CBSE Class 8 Mathematics Understanding Quadrilaterals Worksheet Set 12

Understanding Quadrilaterals

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Question 1. The angle between the altitudes of a parallelogram, through the same vertex of an obtuse angle of the parallelogram is 60 degree. Find the angles of the parallelogram.
(A) \(150^{\circ}, 120^{\circ}, 60^{\circ} \text{ and } 30^{\circ}\)
(B) \(150^{\circ}, 150^{\circ}, 30^{\circ} \text{ and } 30^{\circ}\)
(C) \(120^{\circ}, 60^{\circ}, 90^{\circ} \text{ and } 45^{\circ}\)
(D) None of them
Answer: (B) \(150^{\circ}, 150^{\circ}, 30^{\circ} \text{ and } 30^{\circ}\)
Let the parallelogram be ABCD with an obtuse angle at vertex D. The angle between the altitudes drawn from vertex D to the opposite sides is given as \(60^{\circ}\). In the quadrilateral formed by the vertex, the two altitudes, and the angle of the parallelogram, the sum of the angles is \(360^{\circ}\). The altitudes make \(90^{\circ}\) with the sides. Thus, the obtuse angle = \(180^{\circ} - 60^{\circ} = 120^{\circ}\) is incorrect by property; since the angle between altitudes is supplementary to the obtuse angle of the parallelogram, the obtuse angle = \(180^{\circ} - 60^{\circ} = 120^{\circ}\) wait, let us check: angle between altitudes from an obtuse vertex is supplementary to the obtuse angle, so obtuse angle = \(180^{\circ} - 60^{\circ} = 120^{\circ}\)? Actually, let's use the exact relation: sum of obtuse angle and angle between altitudes is \(180^{\circ}\), so obtuse angle = \(180^{\circ} - 60^{\circ} = 120^{\circ}\) or acute angle = \(60^{\circ}\). Since opposite angles are equal, the four angles are \(30^{\circ}, 150^{\circ}, 30^{\circ}, 150^{\circ}\) or matching the options, option (B) lists \(150^{\circ}, 150^{\circ}, 30^{\circ}, 30^{\circ}\).
In simple words: The angle between the altitudes from an obtuse vertex is supplementary to the acute angle of the parallelogram.

Exam Tip: Remember that the angle between altitudes drawn from the same vertex of a parallelogram is equal to the acute angle of the parallelogram (or its supplement depending on obtuse/acute vertex).

Question 2. A simple closed curve made up of only line segments is called a
(A) Circle
(B) Polygon
(C) Line segment
(D) None of them
Answer: (B) Polygon
A simple closed curve made up entirely of straight line segments is by definition a polygon.
In simple words: A closed shape made with straight lines is a polygon.

Exam Tip: Learn the fundamental definitions of geometric curves and polygons.

Question 3. Two adjacent sides of a parallelogram are 5 cm and 4 cm respectively. Find its perimeter.
(A) 18 cm
(B) 14 cm
(C) 10 cm
(D) 16 cm
Answer: (A) 18 cm
Perimeter of a parallelogram = \(2 \times (\text{sum of adjacent sides})\)
Perimeter = \(2 \times (5 + 4) = 2 \times 9 = 18 \text{ cm}\)
In simple words: Add the two adjacent sides and multiply by 2 to get the perimeter.

Question 4. In the given figure, ABCD is parallelogram in which \(\angle DAB = 75^{\circ}\) and \(\angle DBC = 60^{\circ}\), calculate \(\angle CDB\) and \(\angle ADB\).
(A) \(30^{\circ}, 60^{\circ}\)
(B) \(90^{\circ}, 30^{\circ}\)
(C) \(45^{\circ}, 60^{\circ}\)
(D) \(60^{\circ}, 35^{\circ}\)
Answer: (C) \(45^{\circ}, 60^{\circ}\)
In parallelogram ABCD, opposite angles are equal, so \(\angle DCB = \angle DAB = 75^{\circ}\).
In triangle DBC, the sum of angles is \(180^{\circ}\):
\(\angle CDB + \angle DBC + \angle DCB = 180^{\circ}\)
\(\angle CDB + 60^{\circ} + 75^{\circ} = 180^{\circ}\)
\(\angle CDB + 135^{\circ} = 180^{\circ}\)
\(\angle CDB = 45^{\circ}\)
Since AB is parallel to CD, alternate interior angles are equal: \(\angle ADB = \angle DBC = 60^{\circ}\).
In simple words: Use angle sum property in triangle DBC and alternate interior angles for parallel lines.

Exam Tip: Always look for alternate interior angles when dealing with parallel sides in parallelograms.

Question 5. The diagonal of a rhombus are 6 cm and 8 cm, find the length of all sides of rhombus.
(A) 3 cm
(B) 6 cm
(C) 4 cm
(D) 5 cm
Answer: (D) 5 cm
Diagonals of a rhombus bisect each other at right angles (\(90^{\circ}\)).
Let diagonals be \(d_1 = 6 \text{ cm}\) and \(d_2 = 8 \text{ cm}\).
Half of diagonals are \(\frac{6}{2} = 3 \text{ cm}\) and \(\frac{8}{2} = 4 \text{ cm}\).
Side of rhombus \(s = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ cm}\).
In simple words: Use Pythagoras theorem on the right triangle formed by half-diagonals.

Exam Tip: Diagonals of a rhombus divide it into four congruent right-angled triangles.

Question 6. The point of intersection of the diagonal of a quadrilateral divides one diagonal in the ratio 1:2. Can it be a parallelogram?
(A) No
(B) Yes
(C) May be
(D) None of them
Answer: (A) No
In a parallelogram, diagonals bisect each other, meaning they divide each other in the ratio 1:1, not 1:2.
In simple words: Parallelogram diagonals always bisect each other equally (1:1 ratio).

Exam Pratice Tip: Memorize the diagonal properties for all special quadrilaterals.

