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

## 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,

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

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

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

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

Therefore, area of rhombus

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