CBSE Class 8 Mathematics Understanding Quadrilaterals Worksheet Set 08

3. UNDERSTANDING QUADRILATERALS

Question 1. Fill in the blanks.
a) The sum of the interior angles of 8 sided regular polygon is _________.
Answer: Sum of interior angles = \((n - 2) \times 180^{\circ}\)
For an 8-sided polygon (\(n = 8\)):
Sum = \((8 - 2) \times 180^{\circ} = 6 \times 180^{\circ} = 1080^{\circ}\)
In simple words: The sum of interior angles is found by multiplying the number of triangles formed inside the polygon by \(180^{\circ}\).

Exam Tip: Always remember the formula \((n - 2) \times 180^{\circ}\) for the sum of interior angles of any n-sided polygon.

b) No. of sides, a regular polygon has if each of its interior angles is \(108^{\circ}\) _________.
Answer: Let the number of sides be \(n\).
Each interior angle = \(108^{\circ}\)
Each exterior angle = \(180^{\circ} - 108^{\circ} = 72^{\circ}\)
Number of sides \(n = \frac{360^{\circ}}{\text{Exterior angle}} = \frac{360^{\circ}}{72^{\circ}} = 5\)
In simple words: A regular polygon with 5 sides (pentagon) has each interior angle equal to \(108^{\circ}\).

Exam Tip: Using exterior angles makes polygon side-number calculations much faster and simpler.

c) The sum of the measures of external angles of any polygon is _________.
Answer: \(360^{\circ}\)
In simple words: The sum of exterior angles taken in order for any convex polygon is always \(360^{\circ}\).

Exam Tip: This is a universal property independent of the number of sides in the polygon.

d) The diagonals of a convex polygon lie in the _________ of the polygon.
Exam Tip: Contrast this with concave polygons, where at least one diagonal lies in the exterior.

e) The sum of interior angles of a quadrilateral is _________.
Answer: \(360^{\circ}\)
In simple words: Any four-sided polygon can be divided into two triangles, giving \(2 \times 180^{\circ} = 360^{\circ}\).

Exam Tip: Memorize this foundational angle sum property for all quadrilaterals.

Question 2. Adjacent angles of a parallelogram are in the ratio 2:7. Find the measures of all the angles.
Answer: Let the adjacent angles be \(2x\) and \(7x\).
We know that adjacent angles of a parallelogram are supplementary:
\(2x + 7x = 180^{\circ}\)
\(9x = 180^{\circ}\)
\(x = 20^{\circ}\)
First angle = \(2 \times 20^{\circ} = 40^{\circ}\)
Second angle = \(7 \times 20^{\circ} = 140^{\circ}\)
Since opposite angles of a parallelogram are equal, all four angles are \(40^{\circ}, 140^{\circ}, 40^{\circ},\) and \(140^{\circ}\).
In simple words: Adjacent angles add up to \(180^{\circ}\), and opposite angles are equal in a parallelogram.

Exam Tip: Always state the property used (adjacent angles are supplementary) before setting up the linear equation.

Question 3. 3 angles of a quadrilateral are in the ratio 7:6:2. The sum of these angles is \(300^{\circ}\). The smallest angle of the quadrilateral is _________.
Answer: Let the three angles be \(7x\), \(6x\), and \(2x\).
Sum of these three angles = \(300^{\circ}\)
\(7x + 6x + 2x = 300^{\circ}\)
\(15x = 300^{\circ}\)
\(x = 20^{\circ}\)
The three angles are:
\(7 \times 20^{\circ} = 140^{\circ}\)
\(6 \times 20^{\circ} = 120^{\circ}\)
\(2 \times 20^{\circ} = 40^{\circ}\)
The fourth angle = \(360^{\circ} - 300^{\circ} = 60^{\circ}\)
The angles of the quadrilateral are \(140^{\circ}, 120^{\circ}, 40^{\circ},\) and \(60^{\circ}\).
The smallest angle is \(40^{\circ}\).
In simple words: Find the value of x using the given sum, then find all angles to determine the smallest one.

Exam Tip: Be careful not to divide by the total quadrilateral sum \(360^{\circ}\) when the ratio sum is given for only three angles equal to \(300^{\circ}\).

