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Understanding Quadrilaterals - Types of Quadrilaterals: Trapezium, Kite, Parallelogram

Grade 8ICSE

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

🔑Concepts

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A Trapezium is a quadrilateral with at least one pair of parallel sides. In the figure, side ABAB is parallel to side DCDC. The non-parallel sides are called legs. If the non-parallel sides are equal, it is called an Isosceles Trapezium.

Trapezium ABCD with parallel sides AB and CD
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A Kite is a quadrilateral with two distinct pairs of equal adjacent sides. The diagonals of a kite intersect at right angles (90∘90^{\circ}), and one diagonal bisects the other.

Kite PQRS showing perpendicular diagonals
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A Parallelogram is a quadrilateral where both pairs of opposite sides are parallel and equal. Its opposite angles are equal, and its diagonals bisect each other.

Parallelogram ABCD showing diagonals bisecting each other
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In a parallelogram, consecutive angles are supplementary, meaning their sum is 180∘180^{\circ}. Also, the sum of all interior angles of any quadrilateral is 360∘360^{\circ}.

📐Formulae

Sum of interior angles: ∠A+∠B+∠C+∠D=360∘\angle A + \angle B + \angle C + \angle D = 360^{\circ}

Area of a Parallelogram: Area=Base×HeightArea = Base \times Height

Perimeter of a Parallelogram: P=2(a+b)P = 2(a + b), where aa and bb are lengths of adjacent sides

Area of a Trapezium: Area=12×(sum of parallel sides)×height=12×(a+b)×hArea = \frac{1}{2} \times (sum\ of\ parallel\ sides) \times height = \frac{1}{2} \times (a + b) \times h

Area of a Kite: Area=12×d1×d2Area = \frac{1}{2} \times d_1 \times d_2, where d1d_1 and d2d_2 are the lengths of the diagonals

In a Parallelogram: Adjacent angles ∠A+∠B=180∘\angle A + \angle B = 180^{\circ}

💡Examples

Problem 1:

In a parallelogram ABCDABCD, if ∠A=(2x+10)∘\angle A = (2x + 10)^{\circ} and ∠B=(3x−40)∘\angle B = (3x - 40)^{\circ}, find the measure of all the angles of the parallelogram.

Solution:

  1. In a parallelogram, adjacent angles are supplementary. Therefore, ∠A+∠B=180∘\angle A + \angle B = 180^{\circ}.
  2. Substitute the expressions: (2x+10)+(3x−40)=180(2x + 10) + (3x - 40) = 180.
  3. Simplify: 5x−30=1805x - 30 = 180.
  4. Add 3030 to both sides: 5x=2105x = 210.
  5. Divide by 55: x=42x = 42.
  6. Calculate ∠A\angle A: 2(42)+10=84+10=94∘2(42) + 10 = 84 + 10 = 94^{\circ}.
  7. Calculate ∠B\angle B: 3(42)−40=126−40=86∘3(42) - 40 = 126 - 40 = 86^{\circ}.
  8. Since opposite angles are equal: ∠C=∠A=94∘\angle C = \angle A = 94^{\circ} and ∠D=∠B=86∘\angle D = \angle B = 86^{\circ}.

Explanation:

This problem uses the property that consecutive (adjacent) angles in a parallelogram add up to 180∘180^{\circ} and opposite angles are equal.

Problem 2:

The diagonals of a kite are 12 cm12\ cm and 18 cm18\ cm long. Find the area of the kite. Also, if one of the interior angles formed by the intersection of diagonals is given, what is its value?

Solution:

  1. The formula for the area of a kite is Area=12×d1×d2Area = \frac{1}{2} \times d_1 \times d_2.
  2. Substitute the given values: Area=12×12×18Area = \frac{1}{2} \times 12 \times 18.
  3. Calculate: Area=6×18=108 cm2Area = 6 \times 18 = 108\ cm^2.
  4. By property, the diagonals of a kite always intersect at right angles.
  5. Therefore, the angle formed by the intersection of diagonals is 90∘90^{\circ}.

Explanation:

The solution applies the specific area formula for kites using diagonals and utilizes the geometric property that kite diagonals are perpendicular.

Problem 3:

In the given trapezium PQRSPQRS, PQ∥SRPQ \parallel SR. If ∠P=110∘\angle P = 110^{\circ} and ∠Q=120∘\angle Q = 120^{\circ}, find the measures of ∠S\angle S and ∠R\angle R.

Trapezium PQRS with given angles at P and Q

Solution:

  1. Since PQ∥SRPQ \parallel SR, the consecutive interior angles are supplementary.
  2. ∠P+∠S=180∘\angle P + \angle S = 180^{\circ}
  3. 110∘+∠S=180∘  ⟹  ∠S=180∘−110∘=70∘110^{\circ} + \angle S = 180^{\circ} \implies \angle S = 180^{\circ} - 110^{\circ} = 70^{\circ}
  4. Similarly, ∠Q+∠R=180∘\angle Q + \angle R = 180^{\circ}
  5. 120∘+∠R=180∘  ⟹  ∠R=180∘−120∘=60∘120^{\circ} + \angle R = 180^{\circ} \implies \angle R = 180^{\circ} - 120^{\circ} = 60^{\circ}

Explanation:

Because the top and bottom sides are parallel, the angles on the same side of the transversal (the legs PSPS and QRQR) add up to 180∘180^{\circ}.

Problem 4:

In parallelogram ABCDABCD, the ratio of two adjacent angles is 2:32:3. Find the measure of all four angles.

Parallelogram ABCD with adjacent angles marked as 2x and 3x

Solution:

  1. Let the adjacent angles be 2x2x and 3x3x.
  2. Adjacent angles in a parallelogram are supplementary: 2x+3x=180∘2x + 3x = 180^{\circ}
  3. 5x=180∘  ⟹  x=36∘5x = 180^{\circ} \implies x = 36^{\circ}
  4. First angle: 2×36∘=72∘2 \times 36^{\circ} = 72^{\circ}
  5. Second angle: 3×36∘=108∘3 \times 36^{\circ} = 108^{\circ}
  6. Opposite angles are equal, so the four angles are 72∘,108∘,72∘,108∘72^{\circ}, 108^{\circ}, 72^{\circ}, 108^{\circ}.

Explanation:

In any parallelogram, the sum of angles sharing a common side is always 180∘180^{\circ}.