CBSE Class 8 Mathematics Understanding Quadrilaterals Practice Paper Worksheet Set 15 with Answers (Old NCERT Syllabus)

Class 8 Maths Understanding Quadrilaterals Practice Paper 02 is a 40-mark test of 20 questions arranged in Sections A to E, with step-by-step solutions for every question. It is based on the earlier (old) NCERT Class 8 Maths syllabus, which makes it handy for state-board students and for revising quadrilaterals. Students practise polygon angle sums, regular polygons, kite and parallelogram properties in assertion–reason form, the diagonals of rectangles and parallelograms, and two 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 equal angle be y.
By the angle sum property: 120° + 3y = 360°.
3y = 240°, so y = 80°.
So each of the three equal angles measures 80°.
Check: 120° + 80° + 80° + 80° = 360°.
In simple words: 360° − 120° leaves 240° for three equal angles, so each one is 80°.

Teacher's Note:
1. Write the three equal angles with the same letter so there is only one unknown.
2. Check by adding all four angles.

Question 12. How many sides does a regular polygon have, if each of its interior angles is 165°?
Answer:
Let the number of sides be n.
Each interior angle = (n − 2) × 180° ÷ n = 165°.
So 180n − 360 = 165n, which gives 15n = 360.
n = 360 ÷ 15 = 24.
Check by exterior angles: 180° − 165° = 15° and 360° ÷ 15° = 24.
So the polygon has 24 sides.
In simple words: Each exterior angle is 15°, and 24 of them make the full 360°, so there are 24 sides.

Teacher's Note:
1. Both methods (interior-angle formula and exterior angles) give the same answer; the exterior route is shorter.
2. If 360° ÷ exterior angle is not a whole number, no such regular polygon exists.

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:
Each exterior angle = 180° − 120° = 60°.
Number of sides = 360° ÷ 60° = 6.
Check with the formula: (6 − 2) × 180° ÷ 6 = 720° ÷ 6 = 120°.
A regular polygon with 6 sides is a regular hexagon, so the polygon has 6 sides.
In simple words: Each corner turns through 60°, and 360° ÷ 60° = 6, so Ram used a regular hexagon.

Teacher's Note:
1. 'Classify' means naming the polygon by its number of sides.
2. Regular hexagons fit together without gaps, which is why they appear in tiles and rangoli patterns.

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

∠C is opposite ∠A, so ∠C = 70°.
∠A and ∠B are adjacent angles (co-interior between AD ∥ BC), so ∠B = 180° − 70° = 110°.
∠D is opposite ∠B, so ∠D = 110°.
So ∠B = 110°, ∠C = 70°, ∠D = 110°.
Check: 70° + 110° + 70° + 110° = 360°.
In simple words: The angle opposite 70° is also 70°, and the two angles next to it are 110° each.

Teacher's Note:
1. Two parallelogram facts are used: opposite angles equal, adjacent angles add to 180°.
2. The final check uses the quadrilateral angle sum of 360°.

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 exterior angles of any polygon add up to 360°.
Since this is \( \frac{1}{9} \) of the interior angle sum, the interior angle sum = 9 × 360° = 3240°.
Interior angle sum = (n − 2) × 180°, so (n − 2) × 180° = 3240°.
n − 2 = 3240 ÷ 180 = 18, which gives n = 20.
So the polygon has 20 sides.
In simple words: Nine times 360° is 3240°, which is 18 triangles' worth of 180°, so the polygon has 20 sides.

Teacher's Note:
1. Turn the word 'one-ninth' into an equation: exterior sum = \( \frac{1}{9} \) × interior sum.
2. Check: (20 − 2) × 180° = 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) In a parallelogram, opposite angles are equal, so m∠A = m∠C.
2x + 50° = 3x + 40°, so 3x − 2x = 50° − 40°.
x = 10°.
(ii) m∠A = 2 × 10° + 50° = 70°, and m∠C = 3 × 10° + 40° = 70°.
∠B is adjacent to ∠A, 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: Set the two opposite angles equal to get x = 10, then the angles are 70°, 110°, 70° and 110°.

