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Quantitative Aptitude

Quadrilaterals & Polygons

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Concepts (2)

Master polygon interior/exterior angles and diagonals for SSC CGL. Use `(n-2)*180°` for sum of interior angles, `360°/n` for exterior angles, and `n(n-3)/2` for diagonals.

Core Formula

Polygons are closed figures made of straight line segments. For an n-sided polygon:

  1. Sum of Interior Angles: The sum of all interior angles of any n-sided polygon is given by: Sum = (n - 2) * 180°

  2. Each Interior Angle (for a Regular Polygon): If the polygon is regular (all sides and angles equal), each interior angle is: Each Interior Angle = (n - 2) * 180° / n

  3. Sum of Exterior Angles: The sum of all exterior angles of any convex polygon is always: Sum = 360°

  4. Each Exterior Angle (for a Regular Polygon): For a regular polygon, each exterior angle is: Each Exterior Angle = 360° / n

  5. Relationship between Interior and Exterior Angles: For any vertex of a polygon, the interior angle and its corresponding exterior angle are supplementary: Interior Angle + Exterior Angle = 180°

  6. Number of Diagonals: The number of diagonals in an n-sided polygon is given by: Number of Diagonals = n * (n - 3) / 2

Worked Example 1

Question: Each interior angle of a regular polygon is 150°. The number of sides of the polygon is:

Solution:

  • Method 1 (Using Interior Angle Formula): Each Interior Angle = (n - 2) * 180° / n 150 = (n - 2) * 180 / n 150n = 180n - 360 30n = 360 n = 360 / 30 n = 12

  • Method 2 (Using Exterior Angle - Shortcut): Exterior Angle = 180° - Interior Angle Exterior Angle = 180° - 150° = 30° Each Exterior Angle = 360° / n 30 = 360 / n n = 360 / 30 n = 12

Worked Example 2

Question: The number of diagonals in a decagon (10-sided polygon) is:

Solution:

  • A decagon has n = 10 sides.
  • Using the formula for the number of diagonals: Number of Diagonals = n * (n - 3) / 2 Number of Diagonals = 10 * (10 - 3) / 2 Number of Diagonals = 10 * 7 / 2 Number of Diagonals = 70 / 2 Number of Diagonals = 35

Shortcuts & Tricks

  1. Exterior Angle First: When asked to find the number of sides (n) given an interior angle, always calculate the exterior angle first (180° - Interior Angle). Then use n = 360° / Exterior Angle. This is usually faster and less prone to calculation errors than using the interior angle formula directly.
  2. Memorize Common Polygons: Quickly recall properties for common polygons:
    • Triangle (n=3): Sum Int. = 180°, Each Int. = 60°
    • Square (n=4): Sum Int. = 360°, Each Int. = 90°
    • Pentagon (n=5): Sum Int. = 540°, Each Int. = 108°
    • Hexagon (n=6): Sum Int. = 720°, Each Int. = 120°
    • Octagon (n=8): Sum Int. = 1080°, Each Int. = 135°
  3. Diagonals Check: For small n, you can visualize. For n=3 (triangle), 3*(3-3)/2 = 0 diagonals. Correct.

Common Mistakes

  1. Confusing Interior and Exterior Angles: Always remember Interior Angle + Exterior Angle = 180°. Mixing them up leads to incorrect calculations.
  2. Incorrect n in Sum of Interior Angles: Using n * 180° instead of (n - 2) * 180°. The (n-2) factor is crucial as it represents the number of triangles a polygon can be divided into from one vertex.
  3. Calculation Errors for Diagonals: Forgetting the (n-3) or dividing by 2 in the diagonals formula n * (n - 3) / 2.
  4. Applying Regular Polygon Formulas to Irregular Polygons: Formulas for each interior/exterior angle ((n-2)*180/n and 360/n) only apply to regular polygons. The sum formulas ((n-2)*180 and 360) apply to any convex polygon.

Derivation (brief)

Sum of Interior Angles: Imagine an n-sided polygon. Pick any vertex and draw diagonals to all other non-adjacent vertices. This divides the polygon into (n-2) triangles. Since the sum of angles in each triangle is 180°, the total sum of interior angles of the polygon is (n-2) * 180°.

Sum of Exterior Angles: Consider walking around the perimeter of a polygon. At each vertex, you turn by the exterior angle. By the time you complete one full circuit and return to your starting point facing the original direction, you would have made a full 360° turn. Hence, the sum of all exterior angles is 360°.

