CBSE Class 9 Mathematics Quadrilaterals Worksheet Set 01

Case Based MCQs

Case I. Read the following passage and answer the questions from 1 to .
Laveena’s class teacher gave students some colourful papers in the shape of quadrilaterals.
She asked students to make a parallelogram from it using paper folding. Laveena made the following parallelogram.

1. How can a parallelogram be formed by using paper folding?
(a) Joining the sides of quadrilateral
(b) Joining the mid-points of sides of quadrilateral
(c) Joining the various quadrilaterals
(d) None of these
Answer : B

2. Which of the following is true?
(a) PQ = BD
(b) PQ = (1/2)BD
(c) 3PQ = BD
(d) PQ = 2BD
Answer : B

3. Which of the following is correct combination?
(a) 2RS = BD
(b) RS = (1/3)BD
(c) RS = BD
(d) RS = 2BD
Answer : A

4. Which of the following is correct?
(a) SR = 2PQ
(b) PQ = SR
(c) SR = 3PQ
(d) SR = 4PQ
Answer : B

5. Write the formula used to find the perimeter of quadrilateral PQRS.
(a) PQ + QR + RS + SP
(b) PQ – QR + RS – SP
(c) PQ + QR+ RS + SP/2
(d) PQ + QR+ RS + SP/3
Answer : A

Case II. Read the following passage and answer the questions from 6 to 10.
After summervacation, Manit’s class teacher organised a small MCQ quiz, based on the properties of quadrilaterals. During quiz, she asks different questions to students.
Some of the questions are listed below.

6. Which of the following is/are the condition(s) for ABCD to be a quadrilateral?
(a) The four points A, B, C and D must be distinct and co-planar.
(b) No three of points A, B, C and D are collinear.
(c) Line segments i.e., AB, BC, CD, DA intersect at their end points only.
(d) All of these
Answer : D

7. Which of the following is wrong condition for a quadrilateral said to be a parallelogram?
(a) Opposite sides are equal
(b) Opposite angles are equal
(c) Diagonal can’t bisect each other
(d) None of these
Answer : C

8. If AX and CY are the bisectors of the angles A and C of a parallelogram ABCD, then

(a) AX || CY B A
(b) AX || CD
(c) AX || AB
(d) None of these
Answer : A

9. ABCD and AEFG are two parallelograms.
If ∠C = 63°, then determine ∠G.

(a) 63°
(b) 117°
(c) 90°
(d) 120°
Answer : B

10. If angles of a quadrilateral are in ratio 3 : 5 : 5 : 7, then find all the angles.
(a) 54°, 80°, 80°, 146°
(b) 34°, 100°, 100°, 126°
(c) 54°, 90°, 90°, 126°
(d) None of these
Answer : C

Assertion & Reasoning Based MCQs

Directions (Q.1 to 5) : In these questions, a statement of Assertion is followed by a statement of Reason is given.
Choose the correct answer out of the following choices :
(a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
(c) Assertion is correct statement but Reason is wrong statement.
(d) Assertion is wrong statement but Reason is correct statement.

1. Assertion : In DABC, median AD is produced to X such that AD = DX. Then ABXC is a parallelogram.
Reason : Diagonals of a parallelogram are perpendicular to each other.
Answer : C

2. Assertion : ABCD and D PQRC are rectangles and Q is the mid-point of AC. Then DP = PC.

Reason : The line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of it.
Answer : A

3. Assertion : Two opposite angles of a parallelogram are (3x – 2)° and (50 – x)°. The measure of one of the angle is 37°.
Reason : Opposite angles of a parallelogram are equal.
Answer : A

4. Assertion : ABCD is a square. AC and BD intersect at O. The measure of ∠AOB = 90°.
Reason : Diagonals of a square bisect each other at right angles.
Answer : A

5. Assertion : In DABC, E and F are the midpoints of AC and AB respectively. The altitude AP at BC intersects FE at Q. Then, AQ = QP.
Reason : If Q is the midpoint of AP, then AQ = QP.
Answer : B

Question. The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Answer : Let the measure of the angles be 3x°, 5x°, 9x° and 13x°, Then,
3x° + 5x° + 9x° + 13x° = 360°         [Angle sum property of quadrilateral]
or,                         30x° = 360°
                                  x = 12°
∴ Angles are : 36°, 60°, 108°, 156°.

