CBSE Class 8 Mathematics Understanding Quadrilaterals Practice Paper Assignment Set 11 with Answers (Old NCERT Syllabus)

This Class 8 Maths Understanding Quadrilaterals Practice Paper 01 is a 40-mark paper with 17 questions in Sections A to E, and every question has a clear step-by-step answer. It follows the earlier (old) NCERT Class 8 Maths syllabus, so it is useful for state-board students and for revising this chapter. The paper tests the angle sum of quadrilaterals and polygons, interior and exterior angles of regular polygons, and the angle, side and diagonal properties of parallelograms, rectangles and trapeziums, including a ship figure and a trapezium with angle bisectors.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 7. 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 measure x.
Angle sum of a quadrilateral: 120° + x + x + x = 360°.
3x = 360° − 120° = 240°.
x = 240° ÷ 3 = 80°.
So each of the three equal angles is 80°.
Check: 120° + 3 × 80° = 120° + 240° = 360°.
In simple words: After removing 120° from 360°, the remaining 240° is shared equally by three angles, so each is 80°.

Teacher's Note:
1. Use one variable for all angles that are equal.
2. Finish by checking that the four angles add to 360°.

Question 8. How many sides does a regular polygon have, if each of its interior angles is 165°?
Answer:
Let the regular polygon have n sides.
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.
So n = 360° ÷ 15° = 24.
Check: (24 − 2) × 180° = 3960°, and 3960° ÷ 24 = 165°.
The polygon has 24 sides.
In simple words: Each outside turn is 15°, and 360° ÷ 15° = 24, so the polygon has 24 sides.

Teacher's Note:
1. Number of sides = 360° ÷ exterior angle, for a regular polygon.
2. The equation (n − 2) × 180 = 165n leads to the same result: 180n − 165n = 360, so n = 24.

Question 9. 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 with AB as a transversal, so ∠A + ∠B = 180° (co-interior angles). Hence ∠B = 180° − 70° = 110°.
∠D is opposite ∠B, so ∠D = 110°.
So the remaining angles are ∠B = 110°, ∠C = 70°, ∠D = 110°.
Check: 70° + 110° + 70° + 110° = 360°.
In simple words: The opposite angle is also 70°, and each of the other two is 180° − 70° = 110°.

Teacher's Note:
1. Opposite angles equal, adjacent angles supplementary: these two facts solve most parallelogram angle questions.
2. Name the parallelogram (ABCD) so that you can say which angle is opposite or adjacent.

Section C: Short Answer Questions (3 Marks Each)

Question 10. 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.
Sum of the exterior angles of any polygon = 360°.
Given: 360° = \( \frac{1}{9} \) × (sum of interior angles), so sum of interior angles = 9 × 360° = 3240°.
Sum of interior angles of an n-sided polygon = (n − 2) × 180°.
(n − 2) × 180° = 3240°, so n − 2 = 3240 ÷ 180 = 18.
n = 18 + 2 = 20.
So the polygon has 20 sides.
In simple words: The interior angles total 9 × 360° = 3240°, which is 18 × 180°, so the polygon has 18 + 2 = 20 sides.

Teacher's Note:
1. The exterior-angle sum is 360° for every polygon, so it is the known quantity here.
2. Check: (20 − 2) × 180° = 3240°, and one-ninth of 3240° is 360°.

Question 11. 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 m∠A = m∠C.
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 (co-interior between AD ∥ BC), 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: Equal opposite angles give 2x + 50 = 3x + 40, so x = 10; then ∠A = ∠C = 70° and ∠B = ∠D = 110°.

Teacher's Note:
1. Setting the two expressions equal works because ∠A and ∠C are opposite angles.
2. Getting 70° from both expressions confirms the value of x.

Question 12. In the below figure, ABCD is a rectangle. Its diagonals meet at O. Find x, if OA = 3x + 1 and OB = 2x + 4.

Answer:
The diagonals of a rectangle are equal: AC = BD.
The diagonals of a rectangle bisect each other, so OA = \( \frac{1}{2} \) AC and OB = \( \frac{1}{2} \) BD.
Hence OA = OB, which gives 3x + 1 = 2x + 4.
3x − 2x = 4 − 1, so x = 3.
Check: OA = 3(3) + 1 = 10 and OB = 2(3) + 4 = 10.
In simple words: Half of one diagonal equals half of the other, so 3x + 1 = 2x + 4 and x = 3.

Teacher's Note:
1. In a rectangle, OA = OB = OC = OD because the diagonals are equal and bisect each other.
2. This equation would not hold in a general parallelogram, where OA and OB can differ.

Question 13. In the given figure, ABCD and BDCE are parallelograms with common base DC. If BC ⊥ BD, then find ∠BEC.