Advanced Examples

Question: If the ratio of an interior angle to an exterior angle of a regular polygon is 5:1, find the number of sides of the polygon.

Solution:

  • Let the interior angle be I and the exterior angle be E.
  • Given I : E = 5 : 1, so I = 5E.
  • We know that I + E = 180° (for any regular polygon).
  • Substitute I = 5E into the equation: 5E + E = 180°
  • 6E = 180°
  • E = 180° / 6 = 30°
  • Now, use the formula for the exterior angle of a regular polygon: E = 360° / n
  • 30° = 360° / n
  • n = 360° / 30°
  • n = 12 The polygon has 12 sides (a dodecagon).

Question: The sum of interior angles of a regular octagon is:

Solution:

  • An octagon has n = 8 sides.
  • Using the formula for the sum of interior angles: Sum = (n - 2) * 180° Sum = (8 - 2) * 180° Sum = 6 * 180° Sum = 1080°

Variation Types

  1. Finding n from Diagonals: You might be given the number of diagonals and asked to find n. This involves solving a quadratic equation: n(n-3)/2 = D => n^2 - 3n - 2D = 0. For SSC, often the numbers are small enough to test options or factorize easily.
  2. Irregular Polygons: For irregular polygons, you cannot find each angle without more information, but the sum of interior angles (n-2)*180° and sum of exterior angles 360° still hold true. Questions might involve finding an unknown angle given other angles.
  3. Combined Problems: Questions combining two or more formulas, like the ratio problem above, or finding the number of diagonals given an interior angle.

Time-Saving Methods

  • Practice Mental Math: Quickly calculate (n-2)*180 or 360/n for common n values.
  • Option Elimination: For questions asking for n, if options are provided, you can often plug them into the formula (especially the exterior angle shortcut) to quickly find the correct answer without full algebraic manipulation.
  • Visualisation: For small n, quickly sketch the polygon to confirm the number of diagonals or approximate angle sizes.
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Quadrilaterals sum to 360°. Master properties of parallelogram, rectangle, rhombus, square, trapezium, kite, especially diagonal relations and area formulas for quick SSC CGL problem-solving.

Core Formulas

Quadrilaterals are four-sided polygons. The sum of their interior angles is always 360°.

Key Quadrilaterals and their essential properties:

  1. Parallelogram:

    • Opposite sides are parallel and equal.
    • Opposite angles are equal. Adjacent angles are supplementary (sum to 180°).
    • Diagonals bisect each other.
    • Area = base × height

  2. Rectangle:

    • A parallelogram with all angles equal to 90°.
    • Diagonals are equal and bisect each other.
    • Area = length × width
    • Diagonal d = sqrt(l^2 + w^2)

  3. Rhombus:

    • A parallelogram with all four sides equal.
    • Diagonals bisect each other at 90°.
    • Diagonals bisect the vertex angles.
    • Area = (1/2) × d1 × d2 (where d1, d2 are diagonals)
    • Side s = (1/2) × sqrt(d1^2 + d2^2) (Pythagorean relation with half-diagonals)

  4. Square:

    • A rectangle with all sides equal, or a rhombus with all angles 90°.
    • All sides equal, all angles 90°.
    • Diagonals are equal, bisect each other at 90°, and bisect vertex angles.
    • Area = side^2 or (1/2) × d^2
    • Diagonal d = s × sqrt(2)

  5. Trapezium (Trapezoid):

    • Only one pair of opposite sides is parallel.
    • Area = (1/2) × (sum of parallel sides) × height

  6. Kite:

    • Two pairs of adjacent sides are equal.
    • Diagonals intersect at 90°.
    • One diagonal bisects the other diagonal.
    • Area = (1/2) × d1 × d2

Worked Example 1

Question: The diagonals of a rhombus are 12 cm and 16 cm. The side of the rhombus is: Solution:

  1. In a rhombus, diagonals bisect each other at 90°. This forms four right-angled triangles.
  2. The half-diagonals are 12/2 = 6 cm and 16/2 = 8 cm.
  3. The side of the rhombus is the hypotenuse of these right-angled triangles.
  4. Using the Pythagorean theorem: side^2 = (6)^2 + (8)^2
  5. side^2 = 36 + 64 = 100
  6. side = sqrt(100) = 10 cm.

Worked Example 2

Question: In a parallelogram ABCD, angle A = 70°. What is the measure of angle C? Solution:

  1. In a parallelogram, opposite angles are equal.
  2. Angle A and Angle C are opposite angles.
  3. Therefore, Angle C = Angle A = 70°.

