Question 6 – Name the Type of Quadrilateral
Part (i)
🖼️ Diagram
Figure 1(i): Quadrilateral ABCD with vertices A(-1,-2), B(1,0), C(-1,2), D(-3,0)
📋 Given Information
- Four points: \(A(-1, -2)\), \(B(1, 0)\), \(C(-1, 2)\), \(D(-3, 0)\)
- We need to identify the type of quadrilateral formed by these points
🎯 To Find
The type of quadrilateral formed by the given points and provide reasons using the distance formula.
📐 Formula Used
Properties of Quadrilaterals:
- Square: All sides equal, diagonals equal
- Rectangle: Opposite sides equal, diagonals equal
- Rhombus: All sides equal, diagonals NOT equal
- Parallelogram: Opposite sides equal
📝 Step-by-Step Solution
Using points \(A(-1, -2)\) and \(B(1, 0)\):
\[ AB = \sqrt{(1 – (-1))^2 + (0 – (-2))^2} \] \[ AB = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \text{ units} \]Using points \(B(1, 0)\) and \(C(-1, 2)\):
\[ BC = \sqrt{(-1 – 1)^2 + (2 – 0)^2} \] \[ BC = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \text{ units} \]Using points \(C(-1, 2)\) and \(D(-3, 0)\):
\[ CD = \sqrt{(-3 – (-1))^2 + (0 – 2)^2} \] \[ CD = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \text{ units} \]Using points \(D(-3, 0)\) and \(A(-1, -2)\):
\[ DA = \sqrt{(-1 – (-3))^2 + (-2 – 0)^2} \] \[ DA = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \text{ units} \]Using points \(A(-1, -2)\) and \(C(-1, 2)\):
Using points \(B(1, 0)\) and \(D(-3, 0)\):
We have:
- \(AB = BC = CD = DA = 2\sqrt{2}\) units (all sides are equal)
- \(AC = BD = 4\) units (both diagonals are equal)
Since all four sides are equal AND both diagonals are equal, the quadrilateral ABCD is a square.
Reason: All sides equal (\(2\sqrt{2}\) units) and diagonals equal (4 units)
Part (ii)
🖼️ Diagram
Figure 1(ii): Quadrilateral ABCD with vertices A(-3,5), B(3,1), C(0,3), D(-1,-4)
📋 Given Information
- Four points: \(A(-3, 5)\), \(B(3, 1)\), \(C(0, 3)\), \(D(-1, -4)\)
- We need to identify the type of quadrilateral formed by these points
📝 Step-by-Step Solution
Using points \(A(-3, 5)\) and \(B(3, 1)\):
\[ AB = \sqrt{(3 – (-3))^2 + (1 – 5)^2} \] \[ AB = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \text{ units} \]Using points \(B(3, 1)\) and \(C(0, 3)\):
\[ BC = \sqrt{(0 – 3)^2 + (3 – 1)^2} \] \[ BC = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13} \text{ units} \]Using points \(C(0, 3)\) and \(D(-1, -4)\):
\[ CD = \sqrt{(-1 – 0)^2 + (-4 – 3)^2} \] \[ CD = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2} \text{ units} \]Using points \(D(-1, -4)\) and \(A(-3, 5)\):
\[ DA = \sqrt{(-3 – (-1))^2 + (5 – (-4))^2} \] \[ DA = \sqrt{(-2)^2 + 9^2} = \sqrt{4 + 81} = \sqrt{85} \text{ units} \]We have:
- \(AB = 2\sqrt{13}\) units
- \(BC = \sqrt{13}\) units
- \(CD = 5\sqrt{2}\) units
- \(DA = \sqrt{85}\) units
All four sides have different lengths. Therefore, this quadrilateral does not form any special type of quadrilateral.
Reason: All sides have different lengths
Part (iii)
🖼️ Diagram
Figure 1(iii): Quadrilateral ABCD with vertices A(4,5), B(7,6), C(4,3), D(1,2)
📋 Given Information
- Four points: \(A(4, 5)\), \(B(7, 6)\), \(C(4, 3)\), \(D(1, 2)\)
- We need to identify the type of quadrilateral formed by these points
📝 Step-by-Step Solution
Using points \(A(4, 5)\) and \(B(7, 6)\):
\[ AB = \sqrt{(7 – 4)^2 + (6 – 5)^2} \] \[ AB = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \text{ units} \]Using points \(B(7, 6)\) and \(C(4, 3)\):
\[ BC = \sqrt{(4 – 7)^2 + (3 – 6)^2} \] \[ BC = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \text{ units} \]Using points \(C(4, 3)\) and \(D(1, 2)\):
\[ CD = \sqrt{(1 – 4)^2 + (2 – 3)^2} \] \[ CD = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \text{ units} \]Using points \(D(1, 2)\) and \(A(4, 5)\):
\[ DA = \sqrt{(4 – 1)^2 + (5 – 2)^2} \] \[ DA = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \text{ units} \]Using points \(A(4, 5)\) and \(C(4, 3)\):
Using points \(B(7, 6)\) and \(D(1, 2)\):
We have:
- \(AB = CD = \sqrt{10}\) units (opposite sides are equal)
- \(BC = DA = 3\sqrt{2}\) units (opposite sides are equal)
- \(AC = 2\) units and \(BD = 2\sqrt{13}\) units (diagonals are NOT equal)
Since opposite sides are equal but diagonals are not equal, the quadrilateral ABCD is a parallelogram.
Reason: Opposite sides are equal (\(AB = CD = \sqrt{10}\), \(BC = DA = 3\sqrt{2}\))
⚠️ Common Mistakes
❌ Mistake 1: Confusing rhombus with square. Both have all sides equal, but only a square has equal diagonals.
✅ Correct Approach: Always check BOTH sides and diagonals to correctly identify the quadrilateral type.
❌ Mistake 2: Not checking opposite sides in part (iii) and concluding it’s a general quadrilateral.
✅ Correct Approach: Even if all four sides are not equal, check if opposite sides are equal (parallelogram property).
❌ Mistake 3: Assuming that if diagonals are equal, it must be a square.
✅ Correct Approach: A rectangle also has equal diagonals. Check if all sides are equal to confirm it’s a square.
💡 Key Points to Remember
- Square: All 4 sides equal + Diagonals equal
- Rectangle: Opposite sides equal + Diagonals equal
- Rhombus: All 4 sides equal + Diagonals NOT equal
- Parallelogram: Opposite sides equal (diagonals may or may not be equal)
- Always calculate all four sides and both diagonals to correctly identify the quadrilateral
- Use the distance formula systematically for each calculation
- Simplify all square roots to their simplest form
- A rhombus is NOT a square unless its diagonals are also equal