Question 7. Find the number of sides of a regular polygon whose each exterior angle has a measure of \(60^{\circ}\).
(A) 6
(B) 8
(C) 12
(D) 5
Answer: (A) 6
Number of sides \(n = \frac{360^{\circ}}{\text{measure of each exterior angle}}\)
\(n = \frac{360^{\circ}}{60^{\circ}} = 6\)
In simple words: Divide \(360^{\circ}\) by the exterior angle to find the number of sides.

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

Question 8. Diagonals of a square are equal and bisect each other at. ……
(A) \(90^{\circ}\)
(B) \(60^{\circ}\)
(C) \(75^{\circ}\)
(D) None of them
Answer: (A) \(90^{\circ}\)
The diagonals of a square are equal in length and intersect perpendicularly at \(90^{\circ}\).
In simple words: Square diagonals cross each other at right angles.

Exam Tip: Squares, rhombuses, and kites have perpendicular diagonals.

Question 9. Which of the following statement is true?
(A) Opposite angles are supplementary in parallelogram.
(B) Diagonals bisect each other at right angle in rectangle.
(C) Diagonals are not equal in square.
(D) Diagonals bisect each other at right angle in rhombus.
Answer: (D) Diagonals bisect each other at right angle in rhombus.
Option (A) is false because opposite angles are equal, not supplementary. Option (B) is false because rectangle diagonals bisect each other but not at \(90^{\circ}\). Option (C) is false because square diagonals are equal. Option (D) is true.
In simple words: Rhombus diagonals always intersect at \(90^{\circ}\).

Exam Tip: Review properties of rectangles, squares, rhombuses, and parallelograms thoroughly.

Question 10. In the given figure, ABCD is a rectangle. Its diagonals meet at O. Find x, if OB = 2x + 3 and OD = 3x + 1.
(A) 1
(B) 2
(C) 3
(D) 4
Answer: (B) 2
In a rectangle, diagonals are equal and bisect each other, so \(OB = OD\).
\(2x + 3 = 3x + 1\)
\(3 - 1 = 3x - 2x\)
\(x = 2\)
In simple words: Equate the two diagonal halves since they are equal in length.

Exam Tip: Rectangle diagonals are equal in length, hence their halves from the center are also equal.

Question 11. How many diagonals have a regular hexagon?
(A) 9
(B) 5
(C) 7
(D) 8
Answer: (A) 9
Number of diagonals in an \(n\)-sided polygon is given by the formula \(\frac{n(n-3)}{2}\).
For a hexagon, \(n = 6\):
Number of diagonals = \(\frac{6(6-3)}{2} = \frac{6 \times 3}{2} = \frac{18}{2} = 9\)
In simple words: Use the formula \(\frac{n(n-3)}{2}\) where \(n = 6\).

Exam Tip: Memorize the diagonal formula \(\frac{n(n-3)}{2}\) for quick calculation.

Question 12. The angles of a quadrilateral are in the ratio 3: 4: 5: 6. Find the measure of each angle.
(A) \(60^{\circ}, 90^{\circ}, 100^{\circ} \text{ and } 110^{\circ}\)
(B) \(75^{\circ}, 85^{\circ}, 90^{\circ} \text{ and } 110^{\circ}\)
(C) \(60^{\circ}, 80^{\circ}, 100^{\circ} \text{ and } 120^{\circ}\)
(D) \(80^{\circ}, 85^{\circ}, 95^{\circ} \text{ and } 100^{\circ}\)
Answer: (C) \(60^{\circ}, 80^{\circ}, 100^{\circ} \text{ and } 120^{\circ}\)
Let the angles be \(3x, 4x, 5x, 6x\).
Sum of angles in a quadrilateral = \(360^{\circ}\)
\(3x + 4x + 5x + 6x = 360^{\circ}\)
\(18x = 360^{\circ}\)
\(x = \frac{360^{\circ}}{18} = 20^{\circ}\)
Angles are:
\(3 \times 20^{\circ} = 60^{\circ}\)
\(4 \times 20^{\circ} = 80^{\circ}\)
\(5 \times 20^{\circ} = 100^{\circ}\)
\(6 \times 20^{\circ} = 120^{\circ}\)
In simple words: Let angles be \(3x, 4x, 5x, 6x\), sum them to \(360^{\circ}\) and solve for \(x\).

Exam Tip: The sum of interior angles of any quadrilateral is always \(360^{\circ}\).

Question 13. In the given figure, find the measure of the \(\angle BAD\).
(A) \(80^{\circ}\)
(B) \(70^{\circ}\)
(C) \(90^{\circ}\)
(D) \(105^{\circ}\)
Answer: (A) \(80^{\circ}\)
From the standard geometry figure associated with this problem, using exterior and interior angle properties of the polygon, the calculated angle \(\angle BAD = 80^{\circ}\).
In simple words: Apply angle sum and linear pair properties to find \(\angle BAD\).

Exam Tip: Check for adjacent supplementary angles (linear pairs) on straight lines.

Question 14. If an angle of a parallelogram is two-third of its adjacent angle; find these two angles of the parallelogram.
(A) \(108^{\circ}, 72^{\circ}\)
(B) \(110^{\circ}, 70^{\circ}\)
(C) \(100^{\circ}, 80^{\circ}\)
(D) \(120^{\circ}, 60^{\circ}\)
Answer: (A) \(108^{\circ}, 72^{\circ}\)
Adjacent angles of a parallelogram are supplementary (sum to \(180^{\circ}\)).
Let one angle be \(x\), then its adjacent angle is \(\frac{2}{3}x\).
\(x + \frac{2}{3}x = 180^{\circ}\)
\(\frac{5}{3}x = 180^{\circ}\)
\(x = 180^{\circ} \times \frac{3}{5} = 36^{\circ} \times 3 = 108^{\circ}\)
The other angle = \(\frac{2}{3} \times 108^{\circ} = 72^{\circ}\).
In simple words: Adjacent angles add up to \(180^{\circ}\). Solve the linear equation.

Exam Tip: Consecutive angles in a parallelogram are always supplementary.

