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Understanding Quadrilaterals - Types of Quadrilaterals

Grade 8CBSE

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

🔑Concepts

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A Parallelogram is a quadrilateral where both pairs of opposite sides are parallel. Key properties include: opposite sides are equal, opposite angles are equal, and diagonals bisect each other.

A parallelogram ABCD showing parallel opposite sides.
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A Rhombus is a special parallelogram where all four sides are equal in length. Crucially, its diagonals bisect each other at right angles (90∘90^\circ).

A rhombus showing diagonals intersecting at 90 degrees.
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A Rectangle is a parallelogram with four right angles. Its diagonals are equal in length and bisect each other.

A rectangle showing equal diagonals.
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A Square is a regular quadrilateral. It has four equal sides and four right angles. It possesses all properties of a rectangle, rhombus, and parallelogram.

A square with all sides equal to s.
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A Trapezium has exactly one pair of parallel sides. If the non-parallel sides are equal, it is called an Isosceles Trapezium.

A trapezium with one pair of parallel sides.
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A Kite has two pairs of equal-length sides that are adjacent to each other. Its diagonals are perpendicular, and one diagonal bisects the other.

📐Formulae

Sum of interior angles of a quadrilateral = 360∘360^\circ

Sum of exterior angles of any convex quadrilateral = 360∘360^\circ

Number of diagonals in a quadrilateral = n(n−3)2=4(4−3)2=2\frac{n(n-3)}{2} = \frac{4(4-3)}{2} = 2

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 d1,d2d_1, d_2 are diagonals)

Area of a Trapezium = 12×(a+b)×h\frac{1}{2} \times (a + b) \times h (where a,ba, b are parallel sides and hh is the height)

Perimeter of a Quadrilateral = Sum of all four sides\text{Sum of all four sides}

💡Examples

Problem 1:

In a parallelogram ABCDABCD, the measure of ∠A\angle A is 70∘70^\circ. Find the measures of the remaining angles ∠B\angle B, ∠C\angle C, and ∠D\angle D.

Solution:

  1. In a parallelogram, adjacent angles are supplementary. Therefore, ∠A+∠B=180∘\angle A + \angle B = 180^\circ.
  2. Substitute the given value: 70∘+∠B=180∘  ⟹  ∠B=180∘−70∘=110∘70^\circ + \angle B = 180^\circ \implies \angle B = 180^\circ - 70^\circ = 110^\circ.
  3. Opposite angles of a parallelogram are equal. Therefore, ∠C=∠A=70∘\angle C = \angle A = 70^\circ and ∠D=∠B=110∘\angle D = \angle B = 110^\circ.

Explanation:

This solution uses the property that consecutive angles in a parallelogram sum to 180∘180^\circ and opposite angles are congruent.

Problem 2:

The diagonals of a rhombus are 1616 cm and 1212 cm. Find the length of each side of the rhombus.

Solution:

  1. Let the diagonals be d1=16d_1 = 16 cm and d2=12d_2 = 12 cm. They bisect each other at 90∘90^\circ.
  2. The half-lengths of the diagonals are 162=8\frac{16}{2} = 8 cm and 122=6\frac{12}{2} = 6 cm.
  3. These half-lengths form the legs of a right-angled triangle where the side of the rhombus (ss) is the hypotenuse.
  4. Using Pythagoras theorem: s2=82+62=64+36=100s^2 = 8^2 + 6^2 = 64 + 36 = 100.
  5. s=100=10s = \sqrt{100} = 10 cm.

Explanation:

Since diagonals of a rhombus are perpendicular bisectors, we can use the Pythagorean theorem on one of the four internal right-angled triangles to find the side length.

Problem 3:

In the given rectangle PQRSPQRS, the diagonals intersect at point OO. If OP=2x+4OP = 2x + 4 and OS=3x+1OS = 3x + 1, find the value of xx.

Rectangle PQRS with diagonals intersecting at O.

Solution:

  1. In a rectangle, diagonals are equal in length and bisect each other.
  2. This means PR=QSPR = QS.
  3. Since diagonals bisect each other, OP=OR=12PROP = OR = \frac{1}{2}PR and OQ=OS=12QSOQ = OS = \frac{1}{2}QS.
  4. Therefore, OP=OSOP = OS.
  5. Equating the expressions: 2x+4=3x+12x + 4 = 3x + 1.
  6. Subtracting 2x2x from both sides: 4=x+14 = x + 1.
  7. Subtracting 11 from both sides: x=3x = 3.

Explanation:

Because the diagonals of a rectangle are equal and bisect each other, the segments from the center to any vertex are equal in length. This allows us to set the two algebraic expressions for the segments equal to each other and solve for xx.

Problem 4:

In the isosceles trapezium ABCDABCD where AB∥DCAB \parallel DC and AD=BCAD = BC, if ∠D=110∘\angle D = 110^\circ, find the measure of ∠A\angle A.

Isosceles trapezium ABCD with angle D marked as 110 degrees.

Solution:

  1. In a trapezium, adjacent angles between parallel lines are supplementary (they add up to 180∘180^\circ).
  2. Since AB∥DCAB \parallel DC, ∠A+∠D=180∘\angle A + \angle D = 180^\circ.
  3. Given ∠D=110∘\angle D = 110^\circ, we have ∠A+110∘=180∘\angle A + 110^\circ = 180^\circ.
  4. Therefore, ∠A=180∘−110∘=70∘\angle A = 180^\circ - 110^\circ = 70^\circ.

Explanation:

Consecutive interior angles formed by a transversal (the non-parallel side ADAD) intersecting two parallel lines (ABAB and DCDC) are always supplementary.