CBSE Class 8 Mathematics Understanding Quadrilaterals Worksheet Set 11

Question 1. Three angles of a quadrilateral are \(100^\circ\), \(92^\circ\) and \(48^\circ\). Find the fourth angle.
Answer: Let the fourth angle be \(x\).
We know that the sum of all four interior angles of a quadrilateral is \(360^\circ\).
\(100^\circ + 92^\circ + 48^\circ + x = 360^\circ\)
\(240^\circ + x = 360^\circ\)
\(x = 360^\circ - 240^\circ\)
\(x = 120^\circ\)
Thus, the fourth angle is \(120^\circ\).
In simple words: Add the three given angles and subtract the sum from \(360^\circ\) to find the remaining angle.

Exam Tip: Always remember that the angle sum property of any quadrilateral is \(360^\circ\).

Question 2. The four angles of a quadrilateral are in the ratio \(3 : 4 : 5 : 6\). Find their measures.
Answer: Let the four angles of the quadrilateral be \(3x\), \(4x\), \(5x\) and \(6x\).
The sum of the angles of a quadrilateral is \(360^\circ\).
\(3x + 4x + 5x + 6x = 360^\circ\)
\(18x = 360^\circ\)
\(x = \frac{360^\circ}{18} = 20^\circ\)
Now, substitute the value of \(x\) to find each angle:
First angle = \(3(20^\circ) = 60^\circ\)
Second angle = \(4(20^\circ) = 80^\circ\)
Third angle = \(5(20^\circ) = 100^\circ\)
Fourth angle = \(6(20^\circ) = 120^\circ\)
Thus, the four angles are \(60^\circ\), \(80^\circ\), \(100^\circ\) and \(120^\circ\).
In simple words: Use variables with the given ratio, equate their sum to \(360^\circ\) to find the common multiplier, and then multiply.

Exam Tip: When given in a ratio, always introduce a common variable like \(x\) and set up an algebraic equation.

Question 3. In a parallelogram \(ABCD\), \(\angle A = (3x - 10)^\circ\) and \(\angle B = (5x + 30)^\circ\), find \(x\) and hence find all the angles of the parallelogram.
Answer: Adjacent angles of a parallelogram are supplementary (their sum is \(180^\circ\)).
\(\angle A + \angle B = 180^\circ\)
\((3x - 10)^\circ + (5x + 30)^\circ = 180^\circ\)
\(8x + 20 = 180\)
\(8x = 180 - 20\)
\(8x = 160\)
\(x = \frac{160}{8} = 20\)
Now, find the angles:
\(\angle A = 3(20) - 10 = 60 - 10 = 50^\circ\)
\(\angle B = 5(20) + 30 = 100 + 30 = 130^\circ\)
Since opposite angles of a parallelogram are equal:
\(\angle C = \angle A = 50^\circ\)
\(\angle D = \angle B = 130^\circ\)
Thus, \(x = 20\) and the angles are \(50^\circ\), \(130^\circ\), \(50^\circ\) and \(130^\circ\).
In simple words: Adjacent angles in a parallelogram add up to \(180^\circ\). Solve for \(x\), find two adjacent angles, and use opposite angle equality for the rest.

Exam Tip: Do not confuse adjacent angles with opposite angles in a parallelogram. Adjacent angles are supplementary, while opposite angles are equal.

Question 4. In the figure, \(ABCD\) is a parallelogram. If \(\angle DCE = 65^\circ\), find \(\angle A\) and \(\angle B\).
Answer: Line \(BC\) is extended to \(E\), forming a straight line \(BCE\).
\(\angle DCB + \angle DCE = 180^\circ\) (Linear pair)
\(\angle DCB + 65^\circ = 180^\circ\)
\(\angle DCB = 180^\circ - 65^\circ = 115^\circ\)
In parallelogram \(ABCD\), opposite angles are equal, so:
\(\angle A = \angle DCB = 115^\circ\)
Also, adjacent angles \(\angle A\) and \(\angle B\) are supplementary:
\(\angle A + \angle B = 180^\circ\)
\(115^\circ + \angle B = 180^\circ\)
\(\angle B = 180^\circ - 115^\circ = 65^\circ\)
Thus, \(\angle A = 115^\circ\) and \(\angle B = 65^\circ\).
In simple words: Use the linear pair property on the straight line to find angle C, then use parallelogram properties to find angles A and B.