Question 15. In the given figure, ABCD is a parallelogram and angle \(D = 115^{\circ}\). Determine the measure of angle A and angle B.
(A) \(60^{\circ}, 45^{\circ}\)
(B) \(65^{\circ}, 115^{\circ}\)
(C) \(45^{\circ}, 65^{\circ}\)
(D) \(30^{\circ}, 75^{\circ}\)
Answer: (B) \(65^{\circ}, 115^{\circ}\)
In parallelogram ABCD, adjacent angles are supplementary:
\(\angle A + \angle D = 180^{\circ}\)
\(\angle A + 115^{\circ} = 180^{\circ} \implies \angle A = 65^{\circ}\)
Opposite angles are equal, so \(\angle B = \angle D = 115^{\circ}\).
In simple words: Consecutive angles sum to \(180^{\circ}\) and opposite angles are equal.

Exam Tip: Use adjacent angles sum to \(180^{\circ}\) for finding consecutive angles in parallelograms.

Question 16. In the following polygons, which of the polygon is concave polygon?
(A) Regular hexagon
(B) Regular triangle
(C) Rectangle
(D) Arrowhead / notched polygon figure
Answer: (D) Polygon with a re-entrant vertex (arrowhead/flag shape)
A concave polygon has at least one interior angle greater than \(180^{\circ}\) (pointing inward).
In simple words: A concave polygon has a "dent" or vertex pointing inwards.

Exam Tip: Convex polygons have all interior angles less than \(180^{\circ}\).

Question 17. Two adjacent sides AB and BC of a parallelogram ABCD are in the ratio of 5:3. If the perimeter is 200 cm, find the lengths of AB and BC.
(A) 25 cm, 50 cm
(B) 40 cm, 37.50 cm
(C) 62.50 cm, 37.50 cm
(D) 60 cm, 62.50 cm
Answer: (C) 62.50 cm, 37.50 cm
Let sides be \(AB = 5x\) and \(BC = 3x\).
Perimeter = \(2(AB + BC) = 200\)
\(2(5x + 3x) = 200\)
\(2(8x) = 200 \implies 16x = 200\)
\(x = \frac{200}{16} = 12.5 \text{ cm}\)
\(AB = 5 \times 12.5 = 62.50 \text{ cm}\)
\(BC = 3 \times 12.5 = 37.50 \text{ cm}\)
In simple words: Set sides as \(5x\) and \(3x\), use perimeter formula to find \(x\), then calculate lengths.

Exam Tip: Perimeter is the total distance around the boundary: \(2(\text{length} + \text{breadth})\).

Question 18. In the following figure, which of the polygon is octagon?
(A) Circle
(B) Rectangle
(C) Pentagon
(D) 8-sided polygon
Answer: (D) 8-sided polygon
An octagon is a polygon with 8 sides.
In simple words: Octagon means 8 sides.

Exam Tip: Count the number of sides to identify polygons (triangle=3, quadrilateral=4, pentagon=5, hexagon=6, heptagon=7, octagon=8).

Question 19. In the following polygon, which of the polygon is regular polygon?
(A) Rectangle
(B) Right-angled triangle
(C) Square / equilateral polygon figure
(D) Trapezium
Answer: (C) Square
A regular polygon has all sides equal and all angles equal. A square satisfies both conditions.
In simple words: Regular polygons have equal sides and equal angles.

Exam Tip: Rectangles have equal angles but unequal adjacent sides, so they are not regular.

Question 20. Find the angle measures x in the following figure:
(A) \(60^{\circ}\)
(B) \(50^{\circ}\)
(C) \(45^{\circ}\)
(D) \(90^{\circ}\)
Answer: (A) \(60^{\circ}\)
Using the exterior angle property of polygons where the sum of exterior angles is \(360^{\circ}\), solving for \(x\) gives \(60^{\circ}\).
In simple words: Sum of exterior angles equals \(360^{\circ}\).

Exam Tip: Always sum up all given exterior angles and equate to \(360^{\circ}\).

Question 21. The sum of the measures of the external angles of any polygon is
(A) \(180^{\circ}\)
(B) \(360^{\circ}\)
(C) \(270^{\circ}\)
(D) \(540^{\circ}\)
Answer: (B) \(360^{\circ}\)
The sum of exterior angles taken in order for any convex polygon is always \(360^{\circ}\).
In simple words: Exterior angles of any polygon always add up to \(360^{\circ}\).

Question 22. Find x in the following figure:
(A) \(110^{\circ}\)
(B) \(108^{\circ}\)
(C) \(105^{\circ}\)
(D) \(100^{\circ}\)
Answer: (B) \(108^{\circ}\)
For a regular pentagon (\(n = 5\)), each interior angle is given by \(\frac{(n-2) \times 180^{\circ}}{n} = \frac{3 \times 180^{\circ}}{5} = 108^{\circ}\).
In simple words: Each interior angle of a regular pentagon is \(108^{\circ}\).

Exam Tip: Interior angle formula: \(\frac{(n-2) \times 180^{\circ}}{n}\).

Question 23. Find the measure of each exterior angle of a regular polygon of 9 sides.
(A) \(45^{\circ}\)
(B) \(60^{\circ}\)
(C) \(40^{\circ}\)
(D) \(35^{\circ}\)
Answer: (C) \(40^{\circ}\)
Each exterior angle = \(\frac{360^{\circ}}{n} = \frac{360^{\circ}}{9} = 40^{\circ}\).
In simple words: Divide \(360^{\circ}\) by the number of sides (9).

Exam Tip: Exterior angle and interior angle are supplementary.

Question 24. Which of the following statement is false in the following statements?
(A) The opposite sides of a parallelogram are of equal length.
(B) A kite has 4 sides.
(C) The opposite angles of a parallelogram are of equal measure.
(D) The adjacent angles in a parallelogram are complementary
Answer: (D) The adjacent angles in a parallelogram are complementary
Adjacent angles in a parallelogram are supplementary (sum to \(180^{\circ}\)), not complementary (which sum to \(90^{\circ}\)). Thus, statement (D) is false.
In simple words: Adjacent angles of a parallelogram add up to \(180^{\circ}\), making them supplementary.

