🔑 Important Formulas, Rules & Key Points

• Quadrilateral: A closed polygon with four sides, four vertices, and four angles.
• Angle Sum Property: The sum of all four angles of a quadrilateral is always 360° (∠A + ∠B + ∠C + ∠D = 360°).
• Parallelogram (∠gm): A quadrilateral in which opposite sides are parallel. Key properties:
  • Opposite sides are equal.
  • Opposite angles are equal.
  • Diagonals bisect each other.
• Rectangle: A parallelogram with one right angle (90°). Diagonals of a rectangle are equal.
• Rhombus: A parallelogram with all sides equal. Diagonals of a rhombus bisect each other at right angles (90°).
• Square: A parallelogram with all sides equal and all angles equal to 90°. Diagonals are equal and bisect each other at right angles.
• Congruence Rules used: SAS (Side-Angle-Side), SSS (Side-Side-Side), and CPCT (Corresponding Parts of Congruent Triangles).

Question 1

The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

Solution:

Let the common ratio multiplier be x.

Therefore, the four angles of the quadrilateral are 3x, 5x, 9x, and 13x.

According to the Angle Sum Property of a quadrilateral:

3x + 5x + 9x + 13x = 360°
30x = 360°
x = 360° / 30
x = 12°

Now, substituting the value of x to find each angle:

• First angle = 3 × 12° = 36°
• Second angle = 5 × 12° = 60°
• Third angle = 9 × 12° = 108°
• Fourth angle = 13 × 12° = 156°

Answer: The angles of the quadrilateral are 36°, 60°, 108°, and 156°.

Question 2

If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Solution:

Given: A parallelogram ABCD in which diagonal AC = diagonal BD.

To Prove: ABCD is a rectangle.

Proof:

In triangles ΔABC and ΔBAD:

• AB = AB (Common side)
• BC = AD (Opposite sides of a parallelogram are equal)
• AC = BD (Given)

Therefore, by SSS congruence criterion:
ΔABC ≅ ΔBAD

By CPCT (Corresponding Parts of Congruent Triangles):
∠ABC = ∠BAD --- (Equation 1)

Since AD || BC (opposite sides of parallelogram) and AB is the transversal line:
∠ABC + ∠BAD = 180° (Co-interior angles are supplementary)
∠ABC + ∠ABC = 180° (Using Equation 1)
2 ∠ABC = 180°
∠ABC = 90°

Since ABCD is a parallelogram with one angle equal to 90°, it is proven that ABCD is a rectangle.

Question 3

Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Solution:

Given: A quadrilateral ABCD in which diagonals AC and BD bisect each other at right angles at point O. (i.e., OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°).

To Prove: ABCD is a rhombus.

Proof:

In triangles ΔAOB and ΔAOD:

• OA = OA (Common side)
• ∠AOB = ∠AOD = 90° (Given)
• OB = OD (Given, diagonals bisect each other)

Therefore, by SAS congruence criterion:
ΔAOB ≅ ΔAOD

By CPCT:
AB = AD --- (Equation 1)

Similarly, by comparing ΔAOB and ΔBOC, we can prove:
AB = BC --- (Equation 2)

By comparing ΔBOC and ΔCOD, we can prove:
BC = CD --- (Equation 3)

From Equations 1, 2, and 3, we get:
AB = BC = CD = DA

Since all four sides of quadrilateral ABCD are equal, it is proven that ABCD is a rhombus.

Question 4

Show that the diagonals of a square are equal and bisect each other at right angles.

Solution:

Given: ABCD is a square where AC and BD are diagonals intersecting at point O.

To Prove:
1. AC = BD (Diagonals are equal)
2. OA = OC and OB = OD (Diagonals bisect each other)
3. ∠AOB = 90° (Diagonals bisect at right angles)

Proof:

Part 1: Diagonals are equal (AC = BD)
In ΔABC and ΔBAD:

• AB = AB (Common)
• BC = AD (Sides of a square are equal)
• ∠ABC = ∠BAD = 90° (Angles of a square are 90°)

By SAS congruence criterion, ΔABC ≅ ΔBAD.
Hence, AC = BD (By CPCT).

Part 2: Diagonals bisect each other
Since a square is also a parallelogram, and the diagonals of a parallelogram always bisect each other, we have:
OA = OC and OB = OD.

Part 3: Diagonals are perpendicular (∠AOB = 90°)
In ΔAOB and ΔAOD:

• AO = AO (Common)
• OB = OD (Proven in Part 2)
• AB = AD (Sides of a square are equal)

By SSS congruence criterion, ΔAOB ≅ ΔAOD.
Therefore, ∠AOB = ∠AOD (By CPCT).
We know that ∠AOB + ∠AOD = 180° (Linear Pair).
2 ∠AOB = 180° ⇒ ∠AOB = 90°.

Hence, the diagonals of a square are equal and bisect each other at right angles.

Question 5

Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Solution:

Given: In quadrilateral ABCD, diagonals AC = BD, OA = OC, OB = OD, and ∠AOB = 90°.

To Prove: ABCD is a square.

Proof:

1. To prove it is a parallelogram: Since diagonals AC and BD bisect each other, ABCD must be a parallelogram.

2. To prove all sides are equal: In ΔAOB and ΔAOD:

• OA = OA (Common)
• ∠AOB = ∠AOD = 90° (Given)
• OB = OD (Given)

By SAS criterion, ΔAOB ≅ ΔAOD.
So, AB = AD (By CPCT).
Since opposite sides of a parallelogram are equal, and adjacent sides are equal, AB = BC = CD = DA. Thus, ABCD is a rhombus.

3. To prove one angle is 90°: In ΔABC and ΔBAD:

• AB = AB (Common)
• BC = AD (Proven above)
• AC = BD (Given)

By SSS criterion, ΔABC ≅ ΔBAD.
So, ∠ABC = ∠BAD (By CPCT).
But ∠ABC + ∠BAD = 180° (Co-interior angles since AD || BC).
2 ∠ABC = 180° ⇒ ∠ABC = 90°.

Since ABCD is a rhombus with an angle of 90°, ABCD is a square.

🚀 Aptitude Corner (For NTSE & Olympiad Aspirants)

Practice Question 1: The angle bisectors of a parallelogram always enclose a:
Option A: Rhombus | Option B: Rectangle | Option C: Trapezium | Option D: Square
Quick Answer Key: Option B (Rectangle). Explanation: The consecutive angles of a parallelogram are supplementary. The sum of half of these angles is 90°, making the triangle's third angle 90°.

Practice Question 2: In a quadrilateral ABCD, ∠A + ∠C = 140°. If the other two angles are in the ratio 5:6, find the measure of the largest angle of the quadrilateral.
Solution Trick: Sum of angles ∠B + ∠D = 360° - 140° = 220°. Ratio is 5x + 6x = 220° ⇒ 11x = 220° ⇒ x = 20°. The largest angle among them is 6 × 20° = 120°.

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