Quadrilaterals — Class 9 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 9 Mathematics chapter "Quadrilaterals" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 9 Mathematics chapter "Quadrilaterals" — 8 important questions with detailed answers for CBSE board exam p…
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NCERT Solutions for Class 9 Mathematics — Quadrilaterals. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Quadrilaterals — All 8 Questions Solved
Q1. State the midpoint theorem for quadrilaterals: In quadrilateral PQRS, if L, M, N, O are the midpoints of sides PQ, QR, RS, SP respectively, prove that LMNO is a parallelogram.
Step 1: Midpoint Theorem for Quadrilaterals: The figure formed by joining the midpoints of the sides of any quadrilateral is a parallelogram.
Step 2: Let PQRS be a quadrilateral with:
L = midpoint of PQ
M = midpoint of QR
N = midpoint of RS
O = midpoint of SP
Step 3: In triangle PQR, L and M are midpoints of PQ and QR respectively.
By midpoint theorem in triangle: LM ∥ PR and LM = PR/2
Step 4: In triangle RPS, N and O are midpoints of RS and SP respectively.
By midpoint theorem in triangle: NO ∥ PR and NO = PR/2
Step 5: From steps 3 and 4:
LM ∥ NO and LM = NO
Step 6: Similarly, in triangle PQS, L and O are midpoints of PQ and PS respectively.
By midpoint theorem: LO ∥ QS and LO = QS/2
Step 7: In triangle QRS, M and N are midpoints of QR and RS respectively.
By midpoint theorem: MN ∥ QS and MN = QS/2
Step 8: From steps 6 and 7:
LO ∥ MN and LO = MN
Step 9: Since LM ∥ NO, LM = NO and LO ∥ MN, LO = MN, the quadrilateral LMNO has both pairs of opposite sides equal and parallel.
Answer: LMNO is a parallelogram.
Q2. Define and differentiate between a parallelogram and a trapezium. If ABCD is a parallelogram with AB = 8 cm and perpendicular distance between AB and CD is 5 cm, find the area.
Step 1: Parallelogram: A quadrilateral with both pairs of opposite sides parallel and equal.
Properties: Opposite sides equal, opposite angles equal, diagonals bisect each other.
Step 2: Trapezium: A quadrilateral with only ONE pair of parallel sides.
Properties: One pair of parallel sides, other sides (legs) are not necessarily equal.
Step 3: Key Difference: Parallelogram has TWO pairs of parallel sides; trapezium has only ONE pair.
Step 4: For the given parallelogram ABCD:
AB = 8 cm (base)
Perpendicular distance between AB and CD = 5 cm (height)
Step 5: Area of parallelogram = base × height
Area = AB × h
Area = 8 × 5
Area = 40 cm²
Step 6: Note: Since ABCD is a parallelogram, CD = AB = 8 cm (opposite sides equal) and the perpendicular distance is the same throughout.
Answer: Area of parallelogram ABCD = 40 cm².
Q3. In rectangle PQRS, the diagonals PR and QS intersect at O. If PQ = 6 cm and QR = 8 cm, find the length of diagonal PR and verify that O divides both diagonals into equal parts.
Step 1: In rectangle PQRS:
PQ = 6 cm (length)
QR = 8 cm (breadth)
All angles are 90°
Step 2: Triangle PQR is a right triangle with right angle at Q.
Step 3: By Pythagoras theorem:
PR² = PQ² + QR²
PR² = 6² + 8²
PR² = 36 + 64
PR² = 100
PR = 10 cm
Step 4: In a rectangle, both diagonals are equal and bisect each other.
Therefore, QS = PR = 10 cm
Step 5: Since diagonals bisect each other at O:
PO = PR/2 = 10/2 = 5 cm
OR = PR/2 = 10/2 = 5 cm
So PO = OR ✓
Step 6: Similarly:
QO = QS/2 = 10/2 = 5 cm
OS = QS/2 = 10/2 = 5 cm
So QO = OS ✓
Step 7: Verification: O divides PR into PO = 5 cm and OR = 5 cm (equal parts). O divides QS into QO = 5 cm and OS = 5 cm (equal parts).
Answer: Length of diagonal PR = 10 cm; O divides both diagonals into equal parts of 5 cm each.
Q4. In rhombus ABCD, the diagonals AC and BD intersect at O. If AC = 12 cm and BD = 16 cm, find the side length of the rhombus.
