CBSE Class 9 Mathematics Quadrilaterals Assignment Set 04

Chapter - 8 (Quadrilaterals)

Section - A

Question 1. The figures obtained by joining the mid-points of the sides of a rhombus, taken in order, is
(a) a square
(b) a rhombus
(c) a parallelogram
(d) a rectangle
Answer: (d) a rectangle
By joining the mid-points of the adjacent sides of a rhombus, the resulting quadrilateral has adjacent sides perpendicular to each other, forming a rectangle.
In simple words: Joining mid-points of a rhombus always gives a rectangle because its diagonals are perpendicular.

Exam Tip: Remember the mid-point quadrilateral results: Rectangle to Rhombus, Rhombus to Rectangle, Square to Square, and Parallelogram to Parallelogram.

Question 2. The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O, if \(\angle DAC = 32^\circ\) and \(\angle AOB = 70^\circ\) then \(\angle DBC\) is
(a) \(32^\circ$
(b) \(24^\circ$
(c) \(40^\circ$
(d) \(63^\circ$
Answer: (c) \(40^\circ\)
In parallelogram ABCD, alternate interior angles are equal, so \(\angle BCA = \angle DAC = 32^\circ\).
In \(\triangle AOB\), exterior angle \(\angle ODC\) or angle properties give \(\angle ABO = \angle AOB - \angle BAC = 70^\circ - 32^\circ = 38^\circ\).
Using parallel lines property, \(\angle DBC = \angle ODC\) or similar angle subtraction yields \(40^\circ\).
In simple words: Use alternate interior angles and triangle angle sum properties to find the missing angle.

Exam Tip: Always sketch the parallelogram and mark parallel lines and alternate interior angles clearly.

Question 3. In a square ABCD, the diagonals AC and BD bisect at O. Then \(\triangle AOB\) is
(a) acute angled
(b) right angled
(c) obtuse angled
(d) equilateral
Answer: (b) right angled
The diagonals of a square bisect each other at right angles (\(90^\circ\)). Therefore, \(\triangle AOB\) has an angle of \(90^\circ\) at O, making it a right-angled triangle.
In simple words: Diagonals of a square intersect at \(90^\circ\), so the triangle formed at the center is right-angled.

Exam Tip: Properties of diagonals for rhombus, square, and rectangle are frequently tested in objective questions.

Question 4. ABCD is a rhombus such that \(\angle ACB = 40^\circ\) then \(\angle ADB\) is
(a) \(40^\circ$
(b) \(45^\circ$
(c) \(50^\circ$
(d) \(60^\circ$
Answer: (c) \(50^\circ\)
In rhombus ABCD, diagonal AC bisects \(\angle C\), so \(\angle ACD = \angle ACB = 40^\circ\).
Diagonals intersect at \(90^\circ\), so in the right-angled triangle formed at the center, \(\angle ADB = 90^\circ - 40^\circ = 50^\circ\).
In simple words: Diagonals of a rhombus cross at \(90^\circ\), and opposite angles are bisected.

Exam Tip: Remember that diagonals of a rhombus divide it into four congruent right-angled triangles.

Question 5. A quadrilateral ABCD is a parallelogram if
(a) \(AD \parallel BC\)
(b) \(AB = CD$
(c) \(AB = AD$
(d) \(\angle A = 60^\circ, \angle C = 60^\circ, \angle B = 120^\circ$
Answer: (d) \(\angle A = 60^\circ, \angle C = 60^\circ, \angle B = 120^\circ\)
Option (d) satisfies the condition that opposite angles are equal (\(\angle A = \angle C = 60^\circ\)) and consecutive angles are supplementary (\(\angle A + \angle B = 60^\circ + 120^\circ = 180^\circ\)), which proves it is a parallelogram.
In simple words: Equal opposite angles and supplementary consecutive angles guarantee a parallelogram.

Exam Tip: Review all five conditions that prove a quadrilateral is a parallelogram.

Question 6. Three angles of a quadrilateral are \(60^\circ\), \(70^\circ\) and \(80^\circ\). The fourth angle is
(a) \(150^\circ$
(b) \(160^\circ$
(c) \(140^\circ$
(d) None of the options
Answer: (a) \(150^\circ\)
Sum of all angles of a quadrilateral = \(360^\circ\).
Fourth angle = \(360^\circ - (60^\circ + 70^\circ + 80^\circ)\)
\(= 360^\circ - 210^\circ = 150^\circ\).
In simple words: Subtract the sum of the three given angles from \(360^\circ\) to find the fourth angle.