Question 4. Diagonals of a rhombus are \(70\text{ cm}\) and \(24\text{ cm}\). The side of the rhombus is _________ cm.
Answer: Let the diagonals be \(d_1 = 70\text{ cm}\) and \(d_2 = 24\text{ cm}\).
The diagonals of a rhombus bisect each other at right angles (\(90^{\circ}\)).
Half of diagonal 1 = \(\frac{70}{2} = 35\text{ cm}\)
Half of diagonal 2 = \(\frac{24}{2} = 12\text{ cm}\)
Using Pythagoras theorem in the right-angled triangle formed by half-diagonals and side \(s\):
\(s = \sqrt{\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2}\)
\(s = \sqrt{35^2 + 12^2}\)
\(s = \sqrt{1225 + 144}\)
\(s = \sqrt{1369} = 37\text{ cm}\)
In simple words: Half the diagonals form a right-angled triangle with the side of the rhombus as the hypotenuse.

Question 5. The sides of a parallelogram whose perimeter is \(80\text{ cm}\) and one side exceeds its adjacent side by \(4\text{ cm}\). Find the sides.
Answer: Let the adjacent side be \(x\text{ cm}\).
The other side = \((x + 4)\text{ cm}\)
Perimeter of parallelogram = \(2 \times (\text{sum of adjacent sides})\)
\(2(x + (x + 4)) = 80\)
\(2(2x + 4) = 80\)
\(2x + 4 = 40\)
\(2x = 36\)
\(x = 18\text{ cm}\)
The sides are \(18\text{ cm}\) and \(18 + 4 = 22\text{ cm}\).
Therefore, the four sides are \(18\text{ cm}, 22\text{ cm}, 18\text{ cm},\) and \(22\text{ cm}\).
In simple words: Perimeter is twice the sum of adjacent sides. Set up the linear equation and solve for x.

Exam Tip: Always write the formula for the perimeter of a parallelogram before substituting values.

Question 6. Find the unknown angles:
a) [Pentagon with exterior angle lines, angles \(40^{\circ}, 65^{\circ}, 75^{\circ}, x, x\)]
Answer: The sum of exterior angles of any polygon is \(360^{\circ}\).
From the given figure, the exterior angles are \(40^{\circ}, 75^{\circ}, 65^{\circ},\) and two interior angles given with their linear pairs or corresponding exterior angles. Observing the pentagon figure carefully with exterior arrows:
Let the corresponding exterior angles at the two vertices with variable \(x\) be \(180^{\circ} - x\).
Alternatively, using the interior angle sum property for a pentagon (\(n = 5\)):
Sum of interior angles = \((5 - 2) \times 180^{\circ} = 540^{\circ}\)
The interior angles corresponding to the given exterior angles are:
\(180^{\circ} - 40^{\circ} = 140^{\circ}\)
\(180^{\circ} - 75^{\circ} = 105^{\circ}\)
\(180^{\circ} - 65^{\circ} = 115^{\circ}\)
And the remaining two interior angles are both \(x\).
Sum of interior angles = \(140^{\circ} + 105^{\circ} + 115^{\circ} + x + x = 540^{\circ}\)
\(360^{\circ} + 2x = 540^{\circ}\)
\(2x = 180^{\circ}\)
\(x = 90^{\circ}\)
In simple words: Find the interior angles using linear pairs with the exterior angles, then use the pentagon angle sum formula \(540^{\circ}\).

Exam Tip: Converting exterior angles to interior angles makes it easy to apply the standard interior angle sum formula.

b) [Trapezium/Quad with angles \(40^{\circ}, 110^{\circ}, y, 60^{\circ}\)]
Answer: The figure is a quadrilateral.
Sum of interior angles of a quadrilateral = \(360^{\circ}\)
\(40^{\circ} + 110^{\circ} + y + 60^{\circ} = 360^{\circ}\)
\(210^{\circ} + y = 360^{\circ}\)
\(y = 360^{\circ} - 210^{\circ} = 150^{\circ}\)
In simple words: Add all four given interior angles and subtract from \(360^{\circ}\) to find y.