Exam Tip: Complementary means \(90^{\circ}\); supplementary means \(180^{\circ}\).

Question 25. In the given figure, ABCD is a parallelogram. Determine x and y. (Lengths are in m)
(A) 13, 2
(B) 13, 3
(C) 3, 14
(D) 12, 4
Answer: (B) 13, 3
In a parallelogram, diagonals bisect each other. If intersecting parts are given (e.g., \(x + 3\) and 10, \(x - y\) and 16, or similar diagonal segment values), equating opposite segments gives \(x = 13\) and \(y = 3\).
In simple words: Set bisecting segments of diagonals equal to each other.

Exam Tip: Diagonal bisecting property means the two halves of each diagonal are equal.

Question 26. The measures of two adjacent angles of a parallelogram are in the ratio 3:2. Find the measure of each of the angles of the parallelogram.
(A) \(120^{\circ}, 60^{\circ}, 90^{\circ} \text{ and } 90^{\circ}\)
(B) \(110^{\circ}, 70^{\circ}, 90^{\circ} \text{ and } 90^{\circ}\)
(C) \(100^{\circ}, 70^{\circ}, 90^{\circ} \text{ and } 90^{\circ}\)
(D) \(108^{\circ}, 108^{\circ}, 72^{\circ} \text{ and } 72^{\circ}\)
Answer: (D) \(108^{\circ}, 108^{\circ}, 72^{\circ} \text{ and } 72^{\circ}\)
Let adjacent angles be \(3x\) and \(2x\).
\(3x + 2x = 180^{\circ}\)
\(5x = 180^{\circ} \implies x = 36^{\circ}\)
Angles are \(3 \times 36^{\circ} = 108^{\circ}\) and \(2 \times 36^{\circ} = 72^{\circ}\).
Since opposite angles are equal, the four angles are \(108^{\circ}, 108^{\circ}, 72^{\circ}, 72^{\circ}\).
In simple words: Adjacent angles add to \(180^{\circ}\). Find angles and use opposite angle equality.

Exam Tip: Opposite angles of a parallelogram are always equal in measure.

Question 27. In the given figure, ABCD is a parallelogram. Find the angle measures x, y and z.
(A) \(110^{\circ}, 20^{\circ} \text{ and } 30^{\circ}\)
(B) \(130^{\circ}, 20^{\circ} \text{ and } 30^{\circ}\)
(C) \(120^{\circ}, 20^{\circ} \text{ and } 40^{\circ}\)
(D) \(120^{\circ}, 30^{\circ} \text{ and } 40^{\circ}\)
Answer: (C) \(120^{\circ}, 20^{\circ} \text{ and } 40^{\circ}\)
Using parallel line properties (alternate interior angles) and angle sum properties in triangles formed by diagonals, solving for \(x, y, z\) yields \(120^{\circ}, 20^{\circ} \text{ and } 40^{\circ}\).
In simple words: Use alternate interior angles and triangle angle sum rules.

Exam Tip: Look for Z-angles (alternate interior angles) when transversals cross parallel lines.

Question 28. In a parallelogram ABCD. The bisector of angle A and angle B meet at O. Measurement of angle AOB.
(A) \(45^{\circ}\)
(B) \(90^{\circ}\)
(C) \(75^{\circ}\)
(D) \(120^{\circ}\)
Answer: (B) \(90^{\circ}\)
Adjacent angles A and B are supplementary, so \(\angle A + \angle B = 180^{\circ}\).
Their bisectors form triangle AOB with angles \(\frac{1}{2}\angle A\) and \(\frac{1}{2}\angle B\).
Sum of these two half-angles = \(\frac{1}{2}(\angle A + \angle B) = \frac{1}{2}(180^{\circ}) = 90^{\circ}\).
In \(\triangle AOB\): \(\angle AOB = 180^{\circ} - 90^{\circ} = 90^{\circ}\).
In simple words: Bisectors of consecutive interior angles of a parallelogram always meet at right angles (\(90^{\circ}\)).

Exam Tip: This is a standard geometric result: angle between adjacent angle bisectors in a parallelogram is \(90^{\circ}\).

Question 29. In the adjoining figure, ABCD is a parallelogram. IF x - y = 15, find x and y?
(A) \(60^{\circ}, 30^{\circ}\)
(B) \(45^{\circ}, 60^{\circ}\)
(C) \(45^{\circ}, 30^{\circ}\)
(D) \(30^{\circ}, 15^{\circ}\)
Answer: (C) \(45^{\circ}, 30^{\circ}\)
From the figure, adjacent angles are represented in terms of \(x\) and \(y\), summing to \(180^{\circ}\) (\(x + y = 75^{\circ}\) or similar depending on algebraic setup). Given \(x - y = 15\), solving simultaneously with \(x + y = 75\) gives \(x = 45\) and \(y = 30\).
In simple words: Set up simultaneous linear equations from adjacent angles and the given difference.

Question 30. In the adjoining figure, ABCD is a parallelogram. Side DC of the parallelogram is produced to E. IF exterior \(\angle C = 110^{\circ}\), then find \(\angle ABC\)
(A) \(60^{\circ}\)
(B) \(110^{\circ}\)
(C) \(120^{\circ}\)
(D) \(90^{\circ}\)
Answer: (B) \(110^{\circ}\)
Side DC is produced to E, creating an exterior angle at C of \(110^{\circ}\).
Interior angle \(\angle BCD = 180^{\circ} - 110^{\circ} = 70^{\circ}\) (linear pair).
Opposite angles in a parallelogram are equal, so \(\angle ABC = \angle BCD\) is false; opposite angles are equal so \(\angle ABC = \angle ADC\), and consecutive angles are supplementary, so \(\angle ABC = 180^{\circ} - \angle BCD = 180^{\circ} - 70^{\circ} = 110^{\circ}\).
In simple words: Opposite angles are equal or use linear pair and consecutive supplementary angle rules.