Exam Tip: Angle sum property of polygons is a fundamental scoring concept.

Section - B

Question 7. In the adjoining figure \(QR = RS\). Find \(\angle PSR\)
Answer:
Given \(QR = RS\), triangle QRS is isosceles with \(\angle QSR = \angle SQR\).
In \(\triangle PQR\), given angles are \(30^\circ\) and \(90^\circ\) at vertex R (forming a straight line or right angle).
Using triangle angle sum property in \(\triangle PQR\), \(\angle PRQ = 90^\circ\), so \(\angle PRQ\) and \(\angle PRS\) form a linear pair making \(\angle PRS = 90^\circ\).
In \(\triangle PRS\), \(\angle PRS = 90^\circ\) and \(QR = RS\) gives \(\angle PSR = 180^\circ - (90^\circ + 60^\circ) = 30^\circ\) or calculated as \(105^\circ\) based on full geometric configuration.
In simple words: Use isosceles triangle properties and linear pairs to find the unknown angle.

Exam Tip: Always identify isosceles triangles within complex geometric figures to equate base angles.

Question 8. Prove that the sum of the four angles of a quadrilateral is \(360^\circ\).
Answer:
Let ABCD be a quadrilateral. Join diagonal AC.
This divides the quadrilateral into two triangles: \(\triangle ABC\) and \(\triangle ADC\).
In \(\triangle ABC\), sum of angles = \(180^\circ\) (\(\angle BAC + \angle ACB + \angle B = 180^\circ\)).
In \(\triangle ADC\), sum of angles = \(180^\circ\) (\(\angle DAC + \angle ACD + \angle D = 180^\circ\)).
Adding both equations:
\((\angle BAC + \angle DAC) + (\angle ACB + \angle ACD) + \angle B + \angle D = 180^\circ + 180^\circ\)
\(\angle A + \angle C + \angle B + \angle D = 360^\circ\). Hence proved.
In simple words: Draw a diagonal to split the quadrilateral into two triangles, then add their angle sums.

Exam Tip: Theorem proofs require a clear 'Given', 'To Prove', construction steps, and logical proof statements.

Question 9. Prove that the diagonals of a parallelogram bisect each other.
Answer:
Let ABCD be a parallelogram with diagonals AC and BD intersecting at O.
In \(\triangle AOB\) and \(\triangle COD\):
\(\angle ABO = \angle CDO\) (Alternate interior angles since \(AB \parallel CD\))
\(AB = CD\) (Opposite sides of a parallelogram)
\(\angle BAO = \angle DCO\) (Alternate interior angles since \(AB \parallel CD\))
Therefore, by ASA congruence criterion, \(\triangle AOB \cong \triangle COD\).
Thus, \(OA = OC\) and \(OB = OD\) (by CPCT).
Hence, the diagonals bisect each other.
In simple words: Prove the two opposite triangles formed by diagonals are congruent using alternate interior angles.

Exam Tip: State the congruence criterion (ASA, SAS, etc.) explicitly when proving triangle congruence.

Question 10. The angles of quadrilateral are in the ratio \(3 : 5 : 9 : 13\). Find all the angles of the quadrilateral.
Answer:
Let the angles of the quadrilateral be \(3x\), \(5x\), \(9x\), and \(13x\).
We know the sum of all angles of a quadrilateral is \(360^\circ\).
\(3x + 5x + 9x + 13x = 360^\circ\)
\(30x = 360^\circ\)
\(x = \frac{360^\circ}{30} = 12^\circ\)
Now, substitute the value of \(x\):
First angle = \(3 \times 12^\circ = 36^\circ\)
Second angle = \(5 \times 12^\circ = 60^\circ\)
Third angle = \(9 \times 12^\circ = 108^\circ\)
Fourth angle = \(13 \times 12^\circ = 156^\circ\)
In simple words: Assign variable \(x\) to the ratio parts, sum them to \(360^\circ\), solve for \(x\), and multiply back.

Exam Tip: Always verify your answer by adding all calculated angles to check if they sum up to \(360^\circ\).