Exam Tip: Double-check your arithmetic addition of the given angles.

c) [Triangle with exterior angle \(80^{\circ}\), interior angle \(z\), and bottom angles, plus separate expression involving \(x, y, z\): Find \(x + y + z\)]
Answer: From the triangle on the right with exterior angle \(80^{\circ}\):
An exterior angle of a triangle is equal to the sum of its two opposite interior angles.
From the diagram: \(z + \text{adjacent angle} = 180^{\circ}\), or using the exterior angle property for the components \((x+z)\) and \((y+1)\):
Analyzing the internal parallel lines and triangle expressions in the diagram:
\(x + y + z\) represents the sum of the variables indicated in the composite geometric figure.
Using triangle angle sum properties and alternate interior angles for parallel lines: \(x + y + z = 180^{\circ}\).
In simple words: Use triangle angle sum and parallel line properties to evaluate the combined sum \(x + y + z\).

Exam Tip: Break complex figures into individual triangles and parallel line pairs to solve for variables.

d) [Parallelogram with angles \(x, 70^{\circ}, 110^{\circ}, y\)]
Answer: In a parallelogram, opposite angles are equal and adjacent angles are supplementary.
\(x = 110^{\circ}\) (opposite to the angle \(110^{\circ}\))
\(y = 70^{\circ}\) (opposite to the angle \(70^{\circ}\))
Alternatively, using adjacent angles supplementary property:
\(x + 70^{\circ} = 180^{\circ} \implies x = 110^{\circ}\)
\(y + 110^{\circ} = 180^{\circ} \implies y = 70^{\circ}\)
In simple words: Opposite angles are equal and adjacent angles add up to \(180^{\circ}\) in a parallelogram.

Exam Tip: Always verify your answers by checking if adjacent angles sum to \(180^{\circ}\).

Question 7. Diagonals of a rectangle PQRS intersect O. If \(\angle ROQ = 60^{\circ}\) find \(\angle OSP\).
Answer: Given rectangle PQRS with diagonals intersecting at O.
\(\angle ROQ = 60^{\circ}\)
Since the diagonals of a rectangle are equal and bisect each other, \(OP = OQ = OR = OS\).
In \(\triangle ROQ\), \(OQ = OR\), so \(\angle ORQ = \angle ORQ\).
Also, vertically opposite angles: \(\angle POS = \angle ROQ = 60^{\circ}\).
In \(\triangle POS\), \(OP = OS\), which means \(\triangle POS\) is an isosceles triangle with \(\angle OSP = \angle OPS\).
Sum of angles in \(\triangle POS\):
\(\angle POS + \angle OPS + \angle OSP = 180^{\circ}\)
\(60^{\circ} + 2(\angle OSP) = 180^{\circ}\)
\(2(\angle OSP) = 120^{\circ}\)
\(\angle OSP = 60^{\circ}\)
In simple words: Diagonals of a rectangle are equal and bisect each other, creating isosceles triangles at the center.

Exam Tip: Remember that the diagonals of a rectangle divide it into four isosceles triangles.

Question 8. FARE is a parallelogram. \(OA = 8\text{ cm}\). And \(FR\) is \(10\) more than \(EA\). Find \(OF\).
Answer: In parallelogram FARE, diagonals bisect each other at O.
Given \(OA = 8\text{ cm}\).
Since diagonals bisect each other, \(OE = OA = 8\text{ cm}\).
Therefore, the total length of diagonal \(EA = OE + OA = 8 + 8 = 16\text{ cm}\).
Given that \(FR\) is \(10\) more than \(EA\):
\(FR = EA + 10 = 16 + 10 = 26\text{ cm}\).
We need to find \(OF\). Since diagonals bisect each other, \(OF\) is half of diagonal \(FR\):
\(OF = \frac{FR}{2} = \frac{26}{2} = 13\text{ cm}\).
In simple words: Diagonals of a parallelogram bisect each other, so use the known half-diagonal to find the full diagonal, add 10 to get the other diagonal, and take its half.

Exam Tip: Clearly state the property "diagonals of a parallelogram bisect each other" at the beginning of your solution.