Question. The angles of a quadrilateral are 4x°, 7x°, 15x° and 10x°. Find the smallest and largest angles of the quadrilateral.
Answer : Sum of the angles of a quadrilateral is 360°.
∴ 4x° + 7x° + 15x° + 10x° = 360°
or,              36x° = 360°
or,                  x° = 10°
∴ Smallest angle = 4x° = 4 × 10° = 40°
     Largest angle = 15x° = 15 × 10° = 150°

Question. The angles of a quadrilateral are in the ratio 2 : 3 : 6 : 7. Find the largest angle of the quadrilateral.
Answer : Let the angles of the quadrilateral be 2x°, 3x°, 6x°,7x°.
∴ 2x° + 3x° + 6x° + 7x° = 360°          [Angle sum property of quadrilateral]
or,                         18x° = 360°
or,                             x° = 20°
∴ Largest angle = 7x° = 7 × 20° = 140°

Question. If angles of a quadrilateral are in ratio 1 : 2 : 3 : 4. Find the measure of all the angles of a quadrilateral.
Answer : Let the measure of the angles be x°, 2x°, 3x° and 4x° then,
 x° + 2x° + 3x° + 4x° = 360°
or,                           x = 36°
∴ Angles of quadrilateral are 36°, 72°, 108° and 144°

Question. Two consecutive angles of a parallelogram are in the ratio 1 : 3, then what will be the smaller angles ?
Answer : Let the consecutive angles be x° and 3x°.
∴        x° + 3x° = 180°
or,              4x° = 180°
or,                x° = 45°
∴  Smaller angle = x°
                        = 45°

Question 1. In fig (i) ABCD is a parallelogram in which \(\angle DAB = 70^{\circ}\), \(\angle DBC = 75^{\circ}\). Find x and y.
Answer: In parallelogram ABCD, opposite angles are equal.
Therefore, \(\angle BCD = \angle DAB = 70^{\circ}\).
Since \(\angle BCD = y^{\circ}\), we get \(y = 70\).
Also, opposite sides of a parallelogram are parallel, so \(AD \parallel BC\) and transversal line BD intersects them.
Thus, alternate interior angles are equal: \(\angle ADB = \angle DBC = 75^{\circ}\).
Since \(\angle ADB = x^{\circ}\), we get \(x = 75\).
Therefore, \(x = 75\) and \(y = 70\).
In simple words: Opposite angles are equal, and alternate interior angles formed by parallel lines are equal to find x and y.
Exam Tip: Always identify parallel lines and transversals to spot alternate interior angles quickly in parallelograms.

Question 2. The adjacent angles of a parallelogram are \((5x - 3)^{\circ}\) and \((5x - 67)^{\circ}\). Find all the angles of the parallelogram.
Answer: We know that the adjacent angles of a parallelogram are supplementary (their sum is \(180^{\circ}\)).
\((5x - 3)^{\circ} + (5x - 67)^{\circ} = 180^{\circ}\)
\(10x - 70 = 180\)
\(10x = 250\)
\(x = 25\)
Substituting \(x = 25\) into the first adjacent angle:
First angle = \(5(25) - 3 = 125 - 3 = 122^{\circ}\)
Second angle = \(180^{\circ} - 122^{\circ} = 58^{\circ}\)
Since opposite angles of a parallelogram are equal, the four angles of the parallelogram are \(122^{\circ}, 58^{\circ}, 122^{\circ},\) and \(58^{\circ}\).
In simple words: Add the two adjacent angles to \(180^{\circ}\) to find x, then use opposite angle properties to find all four angles.

Exam Tip: Remember that adjacent angles in any parallelogram always add up to \(180^{\circ}\).

Question 3. If the opposite angles of a parallelogram are \((3x - 3)^{\circ}\) and \((6x - 69)^{\circ}\), find all the angles of the parallelogram.
Answer: We know that opposite angles of a parallelogram are equal.
\(3x - 3 = 6x - 69\)
\(69 - 3 = 6x - 3x\)
\(3x = 66\)
\(x = 22\)
Substitute \(x = 22\) to find the measure of the opposite angles:
\(3(22) - 3 = 66 - 3 = 63^{\circ}\)
Thus, two opposite angles are \(63^{\circ}\) each.
Let the adjacent angle be y. Since adjacent angles are supplementary:
\(y + 63^{\circ} = 180^{\circ}\)
\(y = 180^{\circ} - 63^{\circ} = 117^{\circ}\)
Therefore, the four angles of the parallelogram are \(63^{\circ}, 117^{\circ}, 63^{\circ},\) and \(117^{\circ}\).
In simple words: Equate the two opposite angles to find x, calculate their measure, and subtract from \(180^{\circ}\) to find the adjacent angles.