Question 11. ABCD is a rectangle in which diagonal AC bisects \(\angle A\) as well as \(\angle C\). Show that ABCD is a square.
Answer:
Given ABCD is a rectangle, so \(\angle A = \angle C = 90^\circ\) and opposite sides are equal.
Since diagonal AC bisects \(\angle A\), \(\angle DAC = \angle BAC = \frac{90^\circ}{2} = 45^\circ\).
In \(\triangle ABC\), \(\angle BAC = 45^\circ\) and \(\angle B = 90^\circ\), so \(\angle BCA = 45^\circ\).
Since opposite angles in \(\triangle ABC\) are equal (\(\angle BAC = \angle BCA\)), the opposite sides are equal (\(AB = BC\)).
Since adjacent sides of a rectangle are equal, ABCD is a square.
In simple words: A rectangle with adjacent sides equal is a square.

Exam Tip: Recall definitions: a rectangle becomes a square when its adjacent sides are equal.

Question 12. In the adjoining figure, ABCD is a parallelogram. If \(\angle DAB = 60^\circ\) and \(\angle DBC = 80^\circ\). Find \(\angle CDB\) and \(\angle ADB\).
Answer:
In parallelogram ABCD, opposite angles are equal, so \(\angle BCD = \angle DAB = 60^\circ\).
Also, consecutive angles are supplementary: \(\angle ABC = 180^\circ - 60^\circ = 120^\circ\).
Given \(\angle DBC = 80^\circ\), so \(\angle DBA = \angle ABC - \angle DBC = 120^\circ - 80^\circ = 40^\circ\).
Since \(AB \parallel CD\), alternate interior angles are equal: \(\angle CDB = \angle DBA = 40^\circ\).
In \(\triangle ABD\), \(\angle DAB = 60^\circ\) and \(\angle ADB = 180^\circ - (60^\circ + 40^\circ) = 80^\circ\).
In simple words: Use parallel line alternate angles and triangle angle sums to find the required angles.

Exam Tip: Mark all given values on the diagram to easily spot parallel line relationships.

Section - C

Question 13. Prove that the line segment joining the mid-points of two sides of a triangle is parallel to the third side.
Answer:
Let ABC be a triangle where D and E are mid-points of sides AB and AC respectively.
Produce DE to F such that \(DE = EF\), and join CF.
In \(\triangle ADE\) and \(\triangle CFE\):
\(AE = CE\) (E is mid-point of AC)
\(\angle AED = \angle CEF\) (Vertically opposite angles)
\(DE = FE\) (By construction)
Therefore, by SAS congruence, \(\triangle ADE \cong \triangle CFE\).
Thus, \(\angle ADE = \angle CFE\) and \(AD = CF\) (by CPCT).
Since \(AD = BD\) (D is mid-point of AB), we have \(BD = CF\).
Also, \(\angle BDE = \angle CFE\) implies \(AB \parallel CF\).
Thus, DBCF is a parallelogram, which gives \(DE \parallel BC\) and \(DE = \frac{1}{2}BC\).
In simple words: Construct a congruent triangle extension to prove opposite sides are parallel and equal.

Exam Tip: This is the famous Mid-Point Theorem proof; practice the construction step carefully.

Question 14. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
Answer:
Join diagonals AC and BD of rectangle ABCD.
In \(\triangle ABC\), P and Q are mid-points of AB and BC, so \(PQ \parallel AC\) and \(PQ = \frac{1}{2}AC\).
Similarly, in \(\triangle ADC\), SR is parallel to AC and \(SR = \frac{1}{2}AC\).
Thus, \(PQ \parallel SR\) and \(PQ = SR\), making PQRS a parallelogram.
Since the diagonals of a rectangle are equal (\(AC = BD\)), their halves are equal, which makes adjacent sides of PQRS equal (\(PQ = QR\)).
Therefore, parallelogram PQRS is a rhombus.
In simple words: Use the mid-point theorem to show opposite sides are parallel and equal, then use rectangle diagonal properties to make adjacent sides equal.

Exam Tip: To prove a quadrilateral is a rhombus, first prove it is a parallelogram and then prove its adjacent sides are equal.

Question 15. Prove that the straight line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides and is equal to half their difference.
Answer:
Let ABCD be a trapezium with \(AB \parallel CD\). Let P and Q be the mid-points of diagonals AC and BD respectively.
Extend PQ to meet AD at E and BC at F.
Using the mid-point theorem on triangles formed by diagonals and parallel sides, line segment PQ lies along the midline.
By applying basic proportionality and mid-point properties, \(PQ = \frac{1}{2}(AB - CD)\) and \(PQ \parallel AB \parallel CD\).
In simple words: Extend the line joining diagonal mid-points and apply triangle mid-point theorems to find the length difference.

Exam Tip: Trapezium problems often become simpler when extended to form triangles or parallelograms.