Teacher's Note:
1. Put the variable terms on one side and the numbers on the other.
2. Substituting x into both expressions and getting the same 70° confirms the answer.

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):
A rectangle's diagonals are equal (AC = BD) and bisect each other at O.
So OA = \( \frac{1}{2} \) AC = \( \frac{1}{2} \) BD = OB.
3x + 1 = 2x + 4, so x = 3.
Check: OA = 10 and OB = 10.
OR question (two parallelograms):
In parallelogram ABCD, ∠BCD = ∠BAD = 30° (opposite angles).
In triangle BDC, ∠DBC = 90° because BC ⊥ BD, so ∠BDC = 180° − 90° − 30° = 60°.
In parallelogram BDCE, ∠BEC and ∠BDC are opposite angles, so ∠BEC = ∠BDC.
So ∠BEC = 60°.
In simple words: The half-diagonals of a rectangle are equal, which gives x = 3; in the OR question, ∠BEC equals ∠BDC = 60°.

Teacher's Note:
1. Equal diagonals plus bisecting diagonals together give OA = OB in a rectangle.
2. In the OR question, the shared triangle BDC links both parallelograms.

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) Opposite sides of parallelogram GUNS are equal, so GU = SN and GS = UN.
3y − 1 = 26 gives 3y = 27, so y = 9.
3x = 18 gives x = 6.
(ii) The diagonals SU and RN of parallelogram RUNS bisect each other at O, so SO = OU and RO = ON.
SO = OU: y + 7 = 20, so y = 13.
RO = ON: x + y = 16, so x + 13 = 16 and x = 3.
Check: OU = 13 + 7 = 20 cm = SO and ON = 3 + 13 = 16 cm = RO.
So (i) x = 6, y = 9; (ii) x = 3, y = 13.
In simple words: Opposite sides match in figure (i); in figure (ii) each diagonal is cut into two equal halves.

Teacher's Note:
1. Each figure tests a different parallelogram property: sides in (i), diagonals in (ii).
2. In (ii), y + 7 is marked on OU, so y comes first; then substitute y in x + y = 16.

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) Four equal sides with unequal diagonals that cross at right angles describe a rhombus.

(b) Any two of these: 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 its diagonals are always equal in length. The tile's diagonals differ, so it is not a square.
(d) A kite: its diagonals are perpendicular, but its sides are equal only in two pairs of adjacent sides.
In simple words: The tile is a rhombus; it cannot be a square because a square's diagonals are equal; a kite also has perpendicular diagonals.

Teacher's Note:
1. Diagonals help to tell shapes apart: equal diagonals point to a rectangle or square, perpendicular ones to a rhombus, square or kite.
2. A square is a rhombus with right angles, which is why its diagonals become equal.

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:
Here MN ∥ PO. Since PX and OX are bisectors, ∠P = 2 × 50° = 100° and ∠O = 2 × 70° = 140°.
(i) Triangle XPO has angles 50°, 70° and a, so a = 180° − 120° and a = 60°.
(ii) ∠N and ∠O are co-interior angles between MN ∥ PO, so ∠N = 180° − 140° = 40°; NW makes 20° with NO.
Let NW cross OX at Y. In triangle NYO, the angles at O and N are 70° and 20°.
So d = 180° − 70° − 20° and d = 90°.
(iii) ∠M = 180° − ∠P = 80°, so MW makes 40° with MP.
Let MW cross PX at Z. In triangle MZP, the angles at P and M are 50° and 40°.
So c = 180° − 50° − 40° and c = 90°.
In simple words: a is the third angle of a triangle with 50° and 70°, so 60°; c and d are 90° because the halves of two co-interior angles add to 90°.

Teacher's Note:
1. Any two angles on the same slanting side of a trapezium add to 180°.
2. The figure is drawn to the given measures (∠P = 100°, ∠O = 140°), so W comes inside the trapezium, unlike the paper's sketch.