CBSE Class 8 Mathematics Quadrilaterals Short Answer Questions with Answers

Short Answer Questions

Question 1. Look at the two figures described below.
Figure P is a closed figure made of four line segments joined end to end.
Figure Q is a closed figure made of three line segments and one curved arc.
State which of them is a quadrilateral and give two reasons for your choice. Also write what is meant by “the angles of a quadrilateral”.
Answer:
Figure P is the quadrilateral.
Reason 1: it is a closed figure.
Reason 2: all its four boundaries are straight line segments, so it has four sides and four vertices.
Figure Q is not a quadrilateral, because one boundary is a curved arc, not a line segment, so it does not have four sides.
The angles of a quadrilateral are the four angles formed inside it, each between two sides meeting at a vertex.
In simple words: A quadrilateral must be closed and made of exactly four straight sides; its angles are the four corners inside.

Teacher's Note:
1. Closed, four sides, all straight: check all three.
2. A single curved edge disqualifies a figure.

Question 2. The chapter finally defines a rectangle using one condition only.
(i) Write that definition.
(ii) Write any two other properties of a rectangle that follow from this definition.
Answer:
(i) A rectangle is a quadrilateral in which all the angles are 90°. This one condition is enough, because four right angles force the opposite sides to be equal.
(ii) Two other properties:
(a) The opposite sides are equal and parallel.
(b) The diagonals are equal in length and bisect each other.
In simple words: Four right angles define a rectangle, and from them we get equal parallel opposite sides and equal bisecting diagonals.

Teacher's Note:
1. A definition is the starting point; properties are deduced.
2. Any two of the four properties are acceptable.

Question 3. Three angles of a quadrilateral ABCD measure 85°, 95° and 110°. Find the fourth angle, stating the property of quadrilaterals you have used.
Answer:
Property used: the sum of all the angles of a quadrilateral is 360°.
Sum of the three given angles = 85° + 95° + 110° = 290°.
Fourth angle = 360° − 290° = 70°.
Check: 85 + 95 + 110 + 70 = 360. So ∠D = 70°.
In simple words: The four angles add to 360°, so the missing one is 360° − 290° = 70°.

Teacher's Note:
1. Always name the property when asked.
2. A quick check by adding all four is good practice.

Question 4. A carpenter joins two thin wooden strips and passes a thread through their four endpoints so that a rectangle is formed. One strip is 8 cm long.
(i) How long must the second strip be?
(ii) At which point must the two strips be joined?
Give the reason behind each answer.
Answer:
(i) The second strip must also be 8 cm long. The strips become the diagonals, and the diagonals of a rectangle are equal.
(ii) They must be joined at the midpoint of each strip, because the diagonals of a rectangle bisect each other.
In simple words: The strips are the diagonals, so they must be equal (8 cm) and cross at their midpoints.

Teacher's Note:
1. The angle between the strips does not matter for a rectangle.
2. Equal + bisecting diagonals guarantee a rectangle.

Question 5. Define a square. Then explain, in two or three sentences, why every square is a rectangle but every rectangle is not a square.
Answer:
A square is a quadrilateral in which all the angles are 90° and all the sides are of equal length.
Every square has all its angles 90°, which is exactly the definition of a rectangle, so every square is a rectangle.
A rectangle only needs right angles; its adjacent sides may be different, as in a 6 cm × 2 cm rectangle. Such a rectangle is not a square.
In simple words: A square meets the rectangle condition, but a rectangle with unequal sides is not a square.

Teacher's Note:
1. One example is enough to show “not every”.
2. A square has one condition more than a rectangle.

Question 6. In a parallelogram KLMN, ∠K = 65°. Find ∠L, ∠M and ∠N, and state the two parallelogram properties you have used.
Answer:
Properties used: adjacent angles add up to 180° and opposite angles are equal.
∠L = 180° − 65° = 115° (adjacent to ∠K).
∠M = ∠K = 65° (opposite angles).
∠N = ∠L = 115° (opposite angles).
Check: 65 + 115 + 65 + 115 = 360.
In simple words: Neighbouring angles make 180° and opposite angles match, giving 115°, 65° and 115°.