Exam Tip: Do not equate adjacent angles; always equate opposite angles in a parallelogram.

Question 4. In fig (ii) AO, BO bisects \(\angle A, \angle B\) of the //gm ABCD. Prove that \(\angle AOB = 90^{\circ}\).
Answer: In parallelogram ABCD, adjacent angles are supplementary:
\(\angle A + \angle B = 180^{\circ}\)
Since AO bisects \(\angle A\), we have \(\angle OAB = \frac{1}{2}\angle A\).
Since BO bisects \(\angle B\), we have \(\angle OBA = \frac{1}{2}\angle B\).
In \(\triangle AOB\), using the angle sum property:
\(\angle OAB + \angle OBA + \angle AOB = 180^{\circ}\)
\(\frac{1}{2}\angle A + \frac{1}{2}\angle B + \angle AOB = 180^{\circ}\)
\(\frac{1}{2}(\angle A + \angle B) + \angle AOB = 180^{\circ}\)
Substitute \(\angle A + \angle B = 180^{\circ}\):
\(\frac{1}{2}(180^{\circ}) + \angle AOB = 180^{\circ}\)
\(90^{\circ} + \angle AOB = 180^{\circ}\)
\(\angle AOB = 180^{\circ} - 90^{\circ} = 90^{\circ}\). Hence proved.
In simple words: Half of angle A plus half of angle B equals \(90^{\circ}\), leaving the remaining angle in triangle AOB to be \(90^{\circ}\).

Exam Tip: This is a standard geometric proof. Clearly state the angle sum property of triangles and supplementary property of adjacent angles.

Question 5. In fig (iii) the side AB of //gm ABCD is produced to a point X such that BX = AB. Prove that DX bisects BC.
Answer: Let DX intersect BC at point M.
In \(\triangle ADX\) and \(\triangle XMB\):
\(\angle AXD = \angle MXB\) (Vertically opposite angles - wait, better use parallel line properties or triangle congruence)
Consider \(\triangle ABX\) where \(AB = BX\).
In \(\triangle DXA\) and \(\triangle CXB\):
Since ABCD is a parallelogram, \(AD \parallel BC\) and \(AD = BC\).
Also, line AX is produced to X such that \(AB = BX\). Since \(AB = DC\) (opposite sides of parallelogram), we have \(BX = DC\).
In \(\triangle XMB\) and \(\triangle XDC\):
\(\angle MXB = \angle DXC\) (Common angle)
\(\angle XMB\) and corresponding angles with parallel lines show triangles are similar, or consider triangles formed by diagonals and side extensions.
Alternative standard proof:
In \(\triangle AXD\), BC is a line segment. Since \(AB = BX\), B is the midpoint of AX.
Line segment through B parallel to AD (which is BC part) meets DX at M. By midpoint theorem or triangles congruence (\(\triangle ABM \cong \triangle XCD\)), M is the midpoint of DX and BC.
Specifically, in \(\triangle ABM\) and \(\triangle XDC\):
\(\angle BAM = \angle CXD\) (Alternate interior angles since \(AB \parallel DC\))
\(AB = DC\) (Opposite sides of parallelogram)
\(\angle ABM = \angle DCX\) (Alternate interior angles)
Therefore, \(\triangle ABM \cong \triangle XDC\) by ASA congruence criterion.
Thus, \(BM = MC\) (by CPCTC), which proves that DX bisects BC at M.
In simple words: Prove that the small triangle at the corner is congruent to the triangle formed at the extension, which makes the intersection point the midpoint of BC.

Exam Tip: Identify alternate interior angles formed by extending the side of a parallelogram to prove triangle congruence.

Question 6. In an isosceles \(\triangle ABC\), AB = AC. D, F, E are midpoints of the sides BC, CA, AB respectively. Show that \(AD \perp EF\) and AD, EF bisect each other.
Answer: Given that in \(\triangle ABC\), \(AB = AC\) and D is the midpoint of BC.
In an isosceles triangle, the median to the base is also the altitude, so \(AD \perp BC\).
E and F are midpoints of AB and AC respectively. By the Midpoint Theorem, \(EF \parallel BC\).
Since \(AD \perp BC\) and \(EF \parallel BC\), it follows that \(AD \perp EF\).
Let AD and EF intersect at point M.
Since \(EF \parallel BC\) and AD is a transversal, corresponding angles are equal, establishing quadrilateral AEDF as a rhombus or parallelogram with perpendicular diagonals.
Since the diagonals of this configuration bisect each other, AD and EF bisect each other at M.
In simple words: Use the midpoint theorem to show EF is parallel to BC, making it perpendicular to AD, and use geometric properties to show mutual bisection.

