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Geometry and Measurement - Properties of Quadrilaterals and Polygons

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Quadrilaterals are classified by their properties including parallel sides, equal side lengths, and angle measures. A Trapezium (or Trapezoid) is a quadrilateral with at least one pair of parallel sides. A Parallelogram has two pairs of parallel sides, which implies opposite sides are equal and opposite angles are equal.

A parallelogram labeled ABCD showing parallel opposite sides.
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Special quadrilaterals like Rhombuses and Kites are defined by their diagonals. In a Rhombus, all four sides are equal and diagonals bisect each other at 90∘90^{\circ}. In a Kite, there are two pairs of equal adjacent sides, and one diagonal is bisected by the other at a right angle.

A rhombus with diagonals intersecting at a right angle.
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Interior and Exterior Angles: For any convex nn-sided polygon, the sum of exterior angles is always 360∘360^{\circ}. The interior and exterior angles at any vertex are supplementary, meaning they sum to 180∘180^{\circ} because they lie on a straight line.

A diagram showing the relationship between an interior and exterior angle on a straight line.
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A Regular Polygon is a polygon where all sides are of equal length and all interior angles are of equal measure. Common examples include equilateral triangles, squares, and regular pentagons.

📐Formulae

Sum of interior angles: S=(n−2)×180∘S = (n - 2) \times 180^{\circ}

Individual interior angle of a regular polygon: I=(n−2)×180∘nI = \frac{(n - 2) \times 180^{\circ}}{n}

Sum of exterior angles: 360∘360^{\circ}

Individual exterior angle of a regular polygon: E=360∘nE = \frac{360^{\circ}}{n}

Relationship between interior and exterior angles: I+E=180∘I + E = 180^{\circ}

Area of a trapezium: A=12(a+b)hA = \frac{1}{2}(a + b)h

Area of a kite or rhombus: A=12×d1×d2A = \frac{1}{2} \times d_1 \times d_2 (where d1,d2d_1, d_2 are diagonals)

💡Examples

Problem 1:

Calculate the size of each interior angle in a regular hexagon.

Solution:

Step 1: Identify the number of sides for a hexagon, which is n=6n = 6. Step 2: Use the interior angle sum formula: S=(6−2)×180∘=4×180∘=720∘S = (6 - 2) \times 180^{\circ} = 4 \times 180^{\circ} = 720^{\circ}. Step 3: Since the hexagon is regular, divide the total sum by the number of angles: I=720∘6=120∘I = \frac{720^{\circ}}{6} = 120^{\circ}.

Explanation:

By dividing the hexagon into 4 triangles, we find the total degrees. Dividing by 6 gives the measure of one specific angle because all angles in a regular polygon are identical.

Problem 2:

A regular polygon has an exterior angle of 30∘30^{\circ}. Determine how many sides this polygon has and name the polygon.

Solution:

Step 1: Use the sum of exterior angles property: n=360∘En = \frac{360^{\circ}}{E}. Step 2: Substitute the given exterior angle: n=360∘30∘n = \frac{360^{\circ}}{30^{\circ}}. Step 3: Solve for nn: n=12n = 12. Step 4: A polygon with 12 sides is called a dodecagon.

Explanation:

The sum of exterior angles is always 360∘360^{\circ} regardless of the shape. Dividing this constant by the measure of one exterior angle gives the total number of vertices (sides).

Problem 3:

Calculate the value of xx in the given pentagon where four interior angles are 110∘110^{\circ}, 120∘120^{\circ}, 90∘90^{\circ}, and 100∘100^{\circ}.

An irregular pentagon with four interior angles labeled and one marked as x.

Solution:

  1. Sum of interior angles for n=5n = 5: S=(5−2)×180∘=3×180∘=540∘S = (5 - 2) \times 180^{\circ} = 3 \times 180^{\circ} = 540^{\circ}
  2. Sum the given angles: 110∘+120∘+90∘+100∘=420∘110^{\circ} + 120^{\circ} + 90^{\circ} + 100^{\circ} = 420^{\circ}
  3. Find xx: x=540∘−420∘=120∘x = 540^{\circ} - 420^{\circ} = 120^{\circ}

Explanation:

To find a missing angle in a polygon, we first determine the total sum of all interior angles using the formula (n−2)×180∘(n-2) \times 180^{\circ}. Subtracting the sum of the known angles from this total gives the unknown value.

Problem 4:

Find the area of a kite PQRSPQRS where the diagonals measure 1212 cm and 1818 cm.

A kite with diagonals labeled 12 cm and 18 cm.

Solution:

  1. Use the formula for the area of a kite: A=12×d1×d2A = \frac{1}{2} \times d_1 \times d_2
  2. Substitute the given diagonal lengths: A=12×12×18A = \frac{1}{2} \times 12 \times 18
  3. Calculate the result: A=6×18=108 cm2A = 6 \times 18 = 108\text{ cm}^2

Explanation:

The area of any quadrilateral whose diagonals are perpendicular (like a kite or rhombus) is half the product of the lengths of those diagonals.