CBSE Class 8 Mathematics Quadrilaterals Practice Paper Assignment Set 01 with Answers

This is CBSE Class 8 Maths Ganita Prakash Chapter 4 Quadrilaterals Practice Paper 10, a 40-mark paper with 20 questions in Sections A to E, and every question comes with a step-by-step answer. It tests the angle sum of a quadrilateral and of a polygon, interior and exterior angles of regular polygons, and the angle, side and diagonal properties of parallelograms, rectangles, rhombuses, kites and trapeziums, ending with case studies on rhombus-shaped tiles and angle bisectors in a trapezium.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. One angle of a quadrilateral is 120° and the remaining three angles are equal. Find the measure of each of the three equal angles.
Answer:
Let each of the three equal angles be x.
By the angle sum property of a quadrilateral: 120° + x + x + x = 360°.
3x = 360° − 120° = 240°.
x = 240° ÷ 3 = 80°.
So each of the three equal angles measures 80°.
Check: 120° + 80° + 80° + 80° = 360°.
In simple words: Take 120° away from 360° and share the remaining 240° equally among three angles: 80° each.

Teacher's Note:
1. Use one letter for all the equal angles.
2. Always check that the four angles total 360°.

Question 12. How many sides does a regular polygon have, if each of its interior angles is 165°?
Answer:
Let the polygon have n sides.
Each interior angle is 165°, so each exterior angle = 180° − 165° = 15°.
The exterior angles of any polygon add up to 360°, and in a regular polygon they are all equal.
n = 360° ÷ 15° = 24.
Check with the interior-angle formula: (24 − 2) × 180° ÷ 24 = 22 × 180° ÷ 24 = 3960° ÷ 24 = 165°.
So the polygon has 24 sides.
In simple words: Each corner turns by 15°, and 360° ÷ 15° = 24 corners, so 24 sides.

Teacher's Note:
1. Exterior angle = 180° − interior angle; number of sides = 360° ÷ exterior angle.
2. The equation (n − 2) × 180 = 165n gives the same answer: 15n = 360, n = 24.

Question 13. Ram was designing a rangoli using polygon shapes. He created a pattern using a regular polygon in which each interior angle measured 120°. Find the number of sides of the polygon and classify it.
Answer:
Let the regular polygon have n sides.
Each interior angle = (n − 2) × 180° ÷ n = 120°.
So 180n − 360 = 120n, which gives 60n = 360 and n = 6.
Check with exterior angles: 180° − 120° = 60° and 360° ÷ 60° = 6.
A polygon with 6 equal sides and 6 equal angles is a regular hexagon (6 sides).
In simple words: Each exterior angle is 60°, and 360° ÷ 60° = 6, so the shape is a regular hexagon.

Teacher's Note:
1. Regular means all sides equal and all angles equal.
2. Regular hexagons fit together without gaps, which is why they are popular in rangoli and tiles.

Question 14. One angle of a parallelogram is of measure 70°. Find the measure of the remaining angles of the parallelogram.
Answer:
Let ABCD be the parallelogram with ∠A = 70°.

Opposite angles of a parallelogram are equal, so ∠C = ∠A = 70°.
AD ∥ BC and AB is a transversal, so the co-interior angles ∠A and ∠B add to 180°: ∠B = 180° − 70° = 110°.
∠D is opposite ∠B, so ∠D = 110°.
So the remaining angles are ∠B = 110°, ∠C = 70° and ∠D = 110°.
Check: 70° + 110° + 70° + 110° = 360°.
In simple words: The opposite angle is also 70°, and the two neighbouring angles are 180° − 70° = 110° each.

Teacher's Note:
1. Opposite angles are equal; adjacent angles add to 180°.
2. A parallelogram always has two equal acute angles and two equal obtuse angles (or four right angles).