Exam Tip: Exterior angles formed by extending a side of a parallelogram are equal to the interior opposite angles.

Question 5. \(ABCD\) is an isosceles trapezium with \(AB \parallel DC\) and \(AD = BC\). If \(\angle A = 130^\circ\), find \(\angle B\), \(\angle C\) and \(\angle D\).
Answer: In an isosceles trapezium, the base angles are equal and the angles on the same side of a non-parallel side are supplementary.
Since \(AB \parallel DC\), the interior angles on the same side of transversal \(AD\) are supplementary:
\(\angle A + \angle D = 180^\circ\)
\(130^\circ + \angle D = 180^\circ\)
\(\angle D = 180^\circ - 130^\circ = 50^\circ\)
Similarly, for transversal \(BC\):
\(\angle B + \angle C = 180^\circ\)
In an isosceles trapezium, base angles are equal (\(\angle A = \angle B\) and \(\angle C = \angle D\)):
\(\angle B = \angle A = 130^\circ\)
\(\angle C = \angle D = 50^\circ\)
Thus, \(\angle B = 130^\circ\), \(\angle C = 50^\circ\) and \(\angle D = 50^\circ\).
In simple words: An isosceles trapezium has equal base angles. Consecutive interior angles add up to \(180^\circ\).

Question 6. In a kite \(ABCD\), \(AB = AD\) and \(BC = DC\). If \(\angle A = 80^\circ\) and \(\angle C = 40^\circ\), find \(\angle B\) and \(\angle D\).
Answer: In a kite, the non-equal sides meet, and the angles between the unequal sides are equal.
Specifically, \(\angle B = \angle D\).
The sum of all interior angles of a quadrilateral is \(360^\circ\):
\(\angle A + \angle B + \angle C + \angle D = 360^\circ\)
\(80^\circ + \angle B + 40^\circ + \angle B = 360^\circ\) (since \(\angle B = \angle D\))
\(120^\circ + 2\angle B = 360^\circ\)
\(2\angle B = 360^\circ - 120^\circ\)
\(2\angle B = 240^\circ\)
\(\angle B = \frac{240^\circ}{2} = 120^\circ\)
Since \(\angle D = \angle B\), we have \(\angle D = 120^\circ\).
Thus, \(\angle B = 120^\circ\) and \(\angle D = 120^\circ\).
In simple words: In a kite, the angles between unequal sides are equal. Use the angle sum of a quadrilateral to find them.

Exam Tip: The diagonal of a kite divides it into congruent triangles, making the opposite angles between unequal sides equal.

Question 7. In the figure, \(ABCD\) is a parallelogram. If \(\angle CAB = 32^\circ\), \(\angle ABC = 110^\circ\), find \(x^\circ\), \(y^\circ\) and \(z^\circ\).
Answer: 1. Finding \(y^\circ\):
Since \(AB \parallel DC\) and diagonal \(AC\) acts as a transversal, alternate interior angles are equal.
\(y^\circ = \angle CAB = 32^\circ\)
2. Finding \(z^\circ\):
In triangle \(ABC\), the sum of angles is \(180^\circ\):
\(\angle CAB + \angle ABC + \angle BCA = 180^\circ\)
\(32^\circ + 110^\circ + z^\circ = 180^\circ\)
\(142^\circ + z^\circ = 180^\circ\)
\(z^\circ = 180^\circ - 142^\circ = 38^\circ\)
3. Finding \(x^\circ\):
Since \(AB \parallel DC\) and diagonal \(AC\) acts as a transversal, alternate interior angles are equal:
\(x^\circ = z^\circ = 38^\circ\)
Thus, \(x^\circ = 38^\circ\), \(y^\circ = 32^\circ\) and \(z^\circ = 38^\circ\).
In simple words: Use alternate interior angles for parallel lines and the angle sum property of triangle ABC.