Teacher's Note:
1. K and M are opposite; L and N are opposite.
2. Angles alternate around a parallelogram: 65, 115, 65, 115.

Question 7. ABCD is a rhombus in which ∠A = 70°.
(i) Find the other three angles.
(ii) Write any two properties of the diagonals of a rhombus.
Answer:
(i) A rhombus is a parallelogram, so adjacent angles add to 180° and opposite angles are equal.
∠B = 180° − 70° = 110°, ∠C = ∠A = 70°, ∠D = ∠B = 110°.
(ii) (a) The diagonals bisect each other at 90°.
(b) The diagonals bisect the angles of the rhombus.
In simple words: The angles are 70°, 110°, 70°, 110°, and the diagonals are perpendicular bisectors that also halve the angles.

Teacher's Note:
1. Every parallelogram rule works for a rhombus.
2. Rhombus diagonals have two extra properties.

Question 8. Write the definition of a kite and the definition of a trapezium as given in the chapter, and state one clear difference between the two shapes.
Answer:
Kite: a quadrilateral that can be labelled ABCD so that AB = BC and CD = DA, that is, two non-overlapping adjacent pairs of equal sides.
Trapezium: a quadrilateral having at least one pair of parallel opposite sides.
Difference: a kite is defined by equal adjacent sides and need not have parallel sides, while a trapezium is defined by parallel sides and need not have equal sides.
In simple words: A kite is about equal neighbouring sides, a trapezium is about parallel opposite sides.

Teacher's Note:
1. Definitions should be stated exactly.
2. Compare what each definition is based on.

Question 9. The four angles of a quadrilateral are in the ratio 2 : 3 : 4 : 6. Find all four angles and verify your answer.
Answer:
Let the angles be 2x, 3x, 4x and 6x.
The angles of a quadrilateral add up to 360°: 2x + 3x + 4x + 6x = 360°.
15x = 360°, so x = 24°.
The angles are 48°, 72°, 96° and 144°.
Verification: 48 + 72 + 96 + 144 = 360, and 48 : 72 : 96 : 144 = 2 : 3 : 4 : 6 (dividing each by 24).
In simple words: 15 parts make 360°, so one part is 24° and the angles are 48°, 72°, 96°, 144°.

Teacher's Note:
1. Verify both the sum and the ratio.
2. x is one part, not an angle.

Question 10. In the figure, the diagonals PR and QS of quadrilateral PQRS are equal in length, they cut each other at their midpoints O, and the angle between them is 50°.
(i) Find the four angles formed at O.
(ii) Find ∠OSP and hence show that ∠PSR = 90°.
Answer:
(i) ∠SOR = 50° (given).
∠POS = 180° − 50° = 130° (linear pair on PR).
∠POQ = ∠SOR = 50° and ∠QOR = ∠POS = 130° (vertically opposite angles).
(ii) The diagonals are equal and bisect each other, so OP = OQ = OR = OS.
In △POS, OP = OS and ∠POS = 130°, so ∠OSP = \( \frac{180^\circ - 130^\circ}{2} \) = 25°.
In △ROS, OR = OS and ∠ROS = 50°, so ∠OSR = \( \frac{180^\circ - 50^\circ}{2} \) = 65°.
Hence ∠PSR = 25° + 65° = 90°.
In simple words: The isosceles triangles at S give 25° and 65°, which add up to a right angle.

Teacher's Note:
1. Equal half-diagonals make every triangle at O isosceles.
2. This is why equal bisecting diagonals always give a rectangle.

