Ex.7.2 Q10 Coordinate Geometry Solution - NCERT Maths Class 10

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Question

Find the area of a rhombus if its vertices are \((3, 0)\), \((4, 5)\), \((-1, 4)\) and \((-2, -1)\) taken in order.

[Hint: Area of a rhombus \(=\) (product of its diagonals)]

 

Text Solution

Reasoning:

A rhombus has all sides of equal length and opposite sides are parallel.

What is the known?

The \(x\) and \(y\) co-ordinates of the vertices of the rhombus.

What is the unknown?

The area of the rhombus

Steps:

From the Figure,

Given,

  • Let \(A(3, 0)\), \(B(4, 5)\), \(C(-1, 4)\) and \(D(-2, -1)\) are the vertices of a rhombus \(ABCD\).

We know that the distance between the two points is given by the Distance Formula,

\[\begin{align}\sqrt {{{\left( {{{{x}}_1} - {{{x}}_2}} \right)}^2} + {{\left( {{{{y}}_1} - {{\text{y}}_2}} \right)}^2}} & & ...\;{\text{Equation}}\;{\text{(1)}}\end{align}\]

Therefore, distance between \(A\;(3, 0)\) and \(C \;(-1, 4)\) is given by

Length of diagonal

\[\begin{align} AC &= \sqrt {{{[3 - ( - 1)]}^2} + {{(0 - 4)}^2}}  \\ &= \sqrt {16 + 16} = 4\sqrt 2 \end{align}\]

Therefore, distance between B (4, 5) and D (-2, -1) is given by

Length of diagonal

\[\begin{align} BD &= \sqrt {{{[4 - ( - 2)]}^2} + \left( {5 - {{( - 1)}^2}} \right.}   \\& = \sqrt {36 + 36} \\&= 6\sqrt 2 \end{align}\]

 \[\begin{align} \text{Area of the rhombus } ABCD &= \frac{1}{2} \times \rm{(Product\; of \;lengths\; of \;diagonals)} \\&= \frac{1}{2} \times {\text{AC}} \times {\text{BD}}\end{align}\]

Therefore, area of rhombus

\[\begin{align}ABCD &= \frac{1}{2} \times 4\sqrt 2 \times 6\sqrt 2 \\ &= 24 \;\rm square\; units\end{align}\]

  

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