Exam Tip: Alternate interior angles formed by a transversal across parallel lines are always equal (Z-angle property).

Question 8. \(ABCD\) is a trapezium with \(AD \parallel BC\), \(\angle A = (3x + 1)^\circ\) and \(\angle B = (2x + 4)^\circ\). Find \(x\).
Answer: In trapezium \(ABCD\) with \(AD \parallel BC\), the angles on the same side of the transversal \(AB\) are interior angles on the same side, hence they are supplementary.
\(\angle A + \angle B = 180^\circ\)
\((3x + 1)^\circ + (2x + 4)^\circ = 180^\circ\)
\(5x + 5 = 180\)
\(5x = 180 - 5\)
\(5x = 175\)
\(x = \frac{175}{5} = 35\)
Thus, \(x = 35\).
In simple words: Consecutive angles between parallel lines in a trapezium add up to \(180^\circ\). Set up the equation and solve for \(x\).

Exam Tip: Always identify which sides are parallel to correctly apply the supplementary angle rule for consecutive interior angles.

Question 9. In a trapezium \(ABCD\), \(AB \parallel DC\). If \(\angle A : \angle D = 3 : 2\) and \(\angle B : \angle C = 4 : 5\), find all the angles of the trapezium.
Answer: Since \(AB \parallel DC\), angles on the same side of transversal \(AD\) are supplementary:
\(\angle A + \angle D = 180^\circ\)
Given \(\angle A : \angle D = 3 : 2\), let \(\angle A = 3y\) and \(\angle D = 2y\).
\(3y + 2y = 180^\circ\)
\(5y = 180^\circ \implies y = 36^\circ\)
So, \(\angle A = 3(36^\circ) = 108^\circ\) and \(\angle D = 2(36^\circ) = 72^\circ\).
Similarly, angles on the same side of transversal \(BC\) are supplementary:
\(\angle B + \angle C = 180^\circ\)
Given \(\angle B : \angle C = 4 : 5\), let \(\angle B = 4k\) and \(\angle C = 5k\).
\(4k + 5k = 180^\circ\)
\(9k = 180^\circ \implies k = 20^\circ\)
So, \(\angle B = 4(20^\circ) = 80^\circ\) and \(\angle C = 5(20^\circ) = 100^\circ\).
Thus, the angles are \(\angle A = 108^\circ\), \(\angle B = 80^\circ\), \(\angle C = 100^\circ\) and \(\angle D = 72^\circ\).
In simple words: Use ratios for pairs of consecutive angles along the parallel sides, setting each sum to \(180^\circ\).

Exam Tip: Remember that consecutive angles between parallel lines add up to \(180^\circ\), but opposite angles do not unless it is a parallelogram.

Question 10. In a parallelogram \(PQRS\), \(\angle P\) is \(30^\circ\) more than \(\angle Q\). Find the measures of all the angles.
Answer: Let \(\angle Q = x\).
Then \(\angle P = x + 30^\circ\).
Since adjacent angles of a parallelogram are supplementary:
\(\angle P + \angle Q = 180^\circ\)
\((x + 30^\circ) + x = 180^\circ\)
\(2x + 30^\circ = 180^\circ\)
\(2x = 150^\circ\)
\(x = 75^\circ\)
So, \(\angle Q = 75^\circ\) and \(\angle P = 75^\circ + 30^\circ = 105^\circ\).
Since opposite angles of a parallelogram are equal:
\(\angle R = \angle P = 105^\circ\)
\(\angle S = \angle Q = 75^\circ\)
Thus, the angles are \(105^\circ\), \(75^\circ\), \(105^\circ\) and \(75^\circ\).
In simple words: Set one angle as \(x\), write the other in terms of \(x\), use their supplementary sum to solve, and find opposite angles.

Exam Tip: Always verify that adjacent angles sum to \(180^\circ\) (\(105^\circ + 75^\circ = 180^\circ\)).