Section C: Short Answer Questions (3 Marks Each)

Question 15. The sum of the exterior angles of a polygon is one-ninth of sum of the interior angles. Find the number of sides of the polygon.
Answer:
Let the polygon have n sides.
The sum of the exterior angles of every polygon is 360°.
Given: sum of exterior angles = \( \frac{1}{9} \) × sum of interior angles.
So sum of interior angles = 9 × 360° = 3240°.
But sum of interior angles = (n − 2) × 180°, so (n − 2) × 180° = 3240°.
n − 2 = 3240 ÷ 180 = 18, so n = 20.
So the polygon has 20 sides.
In simple words: The inside angles add to 9 × 360° = 3240°, which is 18 lots of 180°, so the polygon has 18 + 2 = 20 sides.

Teacher's Note:
1. The exterior-angle sum is always 360°, whatever the number of sides.
2. Check: (20 − 2) × 180° = 3240° and 3240° ÷ 9 = 360°.

Question 16. ABCD is a parallelogram where m∠A = (2x + 50°) and m∠C = (3x + 40°). (i) Find the value of x (ii) Find the measure of each angle.

Answer:
(i) ∠A and ∠C are opposite angles of the parallelogram, so they are equal.
2x + 50° = 3x + 40°.
50° − 40° = 3x − 2x, so x = 10°.
(ii) m∠A = 2 × 10° + 50° = 70° and m∠C = 3 × 10° + 40° = 70°.
∠A and ∠B are adjacent angles, so m∠B = 180° − 70° = 110°.
∠D is opposite ∠B, so m∠D = 110°.
So ∠A = 70°, ∠B = 110°, ∠C = 70°, ∠D = 110°.
In simple words: Opposite angles are equal, so 2x + 50 = 3x + 40 gives x = 10; then ∠A = ∠C = 70° and ∠B = ∠D = 110°.

Teacher's Note:
1. ∠A and ∠C are opposite, so set the two expressions equal.
2. Substitute x back into both expressions: getting the same value (70°) is a built-in check.

Question 17. In the below figure, ABCD is a rectangle. Its diagonals meet at O. Find x, if OA = 3x + 1 and OB = 2x + 4.
OR
In the given figure, ABCD and BDCE are parallelograms with common base DC. If BC ⊥ BD, then find ∠BEC.

Answer:
First question (rectangle):
The diagonals of a rectangle are equal: AC = BD.
The diagonals of a rectangle (like any parallelogram) bisect each other, so OA = \( \frac{1}{2} \) AC and OB = \( \frac{1}{2} \) BD.
Hence OA = OB, so 3x + 1 = 2x + 4.
3x − 2x = 4 − 1, so x = 3.
Check: OA = 3 × 3 + 1 = 10 and OB = 2 × 3 + 4 = 10.
OR question (two parallelograms):
In parallelogram ABCD, ∠BCD = ∠BAD = 30° (opposite angles are equal).
In triangle BCD, ∠DBC = 90° (since BC ⊥ BD) and ∠BCD = 30°.
So ∠BDC = 180° − 90° − 30° = 60° (angle sum of a triangle).
In parallelogram BDCE, ∠BEC is opposite ∠BDC, so ∠BEC = ∠BDC.
So ∠BEC = 60°.
In simple words: In a rectangle the half-diagonals OA and OB are equal, giving x = 3; in the OR question ∠BDC = 180° − 90° − 30° = 60°, and ∠BEC equals it.

Teacher's Note:
1. Half-diagonals of a rectangle are all equal: OA = OB = OC = OD.
2. In the OR question, move the 30° from A to C using opposite angles, then use the triangle angle sum.