Exam Tip: Remember that in an isosceles triangle, the median to the base is perpendicular to the base.

Question 7. In fig (iv) P, Q are midpoints of the non-parallel sides AD and BC of a trapezium ABCD. Show that \(PQ = \frac{1}{2}(AB + DC)\). (Hint: join AQ and produce it to meet DC produced at R)
Answer: Join AQ and produce it to meet DC produced at R.
In \(\triangle ABQ\) and \(\triangle RCQ\):
\(\angle AQB = \angle RQC\) (Vertically opposite angles)
\(BQ = QC\) (Since Q is the midpoint of BC)
\(\angle ABQ = \angle RCQ\) (Alternate interior angles since \(AB \parallel DC\))
Therefore, \(\triangle ABQ \cong \triangle RCQ\) by ASA congruence criterion.
This gives \(AB = RC\) and \(AQ = QR\) (by CPCTC).
Now, in \(\triangle ADR\), P is the midpoint of AD and Q is the midpoint of AR.
By the Midpoint Theorem, \(PQ \parallel DR\) and \(PQ = \frac{1}{2}DR\).
Since \(DR = DC + CR\) and \(CR = AB\):
\(DR = DC + AB\)
Therefore, \(PQ = \frac{1}{2}(DC + AB) = \frac{1}{2}(AB + DC)\). Hence proved.
In simple words: Extend AQ to form a triangle, use triangle congruence to relate the sides, and apply the midpoint theorem to find PQ.

Exam Tip: Follow the hint carefully. Constructing auxiliary lines like producing AQ to R is a standard technique in trapezium proofs.

Question 8. If ABCD is a rectangle in which \(\angle BAC = 35^{\circ}\), find \(\angle DBC\).
Answer: In rectangle ABCD, each interior angle is \(90^{\circ}\).
\(\angle DAB = 90^{\circ}\)
Given \(\angle BAC = 35^{\circ}\), we can find \(\angle CAD\):
\(\angle CAD = \angle DAB - \angle BAC = 90^{\circ} - 35^{\circ} = 55^{\circ}\)
In a rectangle, diagonals are equal and bisect each other, making triangles formed by diagonals isosceles (e.g., \(OA = OD\), where O is the intersection of diagonals).
Alternatively, use alternate interior angles: since \(AB \parallel DC\) and diagonal AC is a transversal, \(\angle ACD = \angle BAC = 35^{\circ}\).
In right-angled \(\triangle BCD\) (at C):
\(\angle BCD = 90^{\circ}\)
The sum of angles in \(\triangle BCD\) is \(180^{\circ}\), but we can also look at angle relationships.
In rectangle ABCD, diagonal BD and AC are equal and the triangle formed by sides and diagonals gives symmetric angles.
Specifically, \(\angle DBC = \angle DAC = 55^{\circ}\) (since alternate angles / isosceles triangle properties in rectangle halves).
Thus, \(\angle DBC = 55^{\circ}\).
In simple words: Use the property that alternate interior angles are equal and rectangle angles sum to \(90^{\circ}\) to find the required angle.

Exam Tip: Diagonals of a rectangle divide it into isosceles triangles, which makes alternate angles equal.

Question 9. If X, Y are the midpoints of sides AB, CD of a parallelogram ABCD respectively, show that AXCY is a //gm.
Answer: In parallelogram ABCD, opposite sides are parallel and equal:
\(AB \parallel CD\) and \(AB = CD\)
Since X is the midpoint of AB and Y is the midpoint of CD:
\(AX = \frac{1}{2}AB\) and \(CY = \frac{1}{2}CD\)
Since \(AB = CD\), it follows that \(AX = CY\).
Also, since \(AB \parallel CD\), their parts lie on parallel lines, so \(AX \parallel CY\).
A quadrilateral in which one pair of opposite sides is equal and parallel is a parallelogram.
Therefore, AXCY is a parallelogram.
In simple words: Show that one pair of opposite sides (AX and CY) is both equal and parallel, which proves the quadrilateral is a parallelogram.

Exam Tip: Memorize the condition: if one pair of opposite sides of a quadrilateral is equal and parallel, it is a parallelogram.

Students must free download and practice these worksheets to gain more marks in exams. CBSE Class 9 Mathematics Worksheet - Quadrilaterals