Question 11. The adjacent angles of a parallelogram are in the ratio \(2 : 7\). Find their measures.
Answer: Let the adjacent angles be \(2x\) and \(7x\).
Since adjacent angles of a parallelogram are supplementary:
\(2x + 7x = 180^\circ\)
\(9x = 180^\circ\)
\(x = \frac{180^\circ}{9} = 20^\circ\)
Now, calculate the angles:
First angle = \(2(20^\circ) = 40^\circ\)
Second angle = \(7(20^\circ) = 140^\circ\)
Since opposite angles are equal, all four angles of the parallelogram are \(40^\circ\), \(140^\circ\), \(40^\circ\) and \(140^\circ\).
In simple words: Assign ratio variables, set their sum to \(180^\circ\), find \(x\), and calculate the angles.

Exam Tip: Adjacent angles in any parallelogram always add up to \(180^\circ\).

Question 12. \(P\) is a point in the interior of \(\angle AOB\). \(PM \perp OA\) and \(PN \perp OB\). If \(\angle AOB = 60^\circ\), find \(\angle MPN\).
Answer: Consider quadrilateral \(OMPN\).
We are given:
\(\angle OMP = 90^\circ\) (since \(PM \perp OA\))
\(\angle ONP = 90^\circ\) (since \(PN \perp OB\))
\(\angle MON = \angle AOB = 60^\circ\)
The sum of all interior angles of quadrilateral \(OMPN\) is \(360^\circ\):
\(\angle MON + \angle OMP + \angle MPN + \angle ONP = 360^\circ\)
\(60^\circ + 90^\circ + \angle MPN + 90^\circ = 360^\circ\)
\(240^\circ + \angle MPN = 360^\circ\)
\(\angle MPN = 360^\circ - 240^\circ = 120^\circ\)
Thus, \(\angle MPN = 120^\circ\).
In simple words: Form a quadrilateral with the two perpendiculars and the origin, then use the \(360^\circ\) angle sum property.

Exam Tip: Perpendicular lines always form \(90^\circ\) angles, which helps in finding unknown angles in quadrilaterals.

Question 13. One of the diagonals of a rhombus is equal to one of its sides. Find the angles of the rhombus.
Answer: Let the rhombus be \(ABCD\) with side length \(s\).
Suppose diagonal \(BD\) is equal to the side length, so \(BD = AB = AD = s\).\br /> Thus, triangle \(ABD\) is an equilateral triangle.
Therefore, \(\angle A = 60^\circ\), \(\angle ABD = 60^\circ\) and \(\angle ADB = 60^\circ\).
Since opposite angles of a rhombus are equal, \(\angle C = \angle A = 60^\circ\).
Also, adjacent angles of a rhombus are supplementary (\(\angle A + \angle B = 180^\circ\)):
\(60^\circ + \angle B = 180^\circ \implies \angle B = 120^\circ\).
Since opposite angles are equal, \(\angle D = \angle B = 120^\circ\).
Thus, the angles of the rhombus are \(60^\circ\), \(120^\circ\), \(60^\circ\) and \(120^\circ\).
In simple words: A diagonal equal to a side splits the rhombus into two equilateral triangles, making it easy to find all angles.

Exam Tip: Recognizing equilateral triangles inside special quadrilaterals simplifies angle calculations drastically.

Question 14. Find the perimeter of a rhombus whose diagonals are \(6\text{ cm}\) and \(8\text{ cm}\).
Answer: Let the diagonals of the rhombus be \(d_1 = 6\text{ cm}\) and \(d_2 = 8\text{ cm}\).
The diagonals of a rhombus bisect each other at right angles (\(90^\circ\)).
Half of the first diagonal = \(\frac{6}{2} = 3\text{ cm}\).
Half of the second diagonal = \(\frac{8}{2} = 4\text{ cm}\).
The side of the rhombus \(s\) forms the hypotenuse of a right-angled triangle with these half-diagonals as its legs:
\(s = \sqrt{3^2 + 4^2}\)
\(s = \sqrt{9 + 16} = \sqrt{25} = 5\text{ cm}\).
Perimeter of a rhombus = \(4 \times \text{side} = 4 \times 5 = 20\text{ cm}\).
Thus, the perimeter is \(20\text{ cm}\).
In simple words: Use Pythagoras theorem on half the diagonals to find the side length, then multiply by 4 for the perimeter.

