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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 Rectilinear figure is a plane figure bounded by straight line segments. A quadrilateral is a four-sided rectilinear polygon where the sum of interior angles is always 360∘360^{\circ}. This can be visualized by drawing a diagonal which divides the quadrilateral into two triangles, each accounting for 180∘180^{\circ}.

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A Parallelogram is a quadrilateral where both pairs of opposite sides are parallel and equal. Visually, opposite sides are like parallel tracks, and opposite angles are congruent (∠A=∠C\angle A = \angle C and ∠B=∠D\angle B = \angle D). Additionally, any two adjacent angles are supplementary, summing to 180∘180^{\circ}.

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The diagonals of a parallelogram bisect each other, meaning they meet at a point that is the midpoint for both diagonal lines. While they cut each other in half, they are not necessarily equal in length unless the figure is a rectangle or a square.

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A Rhombus is a parallelogram with all four sides of equal length. Its unique visual property is that its diagonals bisect each other at right angles (90∘90^{\circ}), effectively partitioning the rhombus into four congruent right-angled triangles.

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A Rectangle is a parallelogram where every interior angle is a right angle (90∘90^{\circ}). Visually, it is perfectly 'upright', and its diagonals are equal in length (AC=BDAC = BD), unlike a general parallelogram.

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A Square is a regular quadrilateral that possesses all the properties of a rhombus and a rectangle. It has four equal sides, four 90∘90^{\circ} angles, and diagonals that are equal, bisect each other at 90∘90^{\circ}, and bisect the vertex angles into 45∘45^{\circ} segments.

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A Trapezium is a quadrilateral with at least one pair of parallel sides. In an Isosceles Trapezium, the non-parallel sides are equal in length, and the base angles are equal, making the figure appear symmetrical along a vertical central axis.

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The Mid-point Theorem states that the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it. In a quadrilateral, joining the mid-points of the four sides in order always results in a parallelogram.

📐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.