Step 1: Properties of rhombus:
- All sides are equal
- Diagonals bisect each other at right angles (90°)
- Diagonals divide the rhombus into four congruent right triangles
Step 2: Given:
AC = 12 cm (first diagonal)
BD = 16 cm (second diagonal)
Step 3: Since diagonals bisect each other:
AO = AC/2 = 12/2 = 6 cm
BO = BD/2 = 16/2 = 8 cm
Step 4: In right triangle AOB (with right angle at O):
AB² = AO² + BO²
AB² = 6² + 8²
AB² = 36 + 64
AB² = 100
AB = 10 cm
Step 5: Since all sides of a rhombus are equal:
AB = BC = CD = DA = 10 cm
Step 6: Verification using triangle BOC:
BC² = BO² + OC²
BC² = 8² + 6²
BC² = 64 + 36
BC² = 100
BC = 10 cm ✓
Answer: Side length of rhombus ABCD = 10 cm.
Q5. In trapezium ABCD, AB ∥ CD, AB = 10 cm, CD = 6 cm, and the perpendicular distance between the parallel sides is 4 cm. Calculate the area of the trapezium.
Step 1: Trapezium ABCD has one pair of parallel sides:
AB = 10 cm (one parallel side)
CD = 6 cm (other parallel side)
Perpendicular distance = 4 cm (height)
Step 2: Formula for area of trapezium:
Area = (1/2) × (sum of parallel sides) × height
Area = (1/2) × (a + b) × h
Where a and b are the parallel sides, and h is the perpendicular distance.
Step 3: Substitute values:
Area = (1/2) × (AB + CD) × height
Area = (1/2) × (10 + 6) × 4
Area = (1/2) × 16 × 4
Area = (1/2) × 64
Area = 32 cm²
Step 4: Verification:
If we imagine the trapezium as two rectangles or use the average of parallel sides:
Average of parallel sides = (10 + 6)/2 = 8 cm
Area = 8 × 4 = 32 cm² ✓
Answer: Area of trapezium ABCD = 32 cm².
Q6. In square PQRS with side length 5 cm, find the length of the diagonal and verify using the Pythagoras theorem.
Step 1: Square PQRS with side length = 5 cm
All sides are equal: PQ = QR = RS = SP = 5 cm
All angles are 90°
Step 2: Consider triangle PQR formed by two adjacent sides and a diagonal:
angle Q = 90°
PQ = 5 cm
QR = 5 cm
Step 3: By Pythagoras theorem:
PR² = PQ² + QR²
PR² = 5² + 5²
PR² = 25 + 25
PR² = 50
PR = √50
PR = √(25 × 2)
PR = 5√2 cm
Step 4: Alternative formula for diagonal of square:
For a square with side a, diagonal = a√2
Diagonal = 5√2 cm ✓
Step 5: Numerical value:
5√2 ≈ 5 × 1.414
5√2 ≈ 7.07 cm
Step 6: Verification: In square, both diagonals are equal.
Second diagonal QS = 5√2 cm ✓
Answer: Length of diagonal = 5√2 cm ≈ 7.07 cm.
Q7. Prove that the sum of the interior angles of a quadrilateral is 360°.
Step 1: Consider quadrilateral ABCD.
Step 2: Draw diagonal AC, which divides the quadrilateral into two triangles: triangle ABC and triangle ACD.
Step 3: In triangle ABC:
angle BAC + angle ABC + angle BCA = 180° ... (by angle sum property of triangles)
Step 4: In triangle ACD:
angle CAD + angle ACD + angle ADC = 180° ... (by angle sum property of triangles)
Step 5: Add the equations from steps 3 and 4:
(angle BAC + angle ABC + angle BCA) + (angle CAD + angle ACD + angle ADC) = 180° + 180°
Step 6: Rearrange the terms:
(angle BAC + angle CAD) + angle ABC + (angle BCA + angle ACD) + angle ADC = 360°
Step 7: Observe that:
angle BAC + angle CAD = angle BAD = angle A of the quadrilateral
angle BCA + angle ACD = angle BCD = angle C of the quadrilateral
angle ABC = angle B of the quadrilateral
angle ADC = angle D of the quadrilateral
Step 8: Therefore:
angle A + angle B + angle C + angle D = 360°
Answer: The sum of all interior angles of a quadrilateral = 360°.
Q8. In parallelogram ABCD, angle A = 65°. Find the measures of angles B, C, and D, and verify that the sum of all angles is 360°.