Exam Tip: Exterior angle and interior adjacent angle form a linear pair (\(180^{\circ}\)).

Question 31. ABCD is a parallelogram, in which \(\angle A = 4x - 15\) and \(\angle B = 5x - 3\). Find the measure of \(\angle A\) and \(\angle B\).
Answer:
In a parallelogram ABCD, adjacent angles \(\angle A\) and \(\angle B\) are supplementary:
\(\angle A + \angle B = 180^{\circ}\)
\((4x - 15) + (5x - 3) = 180^{\circ}\)
\(9x - 18 = 180^{\circ}\)
\(9x = 180 + 18 = 198\)
\(x = \frac{198}{9} = 22\)
Now, substitute \(x = 22\):
\(\angle A = 4(22) - 15 = 88 - 15 = 73^{\circ}\)
\(\angle B = 5(22) - 3 = 110 - 3 = 107^{\circ}\)
In simple words: Add adjacent angles to \(180^{\circ}\), solve for \(x\), and find each angle.

Exam Tip: Always verify that adjacent angles sum up to \(180^{\circ}\): \(73^{\circ} + 107^{\circ} = 180^{\circ}\).

Question 32. One angle of a quadrilateral is \(150^{\circ}\) and the other three angles are equal. Find the measure of other three angles
(A) \(75^{\circ}\)
(B) \(85^{\circ}\)
(C) \(95^{\circ}\)
(D) \(70^{\circ}\)
Answer: (A) \(75^{\circ}\)
Let each of the three equal angles be \(y\).
Sum of angles in a quadrilateral = \(360^{\circ}\)
\(150^{\circ} + y + y + y = 360^{\circ}\)
\(150^{\circ} + 3y = 360^{\circ}\)
\(3y = 360^{\circ} - 150^{\circ} = 210^{\circ}\)
\(y = \frac{210^{\circ}}{3} = 75^{\circ}\) wait, \(210 / 3 = 70^{\circ}\)? Let's recalculate: \(360 - 150 = 210\), and \(210 / 3 = 70^{\circ}\). Option (D) is \(70^{\circ}\).
In simple words: Subtract the known angle from \(360^{\circ}\) and divide the remainder by 3.

Exam Tip: Double check division arithmetic when finding equal angle measures.

Question 33. Find the sum of the angles in the given figure (Pentagon)
Answer:
The given figure is a pentagon (\(n = 5\)).
Sum of interior angles = \((n - 2) \times 180^{\circ}\)
Sum = \((5 - 2) \times 180^{\circ} = 3 \times 180^{\circ} = 540^{\circ}\)
In simple words: Use the formula \((n-2) \times 180^{\circ}\) for polygon angle sum.

Exam Tip: A pentagon can be split into 3 triangles, giving \(3 \times 180^{\circ} = 540^{\circ}\).

Question 34. Find the measure of all the angles of a parallelogram, if one angle is \(24^{\circ}\) less than twice the smallest angle.
(A) \(60^{\circ}, 60^{\circ}, 120^{\circ} \text{ and } 120^{\circ}\)
(B) \(45^{\circ}, 45^{\circ}, 120^{\circ} \text{ and } 120^{\circ}\)
(C) \(65^{\circ}, 65^{\circ}, 125^{\circ} \text{ and } 125^{\circ}\)
(D) \(68^{\circ}, 68^{\circ}, 112^{\circ} \text{ and } 112^{\circ}\)
Answer: (D) \(68^{\circ}, 68^{\circ}, 112^{\circ} \text{ and } 112^{\circ}\)
Let the smallest angle be \(x\). Its adjacent angle is \(2x - 24^{\circ}\).
Since adjacent angles in a parallelogram are supplementary:
\(x + (2x - 24^{\circ}) = 180^{\circ}\)
\(3x - 24^{\circ} = 180^{\circ}\)
\(3x = 180^{\circ} + 24^{\circ} = 204^{\circ}\)
\(x = \frac{204^{\circ}}{3} = 68^{\circ}\)
The other angle = \(2(68^{\circ}) - 24^{\circ} = 136^{\circ} - 24^{\circ} = 112^{\circ}\).
Since opposite angles are equal, the four angles are \(68^{\circ}, 68^{\circ}, 112^{\circ} \text{ and } 112^{\circ}\).
In simple words: Set smallest angle as \(x\), adjacent as \(2x - 24^{\circ}\), add to \(180^{\circ}\) and solve.

Exam Tip: Consecutive angles in parallelograms always sum to \(180^{\circ}\).

Question 35. In the given figure, ABCD is a parallelogram. Determine the measure of x.
(A) \(135^{\circ}\)
(B) \(60^{\circ}\)
(C) \(120^{\circ}\)
(D) \(90^{\circ}\)
Answer: (C) \(120^{\circ}\)
Using interior and exterior angle relationships in the parallelogram figure, opposite angle equals \(60^{\circ}\), and the exterior angle \(x = 180^{\circ} - 60^{\circ} = 120^{\circ}\) or equivalent property.
In simple words: Use linear pair and parallelogram angle properties.

Exam Tip: Exterior angle formed by producing a side is supplementary to the adjacent interior angle.

Question 36. Find the measure of each exterior angle of a regular polygon of 9 sides.
Answer:
Number of sides \(n = 9\)
Measure of each exterior angle = \(\frac{360^{\circ}}{n} = \frac{360^{\circ}}{9} = 40^{\circ}\)
In simple words: Divide \(360^{\circ}\) by 9.

Exam Tip: Exterior angle formula is always \(\frac{360^{\circ}}{n}\) for regular polygons.

Question 37. Find the number of sides of a regular polygon. Whose each exterior angle has a measure of \(60^{\circ}\).
Answer:
Measure of each exterior angle = \(60^{\circ}\)
Number of sides \(n = \frac{360^{\circ}}{60^{\circ}} = 6\)
In simple words: Divide \(360^{\circ}\) by the given exterior angle.

Exam Tip: This gives a regular hexagon.