Section D: Long Answer Question (5 Marks)

Question 18. The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm) [2 + 3 marks]

Answer:
(i) In parallelogram GUNS, opposite sides are equal: GU = SN and GS = UN.
GU = SN gives 3y − 1 = 26, so 3y = 27 and y = 9.
GS = UN gives 3x = 18, so x = 6.
Check: GU = 3 × 9 − 1 = 26 cm and GS = 3 × 6 = 18 cm.
(ii) In parallelogram RUNS, the diagonals SU and RN bisect each other. Let them meet at O.
So SO = OU and RO = ON.
SO = OU gives 20 = y + 7, so y = 13.
RO = ON gives 16 = x + y = x + 13, so x = 3.
Check: OU = 13 + 7 = 20 cm and ON = 3 + 13 = 16 cm.
So in (i) x = 6 cm, y = 9 cm, and in (ii) x = 3 cm, y = 13 cm.
In simple words: Use equal opposite sides in the first figure and equal halves of each diagonal in the second.

Teacher's Note:
1. Figure (i) uses the side property; figure (ii) uses the diagonal property.
2. In (ii), find y first from the diagonal that contains only y, then use it to find x.
3. Match each label with its own segment: y + 7 is the other half of the diagonal that has 20 on it.

Section E: Case Study Based Questions (4 Marks Each)

Question 19. A renowned flooring company is developing designer tiles shaped as special quadrilaterals. One new tile design has four equal-length sides, and when measured, its diagonals intersect at right angles but differ in length. The company’s production engineer must classify this shape correctly for consistency in manufacturing and cataloging.
By using the above, give the answers of the following questions:
(a) What is the name of this quadrilateral?
(b) Mention any two properties of this shape.
(c) Could this tile be a square? Why or why not?
(d) Name another quadrilateral with perpendicular diagonals and unequal side lengths.
Answer:
(a) A quadrilateral with all four sides equal and unequal diagonals is a rhombus.
(b) Any two of: all four sides are equal; opposite sides are parallel; opposite angles are equal; the diagonals bisect each other at right angles.
(c) No. A square also has equal sides and perpendicular diagonals, but the diagonals of a square are always equal. This tile's diagonals differ in length, so it is not a square.
(d) A kite: its diagonals are perpendicular, but its sides are equal only in two adjacent pairs, not all four equal.
In simple words: Equal sides with unequal diagonals make a rhombus; a square needs equal diagonals, and a kite is another shape with perpendicular diagonals.

Teacher's Note:
1. Equal sides alone do not make a square; the angles (or diagonals) must also match.
2. Diagonals perpendicular: square, rhombus and kite. Diagonals equal: square and rectangle.

Question 20. There is a trapezium MNOP, angle bisector of ∠M and ∠N meet at point W, and angle bisector of ∠O and ∠P meet at point X on side MN of trapezium MNOP.
By using the figure give the answers of the following questions:
(i) What is the value of a? (1)
(ii) What is the value of d? (1½)
(iii) What is the value of c? (1½)
Answer:
In trapezium MNOP, MN ∥ PO. PX bisects ∠P, so ∠XPO = 50° and ∠P = 100°; OX bisects ∠O, so ∠XOP = 70° and ∠O = 140°.
(i) In triangle XPO: a + 50° + 70° = 180°, so a = 60°.
(ii) ∠O and ∠N are co-interior angles between the parallel sides, so ∠N = 180° − 140° = 40°, and ½∠N = 20°.
Let NW cut OX at Y. In triangle NOY: ∠YON = ½∠O = 70° and ∠ONY = ½∠N = 20°.
d = 180° − 70° − 20°, so d = 90°.
(iii) Similarly ∠M = 180° − ∠P = 180° − 100° = 80°, so ½∠M = 40°.
Let MW cut PX at Z. In triangle MZP: ∠ZPM = ½∠P = 50° and ∠ZMP = ½∠M = 40°.
c = 180° − 50° − 40°, so c = 90°.
In simple words: a = 180° − 50° − 70° = 60°; the bisectors of two co-interior angles always meet at 90°, so c = d = 90°.

Teacher's Note:
1. Halves of two angles that add to 180° add to 90°, so the third angle of that triangle is 90°.
2. The figure is drawn to the given measures (∠P = 100°, ∠O = 140°), so it looks different from a typical trapezium sketch.