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Find the area of the quadrilaterals, the coordinates of whose vertices are$(1, 2), (6, 2), (5, 3)$ and $(3, 4)$
Given:
Vertices of a quadrilateral are $(1, 2), (6, 2), (5, 3)$ and $(3, 4)$.
To do:
We have to find the area of the quadrilateral.
Solution:
Let $A (1, 2), B (6, 2), C(5,3)$ and $D (3, 4)$ be the vertices of a quadrilateral $ABCD$.
Join $A$ and $C$ to get two triangles $ABC$ and $ADC$.
This implies,
Area of quadrilateral $ABCD=$ Area of triangle $ABC+$ Area of triangle $ADC$.
We know that,
Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by,
Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$
Therefore,
Area of triangle \( ABC=\frac{1}{2}[1(2-3)+6(3-2)+5(2-2)] \)
\( =\frac{1}{2}[1(-1)+6(1)+5(0)] \)
\( =\frac{1}{2}[-1+6] \)
\( =\frac{1}{2} \times 5 \)
\( =\frac{5}{2} \) sq. units.
Area of triangle \( ADC=\frac{1}{2}[1(3-4)+5(4-2)+3(2-3)] \)
\( =\frac{1}{2}[1(-1)+5(2)+3(-1)] \)
\( =\frac{1}{2}[-1+10-3] \)
\( =\frac{1}{2} \times 6 \)
\( =3 \) sq. units.
Therefore,
The area of the quadrilateral $ABCD=\frac{5}{2}+3=\frac{5+3(2)}{2}=\frac{11}{2}$ sq. units.
The area of the given quadrilateral is $\frac{11}{2}$ sq. units.
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