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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 ADC
We 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.

Mathematics
RS Agarwal
Standard X

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