Step 1: Properties of parallelogram:
- Opposite angles are equal
- Adjacent angles are supplementary (sum to 180°)
Step 2: Given: angle A = 65°
Step 3: Find angle C (opposite to angle A):
angle C = angle A = 65° (opposite angles in parallelogram are equal)
Step 4: Find angle B (adjacent to angle A):
angle A + angle B = 180° (adjacent angles are supplementary)
65° + angle B = 180°
angle B = 180° - 65°
angle B = 115°
Step 5: Find angle D (opposite to angle B):
angle D = angle B = 115° (opposite angles in parallelogram are equal)
Step 6: Alternatively, angle D is also adjacent to angle A:
angle A + angle D = 180°
65° + angle D = 180°
angle D = 115° ✓
Step 7: Verify the sum of all angles:
angle A + angle B + angle C + angle D = 65° + 115° + 65° + 115°
= 130° + 230°
= 360° ✓
Step 8: Summary:
angle A = 65°, angle B = 115°, angle C = 65°, angle D = 115°
Answer: angle B = 115°, angle C = 65°, angle D = 115°; Sum = 360°.
Quadrilaterals — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. The sum of the interior angles of a quadrilateral is:
- A. 180°
- B. 270°
- C. 360°
- D. 90°
Answer: C. 360° — According to the angle sum property, the sum of all interior angles of any quadrilateral is always 360 degrees.
MCQ 2. In a parallelogram ABCD, if ∠A = 75°, then ∠B is:
- A. 75°
- B. 105°
- C. 90°
- D. 115°
Answer: B. 105° — In a parallelogram, consecutive angles are supplementary. Therefore, ∠A + ∠B = 180°, so ∠B = 180° - 75° = 105°.
MCQ 3. The diagonals of a rectangle ABCD intersect at O. If AC = 10 cm, then BO is:
- A. 5 cm
- B. 10 cm
- C. 7.5 cm
- D. 8 cm
Answer: A. 5 cm — In a rectangle, diagonals are equal and bisect each other. So, AC = BD = 10 cm. Since diagonals bisect each other, BO = BD/2 = 10/2 = 5 cm.
MCQ 4. A quadrilateral has all its sides equal. What type of quadrilateral is it?
- A. Rectangle
- B. Parallelogram
- C. Rhombus
- D. Square
Answer: C. Rhombus — A rhombus is a quadrilateral in which all four sides are equal in length. A square also has all sides equal, but it's a special type of rhombus with all angles 90 degrees.
MCQ 5. In ΔABC, D, E, F are respectively the mid-points of sides AB, BC, CA. If the perimeter of ΔABC is 24 cm, then the perimeter of ΔDEF is:
- A. 6 cm
- B. 12 cm
- C. 18 cm
- D. 24 cm
Answer: B. 12 cm — By the Midpoint Theorem, DE = AC/2, EF = AB/2, and FD = BC/2. Therefore, the perimeter of ΔDEF = DE + EF + FD = (AC + AB + BC)/2 = (Perimeter of ΔABC)/2 = 24/2 = 12 cm.
MCQ 6. A quadrilateral whose opposite sides are parallel is called a:
- A. Trapezium
- B. Parallelogram
- C. Rhombus
- D. Kite
Answer: B. Parallelogram — By definition, a parallelogram is a quadrilateral where both pairs of opposite sides are parallel.
MCQ 7. If one angle of a parallelogram is 90 degrees and all sides are equal, it is a:
- A. Rhombus
- B. Rectangle
- C. Square
- D. Trapezium
Answer: C. Square — A parallelogram with all sides equal is a rhombus. If one angle is 90 degrees, all angles must be 90 degrees. A rhombus with all angles 90 degrees is a square.
MCQ 8. The angles of a quadrilateral are in the ratio 3:5:9:13. The largest angle is:
- A. 100°
- B. 156°
- C. 130°
- D. 120°
Answer: B. 156° — Let the angles be 3x, 5x, 9x, and 13x. Their sum is 360°. So, 3x + 5x + 9x + 13x = 360° => 30x = 360° => x = 12°. The largest angle is 13x = 13 * 12° = 156°.
MCQ 9. In a rhombus, the diagonals intersect at what angle?
- A. 45°
- B. 60°
- C. 90°
- D. 120°
Answer: C. 90° — A key property of a rhombus is that its diagonals bisect each other at right angles (90 degrees).
MCQ 10. If the diagonals of a quadrilateral bisect each other, then it is a:
- A. Trapezium
- B. Rectangle
- C. Rhombus
- D. Parallelogram
Answer: D. Parallelogram — This is one of the conditions for a quadrilateral to be a parallelogram. If the diagonals bisect each other, the quadrilateral is a parallelogram.
MCQ 11. A quadrilateral in which diagonals are equal and bisect each other at right angles is a:
- A. Rhombus
- B. Rectangle
- C. Square
- D. Parallelogram
Answer: C. Square — Diagonals being equal makes it a rectangle. Diagonals bisecting at right angles makes it a rhombus. A quadrilateral that is both a rectangle and a rhombus is a square.
