# Ex.7.3 Q4 Coordinate Geometry Solution - NCERT Maths Class 10

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## Question

Find the area of the quadrilateral whose vertices, taken in order, are $$(-4, -2), (-3, -5), (3, -2)$$ and $$(2, 3)$$

Reasoning:

Let $$ABC$$ be any triangle whose vertices are $$A$$$$(x_1, y_1),$$ $$B$$$$(x_2, y_2)$$ and $$C$$$$(x_3, y_3).$$

Area of a triangle
\begin{align}=\frac{{{1}}}{{{2}}}\left\{ {{{{x}}_{{1}}}\left( {{{{y}}_{{2}}} - {{{y}}_{{3}}}} \right){{ + }}{{{x}}_{{2}}}\left( {{{{y}}_{{3}}} - {{{y}}_{{1}}}} \right){{ + }}{{{x}}_{{3}}}\left( {{{{y}}_{{1}}} - {{{y}}_{{2}}}} \right)} \right\} \end{align}

What is known?

The $$x$$ and $$y$$ co-ordinates of the vertices of the quadrilateral.

What is  unknown?

Solution:

From the figure,

Given,

• Let the vertices of the quadrilateral be $$A$$$$(-4, -2),$$ $$B$$$$(-3, -5),$$ $$C$$$$(3, -2),$$ and $$D$$$$(2, 3).$$
• Join $$AC$$ to form two triangles $$\Delta$$ABC and $$\Delta$$ACD.

We know that,

Area of a triangle
\begin{align} = \frac{{{1}}}{{{2}}}\left\{ {{{{x}}_{{1}}}\left( {{{{y}}_{{2}}} - {{{y}}_{{3}}}} \right){{ + }}{{{x}}_{{2}}}\left( {{{{y}}_{{3}}} - {{{y}}_{{1}}}} \right){{ + }}{{\text{x}}_{\text{3}}}\left( {{{\text{y}}_{\text{1}}} - {{{y}}_{{2}}}} \right)} \right\}\\...\,{\rm{Equation}}\,\left( {{1}} \right)\end{align}

By substituting the values of vertices, $$A, B, C$$ in the Equation ($$1$$),

Area of $$\Delta {ABC}$$

\begin{align}&= \frac{1}{2}[( - 4)\{ ( - 5) - ( - 2)\} + ( - 3)\{ ( - 2) - ( - 2)\} + 3\{ ( - 2) - ( - 2)\} ]\\& = \frac{1}{2}(12 + 0 + 9)\\& = \frac{{21}}{2}{\text{ Square units }}\end{align}

By substituting the values of vertices, $$A, C, D$$ in the Equation ($$1$$),

Area of $$\Delta ACD$$
\begin{align}\\ &= \frac{1}{2}[( - 4)\{ ( - 2) - ( - 3)\} + 3\{ (3) - ( - 2) + 2\{ ( - 2)\} - ( - 2)\} ]\\ &= \frac{1}{2}\{ 20 + 15 + 0\} \\ &= \frac{{35}}{2}{\text{ Square units}}\end{align}

Area of $$ABCD$$ $$=$$ Area of $$\Delta {{ABC}}$$ $$+$$ Area of $$\Delta {{ACD}}$$
\begin{align}\text{Area of}\;{{ABCD}} &= {\text{ Area of }}\Delta {{ABC}} + {\text{Area of }}\Delta {{ACD}}\\ &= \left( {\frac{{21}}{2} + \frac{{35}}{2}} \right){\text{ Square units }}\\ &= 28{\text{ square units }}\end{align}

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