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Geometry - Rectilinear Figures (Quadrilaterals and Parallelograms)

Grade 9ICSE

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

🔑Concepts

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A quadrilateral is a closed figure with four sides. The Angle Sum Property states that the sum of the interior angles is always 360∘360^{\circ}. For any quadrilateral ABCDABCD, ∠A+∠B+∠C+∠D=360∘\angle A + \angle B + \angle C + \angle D = 360^{\circ}.

A general quadrilateral ABCD showing four vertices.
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A Parallelogram is a quadrilateral where opposite sides are parallel and equal. Key properties include: (i) Opposite angles are equal, (ii) Diagonals bisect each other, and (iii) Adjacent angles are supplementary (sum to 180∘180^{\circ}).

Parallelogram ABCD with diagonals AC and BD bisecting at point O.
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A Rhombus is a parallelogram with all four sides equal. Its diagonals bisect each other at right angles (90∘90^{\circ}).

Rhombus showing diagonals intersecting at 90 degrees.
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A Rectangle is a parallelogram with each angle equal to 90∘90^{\circ}. Its diagonals are equal in length and bisect each other.

Rectangle with equal diagonals.

📐Formulae

Angle Sum Property: ∠A+∠B+∠C+∠D=360∘\angle A + \angle B + \angle C + \angle D = 360^{\circ}

Area of a Parallelogram: Base×Height\text{Base} \times \text{Height}

Area of a Rhombus: 12×d1×d2\frac{1}{2} \times d_1 \times d_2 (where d1d_1 and d2d_2 are lengths of diagonals)

Area of a Trapezium: 12×(sum of parallel sides)×h\frac{1}{2} \times (\text{sum of parallel sides}) \times h

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

Pythagorean relationship in Rhombus side (ss): s2=(d12)2+(d22)2s^2 = (\frac{d_1}{2})^2 + (\frac{d_2}{2})^2

💡Examples

Problem 1:

In a parallelogram ABCDABCD, ∠A:∠B=2:3\angle A : \angle B = 2 : 3. Find the measure of all four angles.

Solution:

  1. Let the angles be 2x2x and 3x3x.
  2. In a parallelogram, adjacent angles are supplementary, so ∠A+∠B=180∘\angle A + \angle B = 180^{\circ}.
  3. 2x+3x=180∘  ⟹  5x=180∘  ⟹  x=36∘2x + 3x = 180^{\circ} \implies 5x = 180^{\circ} \implies x = 36^{\circ}.
  4. Therefore, ∠A=2×36∘=72∘\angle A = 2 \times 36^{\circ} = 72^{\circ} and ∠B=3×36∘=108∘\angle B = 3 \times 36^{\circ} = 108^{\circ}.
  5. Since opposite angles are equal: ∠C=∠A=72∘\angle C = \angle A = 72^{\circ} and ∠D=∠B=108∘\angle D = \angle B = 108^{\circ}.

Explanation:

This approach uses the property that consecutive interior angles between parallel lines (the sides of the parallelogram) sum to 180∘180^{\circ}.

Problem 2:

The diagonals of a rhombus are 24 cm24\text{ cm} and 10 cm10\text{ cm}. Calculate the length of one side of the rhombus.

Solution:

  1. Let the diagonals be d1=24 cmd_1 = 24\text{ cm} and d2=10 cmd_2 = 10\text{ cm}.
  2. Diagonals of a rhombus bisect each other at 90∘90^{\circ}.
  3. Half-lengths of the diagonals are 242=12 cm\frac{24}{2} = 12\text{ cm} and 102=5 cm\frac{10}{2} = 5\text{ cm}.
  4. These halves form the base and height of a right-angled triangle where the side (ss) is the hypotenuse.
  5. Using Pythagoras Theorem: s2=122+52=144+25=169s^2 = 12^2 + 5^2 = 144 + 25 = 169.
  6. s=169=13 cms = \sqrt{169} = 13\text{ cm}.

Explanation:

This solution relies on the property that rhombus diagonals create four right-angled triangles at the center intersection.

Problem 3:

In the given figure, ABCDABCD is a trapezium where AB∥DCAB \parallel DC. If ∠A=55∘\angle A = 55^{\circ} and ∠B=70∘\angle B = 70^{\circ}, find the measures of ∠C\angle C and ∠D\angle D.

Trapezium ABCD with parallel sides AB and DC.

Solution:

  1. Since AB∥DCAB \parallel DC, ∠A\angle A and ∠D\angle D are interior angles on the same side of the transversal ADAD.
  2. Therefore, ∠A+∠D=180∘\angle A + \angle D = 180^{\circ}.
  3. 55∘+∠D=180∘  ⟹  ∠D=180∘−55∘=125∘55^{\circ} + \angle D = 180^{\circ} \implies \angle D = 180^{\circ} - 55^{\circ} = 125^{\circ}.
  4. Similarly, ∠B\angle B and ∠C\angle C are interior angles on the same side of the transversal BCBC.
  5. ∠B+∠C=180∘  ⟹  70∘+∠C=180∘  ⟹  ∠C=110∘\angle B + \angle C = 180^{\circ} \implies 70^{\circ} + \angle C = 180^{\circ} \implies \angle C = 110^{\circ}.

Explanation:

In a trapezium, the pairs of angles between the parallel sides and a non-parallel side (consecutive interior angles) are supplementary.

Problem 4:

In a rectangle PQRSPQRS, the diagonals PRPR and QSQS intersect at OO. If ∠OPQ=35∘\angle OPQ = 35^{\circ}, calculate ∠ORQ\angle ORQ and ∠SQR\angle SQR.

Rectangle PQRS with diagonals intersecting at O and angle OPQ marked.

Solution:

  1. In rectangle PQRSPQRS, diagonals are equal and bisect each other, so OP=OQ=OR=OSOP = OQ = OR = OS.
  2. In △OPQ\triangle OPQ, since OP=OQOP = OQ, it is an isosceles triangle. Thus, ∠OQP=∠OPQ=35∘\angle OQP = \angle OPQ = 35^{\circ}.
  3. In △OQR\triangle OQR, since OQ=OROQ = OR, it is an isosceles triangle. Since ∠PQR=90∘\angle PQR = 90^{\circ}, ∠OQR=90∘−35∘=55∘\angle OQR = 90^{\circ} - 35^{\circ} = 55^{\circ}.
  4. Therefore, ∠ORQ=∠OQR=55∘\angle ORQ = \angle OQR = 55^{\circ}.
  5. Since PQ∥SRPQ \parallel SR, ∠SQR=∠OPQ=35∘\angle SQR = \angle OPQ = 35^{\circ} (Alternate Interior Angles).

Explanation:

This problem uses the property that the diagonals of a rectangle are equal and bisect each other, forming four isosceles triangles with the center point.