Answer:
From the figure, ∠BAD = 30°.
In parallelogram ABCD, ∠BCD is opposite ∠BAD, so ∠BCD = 30°.
In triangle BCD, ∠DBC = 90° (BC ⊥ BD) and ∠BCD = 30°.
By the angle sum of a triangle, ∠BDC = 180° − 90° − 30° = 60°.
In parallelogram BDCE, ∠BEC is opposite ∠BDC, so ∠BEC = ∠BDC.
So ∠BEC = 60°.
In simple words: Move the 30° from A to C, find ∠BDC = 60° from triangle BCD, then ∠BEC equals ∠BDC.

Teacher's Note:
1. Look for which angle is opposite which in each parallelogram before writing equations.
2. Triangle BCD is shared by both parallelograms, which links the two figures.

Section D: Long Answer Questions (4 Marks Each)

Question 14. In the following figure of a ship, ABDH and CEFG are two parallelograms. Find the value of x.

Answer:
Let the dotted lines HD and GC cross at X. Both angles marked x are vertically opposite, so it is enough to find ∠CXD.
In parallelogram ABDH, ∠ABD = 130°. Adjacent angles of a parallelogram are supplementary, so ∠HDB = 180° − 130° = 50°.
In parallelogram CEFG, opposite angles are equal, so ∠GCE = ∠GFE = 30°. As D lies on CE, ∠XCD = 30°.
In triangle XCD: ∠XDC = 50° and ∠XCD = 30°.
x = 180° − 50° − 30° = 100°.
So x = 100°.
In simple words: The parallelograms give angles of 50° at D and 30° at C, so the third angle of the small triangle is 100°.

Teacher's Note:
1. Use the adjacent-angle fact in one parallelogram and the opposite-angle fact in the other.
2. The top angle x equals the bottom angle x because they are vertically opposite.
3. The figure is drawn to the given measures (130° and 30°), so the right side of the hull slopes more gently than in the paper's sketch.

Question 15. ABCD is a trapezium such that AB ∥ CD, ∠A : ∠D = 2 : 1, ∠B : ∠C = 7 : 5, find the angles of the trapezium.

Answer:
AB ∥ DC, so AD and BC are transversals. Co-interior angles on the same side add to 180°: ∠A + ∠D = 180° and ∠B + ∠C = 180°.
Let ∠A = 2k and ∠D = k. Then 3k = 180°, so k = 60°.
∠A = 2 × 60° = 120° and ∠D = 60°.
Let ∠B = 7m and ∠C = 5m. Then 12m = 180°, so m = 15°.
∠B = 7 × 15° = 105° and ∠C = 5 × 15° = 75°.
So ∠A = 120°, ∠B = 105°, ∠C = 75°, ∠D = 60°.
Check: 120° + 105° + 75° + 60° = 360°.
In simple words: Each pair of angles on a slanting side adds to 180°; split 180° in the ratios 2 : 1 and 7 : 5.

Teacher's Note:
1. In a trapezium, the angles on the same non-parallel side are co-interior, so they add to 180°.
2. Use different letters (k and m) for the two ratios, because they are not linked.
3. The figure is drawn with these angles, so the obtuse angles 120° and 105° are at the top, along AB.

Question 16. 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.
3y − 1 = 26, so 3y = 27 and y = 9.
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 at O, so SO = OU and RO = ON.
SO = OU: 20 = y + 7, so y = 13.
RO = ON: 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, y = 9, and in (ii) x = 3, y = 13.
In simple words: In (i) match opposite sides; in (ii) the two halves of each diagonal are equal.

Teacher's Note:
1. Figure (i) uses the side property and figure (ii) uses the diagonal property of a parallelogram.
2. In (ii), the segment OU is marked y + 7, so solve for y first and then use it in x + y = 16.

Section E: Case Study Based Question (4 Marks)

Question 17. 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 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:
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 lie on the same non-parallel side ON, so ∠O + ∠N = 180° and ∠N = 40°. Half of ∠N is 20°.
Let NW meet OX at Y. In triangle NOY: ∠YON = 70° and ∠ONY = 20°.
d = 180° − 70° − 20°, so d = 90°.
(iii) Similarly ∠P + ∠M = 180°, so ∠M = 80° and half of ∠M is 40°.
Let MW meet PX at Z. In triangle MZP: ∠ZPM = 50° and ∠ZMP = 40°.
c = 180° − 50° − 40°, so c = 90°.
In simple words: a = 180° − 50° − 70° = 60°, and the bisectors of two angles that add to 180° always meet at a right angle, so c = d = 90°.

Teacher's Note:
1. If two angles add to 180°, their halves add to 90°, so the third angle of the triangle they form is 90°.
2. The figure is drawn to the given measures (∠P = 100°, ∠O = 140°), so it differs from the paper's sketch, where these angles look acute.