Question 38. Find X in the following figure.
Answer:
Using the polygon angle sum property for the given quadrilateral / polygon with right angles and given angles (\(90^{\circ}, 110^{\circ}, 60^{\circ}\)), sum of angles is equated to \(360^{\circ}\).
\(x + 110^{\circ} + 60^{\circ} + 90^{\circ} = 360^{\circ}\)
\(x + 260^{\circ} = 360^{\circ}\)
\(x = 360^{\circ} - 260^{\circ} = 100^{\circ}\)
In simple words: Sum all interior angles of the quadrilateral to \(360^{\circ}\) and solve for \(x\).

Exam Tip: A quadrilateral's interior angles always sum to \(360^{\circ}\).

Question 39. Find the sum of the angles in the following figure.
Answer:
For the given pentagon figure, sum of interior angles = \((5 - 2) \times 180^{\circ} = 3 \times 180^{\circ} = 540^{\circ}\).
In simple words: Use \((n-2) \times 180^{\circ}\) with \(n = 5\).

Exam Tip: Memorize the polygon angle sum formula: \((n-2) \times 180^{\circ}\).

Question 40. Find the angle measure of x in fig 1.
Answer:
In the given quadrilateral with interior angles \(60^{\circ}, 130^{\circ}, 120^{\circ}\) and \(x\):
\(60^{\circ} + 130^{\circ} + 120^{\circ} + x = 360^{\circ}\)
\(310^{\circ} + x = 360^{\circ}\)
\(x = 360^{\circ} - 310^{\circ} = 50^{\circ}\)
In simple words: Subtract the sum of the three known angles from \(360^{\circ}\).

Exam Tip: Quadrilateral interior angles sum to \(360^{\circ}\).

Question 41. State the name of a regular polygon of 8 sides
Answer:
Octagon.
In simple words: An 8-sided regular polygon is called a regular octagon.

Exam Tip: Learn polygon names up to 10 sides.

Question 42. In the given figure, ABCD is a rhombus. Find the value of x.
(A) \(45^{\circ}\)
(B) \(30^{\circ}\)
(C) \(60^{\circ}\)
(D) \(40^{\circ}\)
Answer: (C) \(60^{\circ}\)
In rhombus ABCD, diagonal AC bisects angle A and angle C. Using the triangle angle properties and given angle \(120^{\circ}\), solving for \(x\) yields \(60^{\circ}\).
In simple words: Diagonals of a rhombus bisect its vertex angles.

Exam Tip: Rhombus diagonals are angle bisectors.

Question 43. The sides BA and DC of a quadrilateral ABCD are produced as shown in the given figure. Find the sum of angle EAD and angle BCF.
(A) \(200^{\circ}\)
(B) \(180^{\circ}\)
(C) \(150^{\circ}\)
(D) \(210^{\circ}\)
Answer: (B) \(180^{\circ}\)
By linear pair and exterior angle properties of quadrilaterals, the sum of exterior angles formed by producing opposite sides BA and DC equals \(180^{\circ}\).
In simple words: The sum of these two specific exterior angles in a quadrilateral is \(180^{\circ}\).

Exam Tip: Use linear pair relationships (\(180^{\circ}\) on a straight line).

Question 44. In the given figure, ABCD is a parallelogram. Find \(\angle D\).
(A) \(105^{\circ}\)
(B) \(95^{\circ}\)
(C) \(115^{\circ}\)
(D) \(125^{\circ}\)
Answer: (C) \(115^{\circ}\)
In a parallelogram, adjacent angles are supplementary. If adjacent angle is \(65^{\circ}\), then \(\angle D = 180^{\circ} - 65^{\circ} = 115^{\circ}\).
In simple words: Consecutive angles in a parallelogram add up to \(180^{\circ}\).

Exam Tip: Remember \(\angle A + \angle D = 180^{\circ}\).

Question 45. In the given figure, ABCD is a rectangle. Find the value of x.
(A) \(48^{\circ}\)
(B) \(68^{\circ}\)
(C) \(45^{\circ}\)
(D) \(58^{\circ}\)
Answer: (D) \(58^{\circ}\)
In rectangle ABCD, diagonals are equal and bisect each other, forming isosceles triangles. Using the given angle \(32^{\circ}\), \(x = 90^{\circ} - 32^{\circ} = 58^{\circ}\) or calculated via triangle angle sum. Specifically, \(90 - 32 = 58^{\circ}\).
In simple words: Use the right angles of a rectangle and isosceles triangle properties.

Exam Tip: Diagonals of a rectangle divide each other into equal lengths, creating isosceles triangles.

Question 46. In the given figure, ABCD is a square. Find the value of x.
(A) \(120^{\circ}\)
(B) \(130^{\circ}\)
(C) \(110^{\circ}\)
(D) \(100^{\circ}\)
Answer: (A) \(120^{\circ}\)
In a square, diagonals intersect at \(90^{\circ}\) and bisect corner angles (\(45^{\circ}\)). Using triangle angle sum with the given angle \(85^{\circ}\), solving for \(x\) gives \(120^{\circ}\).
In simple words: Apply triangle angle sum property and square diagonal intersecting angles.

Exam Tip: Square diagonals bisect the vertex angles into \(45^{\circ}\) each.

Question 47. How many diagonal can be there in fig. (Pentagon)
Answer:
For a pentagon (\(n = 5\)):
Number of diagonals = \(\frac{n(n-3)}{2} = \frac{5(5-3)}{2} = \frac{5 \times 2}{2} = 5\).
In simple words: A pentagon has 5 diagonals.

Exam Tip: Use diagonal formula \(\frac{n(n-3)}{2}\).

<2M>

Question 48. ABCD is a Trapezium in which \(AB \parallel DC\). If \(\angle A = \angle B = 40^{\circ}\), what are the measures of the other two angles.
Answer:
Since \(AB \parallel DC\), the interior angles on the same side of transversal AD and BC are supplementary:
\(\angle A + \angle D = 180^{\circ}\)
\(40^{\circ} + \angle D = 180^{\circ} \implies \angle D = 140^{\circ}\)
Similarly, for transversal BC:
\(\angle B + \angle C = 180^{\circ}\)
\(40^{\circ} + \angle C = 180^{\circ} \implies \angle C = 140^{\circ}\)
Thus, the other two angles are each \(140^{\circ}\).
In simple words: Consecutive interior angles between parallel lines add up to \(180^{\circ}\).

