NCERT Class X Chapter 7: Coordinate Geometry Exercise 7.2 Question 10

NCERT Class X Chapter 7: Coordinate Geometry

Question:

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

Given:

The vertices of the rhombus are:

  • A(3, 0)
  • B(4, 5)
  • C(–1, 4)
  • D(–2, –1)

To Find:

The area of the rhombus.

Formula:

Area of a rhombus = \( \dfrac{1}{2} \times d_1 \times d_2 \), where \( d_1 \) and \( d_2 \) are the lengths of the diagonals.

Also, the area of a quadrilateral with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4) \) is:
\[ \text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right| \]

Solution:

Step 1: Assign the vertices in order as A(3, 0), B(4, 5), C(–1, 4), D(–2, –1).

Step 2: Use the formula for the area of a quadrilateral given its vertices:

$$ \text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right| $$

Step 3: Substitute the coordinates:

  • \( x_1 = 3, \; y_1 = 0 \)
  • \( x_2 = 4, \; y_2 = 5 \)
  • \( x_3 = -1, \; y_3 = 4 \)
  • \( x_4 = -2, \; y_4 = -1 \)
$$ \begin{align*} \text{Area} &= \frac{1}{2} \Big| (3 \times 5) + (4 \times 4) + ((-1) \times (-1)) + ((-2) \times 0) \\ &\quad - [0 \times 4 + 5 \times (-1) + 4 \times (-2) + (-1) \times 3] \Big| \end{align*} $$

Step 4: Calculate each term:

  • \( 3 \times 5 = 15 \)
  • \( 4 \times 4 = 16 \)
  • \( -1 \times -1 = 1 \)
  • \( -2 \times 0 = 0 \)
  • \( 0 \times 4 = 0 \)
  • \( 5 \times -1 = -5 \)
  • \( 4 \times -2 = -8 \)
  • \( -1 \times 3 = -3 \)
$$ \begin{align*} \text{Area} &= \frac{1}{2} \Big| 15 + 16 + 1 + 0 - [0 + (-5) + (-8) + (-3)] \Big| \\ &= \frac{1}{2} \Big| 32 - [-16] \Big| \\ &= \frac{1}{2} \Big| 32 + 16 \Big| \\ &= \frac{1}{2} \times 48 \\ &= 24 \end{align*} $$

Step 5: Therefore, the area of the rhombus is 24 square units.

Result:

The area of the rhombus is 24 square units.

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