Exam Tip: Diagonals of a rhombus bisect perpendicularly, creating four congruent right-angled triangles.

Question 15. Lengths of two sides of a parallelogram are in the ratio \(2 : 3\). Find the sides of the parallelogram if its perimeter is \(120\text{ cm}\).
Answer: Let the adjacent sides of the parallelogram be \(2x\) and \(3x\).
Perimeter of a parallelogram = \(2 \times (\text{Length} + \text{Breadth})\)
\(2 \times (2x + 3x) = 120\)
\(2 \times 5x = 120\)
\(10x = 120\)
\(x = \frac{120}{10} = 12\text{ cm}\).
Now, find the side lengths:
First side = \(2(12) = 24\text{ cm}\)
Second side = \(3(12) = 36\text{ cm}\)
Since opposite sides are equal, the four sides are \(24\text{ cm}\), \(36\text{ cm}\), \(24\text{ cm}\) and \(36\text{ cm}\).
In simple words: Use the perimeter formula with ratio variables to solve for \(x\), then calculate the individual side lengths.

Question 16. Find the measure of an exterior angle of a regular polygon: (i) Hexagon (ii) Octagon (iii) Decagon
Answer: The measure of each exterior angle of a regular polygon with \(n\) sides is given by:
Exterior angle = \(\frac{360^\circ}{n}\)
(i) For a Hexagon (\(n = 6\)):
Exterior angle = \(\frac{360^\circ}{6} = 60^\circ\)
(ii) For an Octagon (\(n = 8\)):
Exterior angle = \(\frac{360^\circ}{8} = 45^\circ\)
(iii) For a Decagon (\(n = 10\)):
Exterior angle = \(\frac{360^\circ}{10} = 36^\circ\)
In simple words: Divide \(360^\circ\) by the number of sides to find the measure of each exterior angle.

Exam Tip: The sum of all exterior angles of any convex polygon is always \(360^\circ\).

Question 17. Find the sum of the interior angles of a polygon: (i) Heptagon (ii) Nonagon (iii) Pentagon
Answer: The sum of the interior angles of a polygon with \(n\) sides is given by the formula:
Sum = \((n - 2) \times 180^\circ\)
(i) For a Heptagon (\(n = 7\)):
Sum = \((7 - 2) \times 180^\circ = 5 \times 180^\circ = 900^\circ\)
(ii) For a Nonagon (\(n = 9\)):
Sum = \((9 - 2) \times 180^\circ = 7 \times 180^\circ = 1260^\circ\)
(iii) For a Pentagon (\(n = 5\)):
Sum = \((5 - 2) \times 180^\circ = 3 \times 180^\circ = 540^\circ\)
In simple words: Subtract 2 from the number of sides and multiply by \(180^\circ\).

Exam Tip: Memorize the formula \((n - 2) \times 180^\circ\) for interior angle sums.

Question 18. Each interior angle of a regular polygon is \(140^\circ\). Find the number of sides of the polygon.
Answer: Interior angle + Exterior angle = \(180^\circ\) (Linear pair)
\(140^\circ + \text{Exterior angle} = 180^\circ\)
Exterior angle = \(180^\circ - 140^\circ = 40^\circ\)
The number of sides \(n\) of a regular polygon is given by:
\(n = \frac{360^\circ}{\text{Exterior angle}}\)
\(n = \frac{360^\circ}{40^\circ} = 9\)
Thus, the polygon has 9 sides (it is a nonagon).
In simple words: Find the exterior angle first by subtracting the interior angle from \(180^\circ\), then divide \(360^\circ\) by it to get the number of sides.