Shortcuts & Tricks

  • Pythagorean Triplets: For rhombus side/diagonal problems or rectangle diagonal problems, recognize common triplets (e.g., 3-4-5, 6-8-10, 5-12-13) to quickly find missing lengths without full calculation.
  • Parallelogram Angles: Remember adjacent angles sum to 180° and opposite angles are equal. This allows for quick mental calculations.
  • Square Diagonals: If side s, diagonal d = s√2. If diagonal d, side s = d/√2. Area can be (1/2)d^2. This saves time in square-related problems.
  • Area Formulas: Memorize (1/2)d1d2 for rhombus/kite and (1/2)(a+b)h for trapezium. These are direct applications.

Common Mistakes

  1. Confusing Properties: Mixing up properties between different quadrilaterals, e.g., assuming diagonals of a rectangle are perpendicular (they are not, unless it's a square).
  2. Incorrect Angle Rules: Forgetting that adjacent angles in a parallelogram are supplementary, or assuming all angles are 90°.
  3. Missing 1/2 in Area Formulas: Frequently forgetting the 1/2 factor in the area formulas for rhombus, kite, and trapezium.

Derivation (brief)

The sum of interior angles of any quadrilateral is 360°. This can be briefly understood by drawing any diagonal within the quadrilateral, dividing it into two triangles. Since the sum of angles in each triangle is 180°, the total sum for the quadrilateral is 180° + 180° = 360°.

Advanced Examples

Question 1: The perimeter of a rhombus is 52 cm. One of its diagonals is 10 cm. Find the area of the rhombus. Solution:

  1. Perimeter of rhombus = 4 × side. So, 4s = 52 cm => s = 13 cm.
  2. Diagonals of a rhombus bisect each other at 90°. Let the given diagonal be d1 = 10 cm. So, half of d1 is 10/2 = 5 cm.
  3. Consider one of the right-angled triangles formed by a side and half-diagonals. The hypotenuse is the side (13 cm), one leg is d1/2 (5 cm), and the other leg is d2/2.
  4. Using Pythagoras: (d2/2)^2 = s^2 - (d1/2)^2 = 13^2 - 5^2 = 169 - 25 = 144.
  5. So, d2/2 = sqrt(144) = 12 cm. This means d2 = 2 × 12 = 24 cm.
  6. Area of rhombus = (1/2) × d1 × d2 = (1/2) × 10 × 24 = 120 cm^2.

Question 2: An isosceles trapezium has parallel sides of length 12 cm and 20 cm. The non-parallel sides are each 5 cm. Find its area. Solution:

  1. Draw perpendiculars from the endpoints of the shorter parallel side (12 cm) to the longer parallel side (20 cm). This forms a rectangle in the middle and two congruent right-angled triangles at the ends.
  2. The base of the rectangle is 12 cm. The remaining length on the longer side is 20 - 12 = 8 cm.
  3. Since it's an isosceles trapezium, this 8 cm is split equally between the two right-angled triangles, so each triangle has a base of 8/2 = 4 cm.
  4. The non-parallel side (5 cm) is the hypotenuse of these right-angled triangles. Let 'h' be the height of the trapezium.
  5. Using Pythagoras: h^2 = 5^2 - 4^2 = 25 - 16 = 9.
  6. So, h = sqrt(9) = 3 cm.
  7. Area of trapezium = (1/2) × (sum of parallel sides) × height = (1/2) × (12 + 20) × 3
  8. Area = (1/2) × 32 × 3 = 16 × 3 = 48 cm^2.

Variation Types

  • Cyclic Quadrilaterals: All four vertices lie on a circle. Opposite angles sum to 180°. (e.g., a rectangle is always cyclic).
  • Tangential Quadrilaterals: All four sides are tangent to an inscribed circle. The sum of opposite sides are equal (a+c = b+d).

Time-Saving Methods

  • Visualisation: Quickly sketch the quadrilateral and label known values. This helps in identifying the relevant properties and formulas.
  • Property Recall: Instantly recall specific properties for each quadrilateral type (e.g., diagonals of a rhombus are perpendicular) rather than deriving them.
  • Formula Application: Directly apply the correct area or diagonal formula. Avoid re-deriving basic relationships during the exam.
  • Option Elimination: Sometimes, properties (like angle sums or side relations) can help eliminate incorrect options quickly without full calculation.
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