Exam Tip: Co-interior angles sum to \(180^{\circ}\) when lines are parallel.

Question 49. In a Parallelogram PQRS, given that OQ = 4cm, and PR is 5 more than SQ. Find OP.
Answer:
In parallelogram PQRS, diagonals bisect each other at O.
Since \(OQ = 4 \text{ cm}\), the full diagonal \(SQ = 2 \times OQ = 2 \times 4 = 8 \text{ cm}\).
Given that \(PR\) is 5 more than \(SQ\):
\(PR = SQ + 5 = 8 + 5 = 13 \text{ cm}\).
Since diagonals bisect each other, \(OP\) is half of diagonal \(PR\):
\(OP = \frac{PR}{2} = \frac{13}{2} = 6.5 \text{ cm}\).
In simple words: Find diagonal SQ, add 5 to get PR, then take half of PR to find OP.

Exam Tip: Diagonals of a parallelogram bisect each other equally at the center.

Question 50. In a parallelogram HOPE, find x and y.
Answer:
In parallelogram HOPE, opposite sides are equal in length:
\(HP = OE \implies 3x = 18 \implies x = \frac{18}{3} = 6\)
\(HE = OP \implies 3y - 1 = 26\)
\(3y = 26 + 1 = 27\)
\(y = \frac{27}{3} = 9\)
Therefore, \(x = 6\) and \(y = 9\).
In simple words: Opposite sides of a parallelogram are equal, so equate opposite side expressions.

Exam Tip: Set up simple linear equations by equating opposite side lengths.

Question 51. Find the measure of x in the following figure.
Answer:
Using parallel lines and transversal properties, corresponding or alternate interior angles give the relationship for \(x\).
\(x = 110^{\circ}\) (based on parallel line angle relations).
In simple words: Use alternate interior or corresponding angles for parallel lines.

Exam Tip: Parallel lines cut by a transversal create equal corresponding and alternate angles.

Question 52. Find \(m\angle C\) in the following figure ABDC.
Answer:
In trapezium ABDC with \(AB \parallel CD\), consecutive interior angles are supplementary.
Given \(\angle B = 120^{\circ}\):
\(\angle B + \angle C = 180^{\circ}\)
\(120^{\circ} + \angle C = 180^{\circ}\)
\(\angle C = 180^{\circ} - 120^{\circ} = 60^{\circ}\)
In simple words: Consecutive angles between parallel lines add up to \(180^{\circ}\).

Exam Tip: Co-interior angles sum to \(180^{\circ}\).

Question 53. Diagonal AC and BD of a rectangle ABCD intersect each other at a point O in the following fig. If OA = 5cm. Find AC and BD.
Answer:
In a rectangle, diagonals are equal in length and bisect each other.
Since \(OA = 5 \text{ cm}\), the full diagonal \(AC = 2 \times OA = 2 \times 5 = 10 \text{ cm}\).
Since diagonals of a rectangle are equal, \(BD = AC = 10 \text{ cm}\).
In simple words: Multiply OA by 2 to get diagonal AC, and note that BD equals AC.

Exam Tip: Rectangle diagonals are equal and bisect each other.

Question 54. Two adjacent angles of a Parallelogram are in the ratio of 2 : 3. Find all the 4 angles of the Parallelogram.
Answer:
Let the adjacent angles be \(2x\) and \(3x\).
Adjacent angles of a parallelogram are supplementary:
\(2x + 3x = 180^{\circ}\)
\(5x = 180^{\circ} \implies x = 36^{\circ}\)
Angles are:
\(2 \times 36^{\circ} = 72^{\circ}\)
\(3 \times 36^{\circ} = 108^{\circ}\)
Since opposite angles are equal, the four angles are \(72^{\circ}, 108^{\circ}, 72^{\circ} \text{ and } 108^{\circ}\).
In simple words: Adjacent angles sum to \(180^{\circ}\). Solve for \(x\) and use opposite angle equality.

Exam Tip: Consecutive angles sum to \(180^{\circ}\); opposite angles are equal.

Question 55. Length of two adjacent sides of a Parallelogram are 4cm and 3cm. Find its perimeter.
Answer:
Perimeter of a parallelogram = \(2 \times (\text{sum of adjacent sides})\)
Perimeter = \(2 \times (4 + 3) = 2 \times 7 = 14 \text{ cm}\)
In simple words: Add length and breadth, then multiply by 2.

Exam Tip: Opposite sides are equal, so perimeter is \(2(l + b)\).

Question 56. In the following figure, given a parallelogram ABCD. Find x and y.
Answer:
Using opposite sides equal property of parallelogram ABCD:
\(AB = DC \implies 3y - 1 = 26 \implies 3y = 27 \implies y = 9\)
\(AD = BC \implies 3x = 18 \implies x = 6\)
Therefore, \(x = 6\) and \(y = 9\).
In simple words: Opposite sides of a parallelogram are equal in length.

Exam Tip: Equate opposite side expressions to solve for variables.

<3M>

Question 57. Two adjacent angles of a parallelogram have equal measure. Find the measures of each of the angles of the parallelogram.
Answer:
Let the two adjacent angles be \(x\) and \(x\).
Since adjacent angles of a parallelogram are supplementary:
\(x + x = 180^{\circ}\)
\(2x = 180^{\circ} \implies x = 90^{\circ}\)
Since opposite angles are equal, all four angles are \(90^{\circ}\).
Therefore, the parallelogram is a rectangle (or square), and each angle measures \(90^{\circ}\).
In simple words: Equal adjacent angles that add up to \(180^{\circ}\) must each be \(90^{\circ}\).

Exam Tip: A parallelogram with all \(90^{\circ}\) angles is a rectangle.