Question 19. The measure of an exterior angle of a regular polygon is \(72^\circ\). What is the name of the polygon?
Answer: Let \(n\) be the number of sides of the regular polygon.
\(n = \frac{360^\circ}{\text{Exterior angle}}\)
\(n = \frac{360^\circ}{72^\circ} = 5\)
A polygon with 5 sides is called a pentagon.
Thus, the polygon is a pentagon.
In simple words: Divide \(360^\circ\) by the exterior angle to find the number of sides, then name the polygon.

Exam Tip: A 5-sided polygon is a pentagon, 6 is a hexagon, 7 is a heptagon, 8 is an octagon, and 9 is a nonagon.

Question 20. Find the measure of an interior angle of a regular hexagon.
Answer: For a regular hexagon, the number of sides \(n = 6\).
First, find the exterior angle:
Exterior angle = \(\frac{360^\circ}{6} = 60^\circ\)
Interior angle = \(180^\circ - \text{Exterior angle}\)
Interior angle = \(180^\circ - 60^\circ = 120^\circ\)
Alternatively, using the interior angle formula:\br /> Sum = \((6 - 2) \times 180^\circ = 4 \times 180^\circ = 720^\circ\)
Each interior angle = \(\frac{720^\circ}{6} = 120^\circ\)
Thus, the measure of an interior angle of a regular hexagon is \(120^\circ\).
In simple words: Find the exterior angle first (\(360/6 = 60\)) and subtract it from \(180^\circ\).

Exam Tip: Exterior angle method is generally quicker than calculating the total interior sum and dividing.

Question 21. The ratio of an interior angle to the exterior angle of a regular polygon is \(5 : 1\). Find the number of sides.
Answer: Let the interior angle be \(5x\) and the exterior angle be \(1x\).
Since an interior angle and its corresponding exterior angle form a linear pair:
\(5x + x = 180^\circ\)
\(6x = 180^\circ \implies x = 30^\circ\)
Exterior angle = \(1(30^\circ) = 30^\circ\).
Now, find the number of sides \(n\):
\(n = \frac{360^\circ}{\text{Exterior angle}} = \frac{360^\circ}{30^\circ} = 12\)
Thus, the polygon has 12 sides.
In simple words: Use the given ratio with \(x\) to find the exterior angle, then divide \(360^\circ\) by it.

Exam Tip: Interior and exterior angles are always supplementary on any vertex of a polygon.

Question 22. The sum of the interior angles of a polygon is \(540^\circ\). What is the name of the polygon?
Answer: Let \(n\) be the number of sides of the polygon.
Using the interior angle sum formula:
\((n - 2) \times 180^\circ = 540^\circ\)
\(n - 2 = \frac{540^\circ}{180^\circ}\)
\(n - 2 = 3\)
\(n = 3 + 2 = 5\)
A polygon with 5 sides is a pentagon.
Thus, the polygon is a pentagon.
In simple words: Divide the total angle sum by \(180^\circ\) and add 2 to find the number of sides.

Question 23. How many diagonals can we draw in a: (i) pentagon (ii) hexagon (iii) triangle (iv) quadrilateral
Answer: The number of diagonals in a polygon with \(n\) sides is given by the formula:
Number of diagonals = \(\frac{n(n - 3)}{2}\)
(i) For a Pentagon (\(n = 5\)):
Diagonals = \(\frac{5(5 - 3)}{2} = \frac{5 \times 2}{2} = 5\)
(ii) For a Hexagon (\(n = 6\)):
Diagonals = \(\frac{6(6 - 3)}{2} = \frac{6 \times 3}{2} = 9\)
(iii) For a Triangle (\(n = 3\)):
Diagonals = \(\frac{3(3 - 3)}{2} = \frac{3 \times 0}{2} = 0\)
(iv) For a Quadrilateral (\(n = 4\)):
Diagonals = \(\frac{4(4 - 3)}{2} = \frac{4 \times 1}{2} = 2\)
In simple words: Substitute the number of sides \(n\) into the diagonal formula \(\frac{n(n - 3)}{2}\).

Exam Tip: Triangles have 0 diagonals because no two non-adjacent vertices exist.