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?

The area of the quadrilateral.

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

Text Solution

  
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