Question 11. In a parallelogram PQRS, ∠P is twice ∠Q. Find all four angles of the parallelogram and show that their sum is 360°.
Answer:
Adjacent angles add up to 180°: ∠P + ∠Q = 180°.
With ∠P = 2∠Q: 2∠Q + ∠Q = 180°, so 3∠Q = 180° and ∠Q = 60°.
∠P = 120°; opposite angles are equal, so ∠R = ∠P = 120° and ∠S = ∠Q = 60°.
Sum = 120° + 60° + 120° + 60° = 360°.
In simple words: ∠Q = 60° and ∠P = 120°, repeated in opposite corners, and they total 360°.

Teacher's Note:
1. P and Q are adjacent, so they add to 180°.
2. Write the equation in one unknown.

Question 12. ABCD is a rhombus and BD is one of its diagonals. If ∠ABD = 35°, find ∠ABC and ∠BCD. Give a reason for each step.
Answer:
The diagonals of a rhombus bisect its angles, so BD bisects ∠ABC.
∠ABC = 2 × ∠ABD = 2 × 35° = 70°.
Adjacent angles of a rhombus add up to 180°, so ∠BCD = 180° − 70° = 110°.
Check: the angles 70°, 110°, 70°, 110° add up to 360°.
In simple words: Doubling 35° gives ∠ABC = 70°, and its neighbour ∠BCD is 110°.

Teacher's Note:
1. Give a reason with every step.
2. ∠B and ∠C are adjacent angles.

Question 13. On a geoboard, two rubber bands of equal length are stretched so that they cross each other at their midpoints at right angles, and their four endpoints are joined.
(i) Name the quadrilateral obtained and justify it.
(ii) If now one of the two bands alone is made longer at both ends by the same amount, what quadrilateral is obtained? Justify.
Answer:
(i) A square. The bands are the diagonals: being equal and bisecting each other, they make the figure a rectangle; since they also cross at 90°, the four small triangles are congruent (SAS), so all four sides are equal. A rectangle with equal sides is a square.
(ii) A rhombus. The diagonals still bisect each other at right angles, so the sides stay equal, but the diagonals are no longer equal, so the angles are no longer 90°.
In simple words: Equal perpendicular bisecting diagonals give a square; making one longer gives a rhombus.

Teacher's Note:
1. Lengthening both ends equally keeps the crossing at the midpoint.
2. Equal diagonals are what make the angles 90°.

Question 14. Two cardboard cut-outs of an equilateral triangle of side 7 cm are placed together along a common side to form a quadrilateral. Find the sides and the angles of this quadrilateral, and name it with justification.
Answer:
Each triangle has all sides 7 cm and all angles 60°; the common side is hidden inside the figure.
Sides: the four outer sides are 7 cm each, so the quadrilateral is a rhombus.
Angles: at each end of the common side, 60° + 60° = 120°; the other two angles stay 60°.
So the angles are 60°, 120°, 60°, 120° (sum 360°). It is a rhombus but not a square, since its angles are not 90°.
In simple words: Four 7 cm sides with angles 60°, 120°, 60°, 120°: a rhombus.

Teacher's Note:
1. Angles at the joined side add up.
2. Equal sides alone make it a rhombus.

Question 15. PQRS is a trapezium in which PQ ∥ SR. If ∠P = 115° and ∠Q = 80°, find ∠S and ∠R. State the property of parallel lines that you used.
Answer:
Property used: when two lines are parallel, the interior angles on the same side of a transversal add up to 180°.
PS is a transversal of PQ ∥ SR: ∠P + ∠S = 180°, so ∠S = 180° − 115° = 65°.
QR is a transversal: ∠Q + ∠R = 180°, so ∠R = 180° − 80° = 100°.
Check: 115 + 80 + 100 + 65 = 360.
In simple words: Each slanting side makes co-interior angles adding to 180°, giving ∠S = 65° and ∠R = 100°.

Teacher's Note:
1. Use each non-parallel side as a transversal.
2. Opposite angles of a trapezium need not be equal.