MCQ 12. ABCD is a quadrilateral. P, Q, R, S are the midpoints of AB, BC, CD, and DA respectively. If AC = 8 cm and BD = 10 cm, then PQRS is a parallelogram with sides PQ and QR being:
- A. 4 cm, 5 cm
- B. 4 cm, 8 cm
- C. 5 cm, 10 cm
- D. 8 cm, 10 cm
Answer: A. 4 cm, 5 cm — By the Midpoint Theorem, in ΔABC, PQ is parallel to AC and PQ = AC/2 = 8/2 = 4 cm. In ΔDBC, QR is parallel to BD and QR = BD/2 = 10/2 = 5 cm. Similarly, RS = AC/2 and SP = BD/2. PQRS is a parallelogram.
MCQ 13. A quadrilateral having only one pair of opposite sides parallel is called a:
- A. Parallelogram
- B. Trapezium
- C. Rhombus
- D. Rectangle
Answer: B. Trapezium — By definition, a trapezium is a quadrilateral with exactly one pair of parallel sides.
MCQ 14. In a parallelogram ABCD, ∠DAB = 60°. The bisector of ∠DAB meets CD at E. Then ∠AEC is:
- A. 120°
- B. 90°
- C. 100°
- D. 110°
Answer: A. 120° — In parallelogram ABCD, ∠DAB = 60°. AE bisects ∠DAB, so ∠DAE = ∠EAB = 30°. Since AB || DC, ∠DEA = ∠EAB = 30° (alternate interior angles). This makes ΔADE isosceles with AD = DE. If ABCD is a rhombus (all sides equal), then AD = BC. If AD=DE, then BC=DE. If AB=BC (rhombus), then CE = CD - DE = AB - AD = 0, meaning E coincides with C. In this case, ∠AEC = ∠ADC. Since ∠ADC = 180° - ∠DAB = 180° - 60° = 120°, then ∠AEC = 120°.
MCQ 15. In parallelogram ABCD, if AB = 12 cm, AD = 8 cm, and the diagonal BD = 10 cm, then the perimeter of ΔABD is:
- A. 20 cm
- B. 30 cm
- C. 24 cm
- D. 36 cm
Answer: B. 30 cm — The perimeter of a triangle is the sum of its sides. So, for ΔABD, the perimeter = AB + AD + BD = 12 cm + 8 cm + 10 cm = 30 cm.
MCQ 16. In ΔABC, M is the midpoint of AB. A line through M parallel to BC intersects AC at N. If BC = 10 cm, then MN is:
- A. 5 cm
- B. 10 cm
- C. 7.5 cm
- D. Cannot be determined
Answer: A. 5 cm — By the Midpoint Theorem, the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half of its length. Since M is the midpoint of AB and MN || BC, N must be the midpoint of AC, and MN = BC/2 = 10/2 = 5 cm.
MCQ 17. Which of the following is true for a square?
- A. Diagonals bisect each other at right angles
- B. Diagonals are equal
- C. All sides are equal
- D. All of the above
Answer: D. All of the above — A square is a regular quadrilateral, meaning all its sides are equal and all its angles are 90 degrees. Its diagonals are equal, bisect each other at right angles, and bisect the vertex angles. So, all options are true.
MCQ 18. Two consecutive angles of a parallelogram are (x + 60)° and (2x + 30)°. The value of x is:
- A. 30
- B. 40
- C. 50
- D. 60
Answer: A. 30 — Consecutive angles of a parallelogram are supplementary. So, (x + 60)° + (2x + 30)° = 180°. This gives 3x + 90 = 180, so 3x = 90, and x = 30.
MCQ 19. In ΔABC, D, E, F are midpoints of AB, BC, CA respectively. If AB=5cm, BC=6cm, CA=7cm, then the perimeter of the quadrilateral ADEF is:
- A. 18 cm
- B. 12 cm
- C. 15 cm
- D. 20 cm
Answer: B. 12 cm — ADEF is a quadrilateral with vertices A, D, E, F. AD = AB/2 = 5/2 = 2.5 cm. FA = CA/2 = 7/2 = 3.5 cm. By Midpoint Theorem, DE = CA/2 = 7/2 = 3.5 cm, and EF = AB/2 = 5/2 = 2.5 cm. The perimeter of ADEF = AD + DE + EF + FA = 2.5 + 3.5 + 2.5 + 3.5 = 12 cm.
MCQ 20. In a rhombus ABCD, if ∠A = 60°, then the diagonal BD is equal to:
- A. AB
- B. AC
- C. BC/2
- D. 2*AD
Answer: A. AB — In a rhombus, all sides are equal (AB = AD). If ∠A = 60°, then ΔABD is an isosceles triangle with an angle of 60° between its equal sides. This makes ΔABD an equilateral triangle, so AB = AD = BD.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.