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Quadrilaterals - Playing with Quadrilaterals

Grade 8CBSE

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

🔑Concepts

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A quadrilateral is a polygon with four sides, four vertices, and four angles. The sum of the interior angles of a quadrilateral is always 360∘360^\circ. This property can be derived by dividing the quadrilateral into two triangles.

Quadrilateral ABCD divided into two triangles by diagonal AC
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A Parallelogram is a special quadrilateral where opposite sides are parallel and equal. Its opposite angles are equal, and its diagonals bisect each other.

Parallelogram with diagonals bisecting at point O
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A Rhombus is a parallelogram with all four sides of equal length. Its diagonals are perpendicular bisectors of each other, meaning they intersect at 90∘90^\circ.

Rhombus showing perpendicular diagonals
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A Trapezium is a quadrilateral with at least one pair of parallel sides. If the non-parallel sides are equal, it is called an Isosceles Trapezium.

📐Formulae

Sum of interior angles of an n-sided polygon=(n−2)×180∘\text{Sum of interior angles of an } n\text{-sided polygon} = (n - 2) \times 180^\circ

Sum of exterior angles of any polygon=360∘\text{Sum of exterior angles of any polygon} = 360^\circ

Number of diagonals in an n-sided polygon=n(n−3)2\text{Number of diagonals in an } n\text{-sided polygon} = \frac{n(n - 3)}{2}

Each exterior angle of a regular n-gon=360∘n\text{Each exterior angle of a regular } n\text{-gon} = \frac{360^\circ}{n}

Each interior angle of a regular n-gon=(n−2)×180∘n\text{Each interior angle of a regular } n\text{-gon} = \frac{(n - 2) \times 180^\circ}{n}

💡Examples

Problem 1:

Three angles of a quadrilateral are 110∘110^\circ, 70∘70^\circ, and 80∘80^\circ. Find the fourth angle xx.

Solution:

Sum of angles = 110∘+70∘+80∘+x=360∘110^\circ + 70^\circ + 80^\circ + x = 360^\circ. 11070+80260\begin{array}{r} 110 \\ 70 \\ + 80 \\ \hline 260 \end{array} So, 260∘+x=360∘260^\circ + x = 360^\circ. x=360∘−260∘=100∘x = 360^\circ - 260^\circ = 100^\circ.

Explanation:

We use the Angle Sum Property of a quadrilateral which states that the sum of all four interior angles is equal to 360∘360^\circ.

Problem 2:

Find the number of sides of a regular polygon whose each exterior angle has a measure of 45∘45^\circ.

Solution:

Let the number of sides be nn. We know that n=360∘Exterior Anglen = \frac{360^\circ}{\text{Exterior Angle}}. n=360∘45∘=8n = \frac{360^\circ}{45^\circ} = 8.

Explanation:

For any regular polygon, the product of the number of sides and the measure of each exterior angle is always 360∘360^\circ.

Problem 3:

In a parallelogram PQRSPQRS, the adjacent angles ∠P\angle P and ∠Q\angle Q are in the ratio 2:32:3. Find the measure of all angles.

Solution:

Let ∠P=2x\angle P = 2x and ∠Q=3x\angle Q = 3x. In a parallelogram, adjacent angles are supplementary: 2x+3x=180∘2x + 3x = 180^\circ. 5x=180∘  ⟹  x=180∘5=36∘5x = 180^\circ \implies x = \frac{180^\circ}{5} = 36^\circ. ∠P=2×36∘=72∘\angle P = 2 \times 36^\circ = 72^\circ. ∠Q=3×36∘=108∘\angle Q = 3 \times 36^\circ = 108^\circ. Opposite angles are equal, so ∠R=∠P=72∘\angle R = \angle P = 72^\circ and ∠S=∠Q=108∘\angle S = \angle Q = 108^\circ.

Explanation:

In a parallelogram, adjacent angles sum to 180∘180^\circ (they are consecutive interior angles between parallel lines) and opposite angles are equal.

Problem 4:

In the given parallelogram ABCDABCD, find the values of xx and yy if the diagonals ACAC and BDBD intersect at OO, given OA=x+yOA = x + y, OC=16OC = 16, OB=x+7OB = x + 7, and OD=20OD = 20.

Parallelogram ABCD with intersecting diagonals AC and BD

Solution:

  1. In a parallelogram, diagonals bisect each other.
  2. Therefore, OA=OCOA = OC and OB=ODOB = OD.
  3. From OB=ODOB = OD, we have: x+7=20x + 7 = 20 x=20−7x = 20 - 7 x=13x = 13
  4. From OA=OCOA = OC, we have: x+y=16x + y = 16 13+y=1613 + y = 16 y=16−13y = 16 - 13 y=3y = 3 Final Answer: x=13x = 13, y=3y = 3.

Explanation:

We use the property that diagonals of a parallelogram bisect each other to set up two linear equations and solve for the unknown variables.

Problem 5:

Find the value of xx in the following quadrilateral where three exterior angles are 100∘100^\circ, 80∘80^\circ, and 70∘70^\circ.

Quadrilateral with extended sides showing exterior angles

Solution:

  1. The sum of the exterior angles of any polygon is always 360∘360^\circ.
  2. Let the fourth exterior angle be xx.
  3. 100∘+80∘+70∘+x=360∘100^\circ + 80^\circ + 70^\circ + x = 360^\circ
  4. 250∘+x=360∘250^\circ + x = 360^\circ
  5. x=360∘−250∘x = 360^\circ - 250^\circ
  6. x=110∘x = 110^\circ Final Answer: x=110∘x = 110^\circ.

Explanation:

This solution applies the Exterior Angle Sum Property, which states that the sum of exterior angles of any convex polygon is 360∘360^\circ.