD is a quadrilateral whose vertices are $$A(-7, 50, B(-6, -7), C(-3, -8)$$ and $$D(2, 3)$$.By joining AC, we get two triangles ABC and ADCWe know that:Area of $$\Delta ABC=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$$Area of triangle ABC:$$=\dfrac{1}{2}[-7(-7-(-8)]+(-6)[(-8)-5]+(-3)[5-(-7)]$$$$=\dfrac{1}{2}[-7\times (1)+(-6)\times (-13)+(-3)\times 12]$$$$=\dfrac{1}{2}[-7+78-36]$$$$=\dfrac{1}{2}\times 35=\dfrac{35}{2}$$ sq. units.Area of triangle ADC.$$=\dfrac{1}{2}[-7(-8-3)+(-3)(3-5)+(2)(-5(-8)]$$$$=\dfrac{1}{2}[(-7)\times (-11)+(-3)\times (-2)+2\times 13]$$$$=\dfrac{1}{2}[77+6+26]=\dfrac{109}{2}$$ sq. units.Now, Area of quadrilateral PQRS$$=$$Area of triangle ABC$$+$$Area of triangle ADC$$=35/2+109/2$$$$=72$$ sq. units.MathematicsRS AgarwalStandard X

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