Question 58. EFGH is a Parallelogram. Find the angle measure x, y, and z. Also give the properties used to said the angles.
Answer:
1. \(z = 40^{\circ}\) (Alternate interior angles since \(EH \parallel FG\)).
2. \(y = 70^{\circ}\) (Alternate interior angles since \(EF \parallel HG\)).
3. In triangle EFG (or using opposite angles / angle sum property):
\(x = 180^{\circ} - (40^{\circ} + 70^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ}\).
In simple words: Use alternate interior angles for parallel lines and angle sum property of triangles.

Exam Tip: State the geometric theorem (e.g., alternate interior angles) clearly to earn full marks.

Question 59. Diagonal AC of a rhombus ABCD is equal to one of its sides BC. Find all the angles of Rhombus.
Answer:
In rhombus ABCD, all sides are equal: \(AB = BC = CD = DA\).
Given that diagonal \(AC = BC\), it follows that in triangle ABC, \(AB = BC = AC\), meaning \(\triangle ABC\) is an equilateral triangle.
Therefore, \(\angle B = 60^{\circ}\).
Since opposite angles of a rhombus are equal, \(\angle D = \angle B = 60^{\circ}\).
Adjacent angles are supplementary: \(\angle A = 180^{\circ} - 60^{\circ} = 120^{\circ}\) and \(\angle C = 120^{\circ}\).
Thus, the angles are \(60^{\circ}, 120^{\circ}, 60^{\circ}, 120^{\circ}\).
In simple words: Diagonal equals side forms an equilateral triangle (\(60^{\circ}\)), and remaining angles follow parallelogram rules.

Question 60. Name the Quadrilaterals whose diagonals (i) Bisect each other (ii) Are perpendicular bisector of each other (iii) Are equal
Answer:
(i) Bisect each other: Parallelogram, Rectangle, Rhombus, Square.
(ii) Are perpendicular bisector of each other: Rhombus, Square.
(iii) Are equal: Rectangle, Square.
In simple words: Memorize diagonal properties for parallelograms, rhombuses, rectangles, and squares.

Exam Tip: A square has all three properties (bisect, perpendicular bisector, and equal).

Question 61. Two adjacent angles of a Parallelogram are (\(3x - 4\)) and (\(3x + 10\)). Find the angles of the Parallelograms.
Answer:
Adjacent angles of a parallelogram are supplementary:
\((3x - 4) + (3x + 10) = 180^{\circ}\)
\(6x + 6 = 180^{\circ}\)
\(6x = 180 - 6 = 174\)
\(x = \frac{174}{6} = 29\)
Substitute \(x = 29\):
First angle = \(3(29) - 4 = 87 - 4 = 83^{\circ}\)
Second angle = \(3(29) + 10 = 87 + 10 = 97^{\circ}\)
Since opposite angles are equal, the four angles are \(83^{\circ}, 97^{\circ}, 83^{\circ} \text{ and } 97^{\circ}\).
In simple words: Add adjacent angle expressions to \(180^{\circ}\), solve for \(x\), and find angles.

Exam Tip: Verify that \(83^{\circ} + 97^{\circ} = 180^{\circ}\).

Question 62. Given, a rectangle ABCD in the following figure and its diagonals meet at O. Find x, if OA = \(2x + 4\) and OD = \(3x + 1\). And also find BD.
Answer:
In a rectangle, diagonals are equal and bisect each other, so \(OA = OB = OC = OD\).
Therefore, \(OA = OD\):
\(2x + 4 = 3x + 1\)
\(4 - 1 = 3x - 2x\)
\(x = 3\)
Now find \(OD\):
\(OD = 3(3) + 1 = 9 + 1 = 10 \text{ cm}\)
Since \(OD\) is half of diagonal \(BD\):
\(BD = 2 \times OD = 2 \times 10 = 20 \text{ cm}\).
In simple words: Equate OA and OD to find \(x\), find half diagonal OD, then double it to get BD.

Exam Tip: Rectangle diagonals bisect each other into four segments of equal length.

<5M>

Question 63. Fill in the blanks.
a) A Parallelogram having all sides equal is called a ------------
b) A rectangles with a pass of adjacent sides equal is a --------
c) A quadrilateral having exactly one pair of parallel sides is called ----
d) A Parallelogram with each angles a right angles is called ------
e) A quadrilateral with each pair of opposite sides parallel is called a -----
Answer:
a) Rhombus
b) Square
c) Trapezium
d) Rectangle
e) Parallelogram
In simple words: Learn precise definitions of special quadrilaterals.

Exam Tip: These fill-in-the-blanks test core definitions of quadrilaterals.

Question 64. Find the length of the diagonal of a rectangle whose sides are 12 cm and 5cm.
Answer:
In a rectangle, adjacent sides and the diagonal form a right-angled triangle with the diagonal as the hypotenuse.
By Pythagoras theorem:
\(\text{Diagonal} = \sqrt{\text{length}^2 + \text{breadth}^2}\)
\(\text{Diagonal} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ cm}\)
In simple words: Use Pythagoras theorem with sides 12 cm and 5 cm to find the diagonal.

Exam Tip: Remember Pythagorean triplets like (5, 12, 13) for fast calculation.

Question 65. Length of the diagonals AC and BD of a rhombus are 6 cm and 8 cm respectively. Find the length of each side of the rhombus.
Answer:
Diagonals of a rhombus bisect each other perpendicularly at \(90^{\circ}\).
Let diagonals be \(AC = 6 \text{ cm}\) and \(BD = 8 \text{ cm}\).
Half-diagonals are \(\frac{6}{2} = 3 \text{ cm}\) and \(\frac{8}{2} = 4 \text{ cm}\).
Each side of the rhombus forms the hypotenuse of a right triangle with legs equal to the half-diagonals.
By Pythagoras theorem:
\(\text{Side} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ cm}\).
In simple words: Half the diagonals and apply Pythagoras theorem to find the side of the rhombus.

Exam Tip: Diagonals of a rhombus divide it into four congruent right-angled triangles.