Question 16. UVWX is an isosceles trapezium with UV ∥ XW and UX = VW. If ∠U = 72°, find ∠V, ∠W and ∠X, giving a reason for each.
Answer:
∠V = ∠U = 72°, because in an isosceles trapezium the base angles (opposite the equal sides) are equal.
UX is a transversal of UV ∥ XW, so ∠U + ∠X = 180° and ∠X = 108°.
Similarly ∠V + ∠W = 180°, so ∠W = 108°.
Check: 72 + 72 + 108 + 108 = 360.
In simple words: The base angles are both 72°, and the top angles are both 180° − 72° = 108°.

Teacher's Note:
1. Isosceles trapezium: equal base angles.
2. Then use co-interior angles for the top.

Question 17. In a quadrilateral, three angles are equal to one another and each of them is 20° less than the fourth angle. Find all four angles and check your answer.
Answer:
Let each of the three equal angles be x; the fourth angle is x + 20°.
Angle sum: x + x + x + (x + 20°) = 360°.
4x + 20° = 360°, so 4x = 340° and x = 85°.
The angles are 85°, 85°, 85° and 105°.
Check: 85 + 85 + 85 + 105 = 360.
In simple words: Three angles of 85° and one of 105° make 360°.

Teacher's Note:
1. “20° less than the fourth” means the fourth is x + 20°.
2. Subtract 20 before dividing by 4.

Question 18. A quadrilateral has four equal sides and one of its angles is 90°. Using the properties of a rhombus, justify that the quadrilateral must be a square.
Answer:
A quadrilateral with four equal sides is a rhombus.
A rhombus is a parallelogram, so adjacent angles add up to 180° and opposite angles are equal.
One angle is 90°, so each angle next to it is 180° − 90° = 90°, and the angle opposite it is also 90°.
All four angles are 90° and all four sides are equal, which is exactly the definition of a square.
In simple words: One right angle in a rhombus forces all four to be right angles, so it is a square.

Teacher's Note:
1. Chain the facts: rhombus, then parallelogram, then angle rules.
2. End by matching the definition of a square.

Question 19. ABCD is a kite in which AB = BC and CD = DA. The diagonals meet at O. Using congruent triangles, justify that
(i) BD bisects ∠ABC, and
(ii) BD is perpendicular to AC and AO = OC.
Answer:
(i) In △ABD and △CBD: AB = CB and AD = CD (given), and BD is common.
By SSS, △ABD ≅ △CBD, so ∠ABD = ∠CBD. Hence BD bisects ∠ABC (similarly it bisects ∠ADC).
(ii) In △AOB and △COB: AB = CB, ∠ABO = ∠CBO (from (i)) and OB is common.
By SAS, △AOB ≅ △COB, so AO = OC and ∠AOB = ∠COB.
These two angles form a linear pair, so each is 90°. Hence BD ⟂ AC.
In simple words: SSS gives equal angles at B, then SAS gives AO = OC and two equal angles on a line, each 90°.

Teacher's Note:
1. Use the result of part (i) inside part (ii).
2. Equal angles forming a linear pair are right angles.

Question 20. In a quadrilateral ABCD, the opposite sides are equal, that is AB = CD and BC = AD. By drawing the diagonal AC and using congruence, justify that ABCD must be a parallelogram.
Answer:
Draw the diagonal AC. In △ABC and △CDA: AB = CD, BC = DA (given) and AC is common.
By SSS, △ABC ≅ △CDA, so ∠BAC = ∠DCA and ∠BCA = ∠DAC.
∠BAC and ∠DCA are alternate angles for AB and DC with transversal AC, so AB ∥ DC.
∠BCA and ∠DAC are alternate angles for BC and AD, so BC ∥ AD.
Both pairs of opposite sides are parallel, so ABCD is a parallelogram.
In simple words: SSS congruence gives equal alternate angles, which make both pairs of opposite sides parallel.

Teacher's Note:
1. Match the vertices carefully: A↔C, B↔D.
2. Equal alternate angles prove lines parallel.