Question 5
🖼️ Diagram
Figure 1: Quadrilateral ABCD with vertices A(3,4), B(6,7), C(9,4), D(6,1) and diagonals AC and BD
📋 Given Information
- Four friends are seated at points: \(A(3, 4)\), \(B(6, 7)\), \(C(9, 4)\), and \(D(6, 1)\)
- Champa claims ABCD is a square
- Chameli disagrees with Champa
- We need to verify who is correct using the distance formula
🎯 To Find
Determine whether ABCD is a square by calculating the lengths of all four sides and both diagonals using the distance formula.
📐 Formula Used
Explanation: The distance formula calculates the distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) in a coordinate plane.
Properties of a Square:
- All four sides are equal in length
- Both diagonals are equal in length
- Diagonals bisect each other at right angles
📝 Step-by-Step Solution
Using points \(A(3, 4)\) and \(B(6, 7)\):
Using points \(B(6, 7)\) and \(C(9, 4)\):
\[ BC = \sqrt{(9 – 6)^2 + (4 – 7)^2} \] \[ BC = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} \] \[ BC = 3\sqrt{2} \text{ units} \]Using points \(C(9, 4)\) and \(D(6, 1)\):
\[ CD = \sqrt{(6 – 9)^2 + (1 – 4)^2} \] \[ CD = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} \] \[ CD = 3\sqrt{2} \text{ units} \]Using points \(D(6, 1)\) and \(A(3, 4)\):
\[ DA = \sqrt{(3 – 6)^2 + (4 – 1)^2} \] \[ DA = \sqrt{(-3)^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} \] \[ DA = 3\sqrt{2} \text{ units} \]Using points \(A(3, 4)\) and \(C(9, 4)\):
Using points \(B(6, 7)\) and \(D(6, 1)\):
We have found:
- \(AB = BC = CD = DA = 3\sqrt{2}\) units (all sides are equal)
- \(AC = BD = 6\) units (both diagonals are equal)
Since all four sides are equal and both diagonals are equal, ABCD satisfies the properties of a square.
All sides: \( AB = BC = CD = DA = 3\sqrt{2} \) units
Both diagonals: \( AC = BD = 6 \) units
💡 Alternative Method: Using One Diagonal and Pythagoras Theorem
\(AB = BC = CD = DA = 3\sqrt{2}\) units
\(AC = 6\) units
In a square, if side = \(a\), then diagonal = \(a\sqrt{2}\)
Here, side = \(3\sqrt{2}\), so diagonal should be:
\[ \text{Diagonal} = 3\sqrt{2} \times \sqrt{2} = 3 \times 2 = 6 \text{ units} \]This matches our calculated diagonal AC = 6 units.
This confirms that angle A is 90°, and ABCD is indeed a square.
⚠️ Common Mistakes
❌ Mistake 1: Calculating only the sides and concluding it’s a square without checking diagonals.
✅ Correct Approach: A rhombus also has all sides equal. To confirm it’s a square, we must verify that both diagonals are equal as well.
❌ Mistake 2: Sign errors when calculating differences like \((4 – 7) = -3\), then squaring to get \((-3)^2 = -9\).
✅ Correct Approach: Remember that \((-3)^2 = 9\) (positive). The square of any real number is always non-negative.
❌ Mistake 3: Not simplifying \(\sqrt{18}\) to \(3\sqrt{2}\).
✅ Correct Approach: Always simplify radicals: \(\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}\).
💡 Key Points to Remember
- The distance formula is: \(\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}\)
- A square has all four sides equal AND both diagonals equal
- A rhombus has all sides equal but diagonals are NOT equal (this is the key difference)
- In a square with side \(a\), the diagonal = \(a\sqrt{2}\)
- Always verify both conditions (sides and diagonals) before concluding a quadrilateral is a square
- When squaring negative numbers: \((-a)^2 = a^2\) (always positive)
- Simplify square roots to their simplest radical form
- The distance formula is derived from the Pythagoras theorem