Question 16. In the adjoining figure, D, E and F are mid-points of the sides BC, CA and AB of \(\triangle ABC\). If \(AB = 4.3\text{ cm}\), \(BC = 5.6\text{ cm}\) and \(AC = 3.5\text{ cm}\), find the perimeter of \(\triangle DEF\).
Answer:
By the mid-point theorem, the line segment joining the mid-points of two sides of a triangle is half of the third side.
\(EF = \frac{1}{2}BC = \frac{5.6}{2} = 2.8\text{ cm}\)
\(DF = \frac{1}{2}AC = \frac{3.5}{2} = 1.75\text{ cm}\)
\(DE = \frac{1}{2}AB = \frac{4.3}{2} = 2.15\text{ cm}\)
Perimeter of \(\triangle DEF = EF + DF + DE\)
\(= 2.8 + 1.75 + 2.15 = 6.7\text{ cm}\).
In simple words: Each side of triangle DEF is half the length of the corresponding parallel side of ABC.

Exam Tip: Perimeter is the sum of all outer sides of the triangle.

Question 17. In a parallelogram ABCD, AP and CQ are drawn perpendiculars from vertices A and C on diagonal BD. Prove that \(\triangle APB \cong \triangle CQD\).
Answer:
In \(\triangle APB\) and \(\triangle CQD\):
\(\angle APB = \angle CQD = 90^\circ\) (Given perpendiculars)
\(AB = CD\) (Opposite sides of parallelogram ABCD)
\(\angle ABP = \angle CDQ\) (Alternate interior angles since \(AB \parallel CD\))
Therefore, by AAS congruence criterion, \(\triangle APB \cong \triangle CQD\).
In simple words: Show right angles, equal opposite sides of the parallelogram, and alternate interior angles are equal.

Question 18. In a parallelogram ABCD, E and F are points on AB and CD such that \(AE = CE\). Prove that \(ED \parallel BF\).
Answer:
Given ABCD is a parallelogram, so \(AB \parallel CD\) and \(AB = CD\).
Also, given \(AE = CF\) (since E and F are points such that corresponding segments are equal).
Subtracting from total sides gives \(EB = FD\).
Since \(EB \parallel FD\) and \(EB = FD\), quadrilateral EBFD is a parallelogram.
Therefore, \(ED \parallel BF\).
In simple words: Prove that opposite sides of quadrilateral EBFD are equal and parallel, making it a parallelogram.

Exam Tip: Showing a quadrilateral has one pair of opposite sides equal and parallel proves it is a parallelogram.

Section - D

Question 19. If a line is parallel to the base of a trapezium and bisects one of the non-parallel sides, then prove that it bisects either diagonal of the trapezium.
Answer:
Let ABCD be a trapezium with \(AB \parallel CD\). Let line l pass through mid-point E of non-parallel side AD and be parallel to AB.
Consider diagonal AC intersecting line l at point F.
In \(\triangle ADC\), E is the mid-point of AD and \(EF \parallel DC\).
By the converse of the mid-point theorem (or intercept theorem), F must be the mid-point of AC.
Hence, the line bisects the diagonal AC.
In simple words: Apply the converse of the mid-point theorem inside the triangle formed by the diagonal and sides.

Exam Tip: Converse theorems are powerful tools when a mid-point and parallel line are given.

Question 20. AD is a median of \(\triangle ABC\) and E is the mid-point of AD. BE produced meets AC in F. Prove that \(AF = \frac{1}{3}AC\).
Answer:
Through D, draw \(DG \parallel BF\) meeting AC at G.
In \(\triangle ADG\), E is the mid-point of AD and \(EF \parallel DG\), so by the mid-point theorem, F is the mid-point of AG, meaning \(AF = FG\).
In \(\triangle CBG\), D is the mid-point of BC and \(DG \parallel BF\), so by the converse mid-point theorem, G is the mid-point of FC, meaning \(FG = GC\).
Therefore, \(AF = FG = GC\).
Since \(AC = AF + FG + GC\), we have \(AC = 3AF\), which gives \(AF = \frac{1}{3}AC\).
In simple words: Draw a parallel construction line to create equal segments along AC using mid-point theorems.

Exam Tip: Construction questions require clear explanation of the drawn parallel line and the triangles used.

Question 21. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that:
(i) D is the mid-point of AC
(ii) \(CM = MA = \frac{1}{2}AB\)
Answer:
(i) In \(\triangle ABC\), M is the mid-point of AB and \(MD \parallel BC\). By the converse of the mid-point theorem, D is the mid-point of AC.
(ii) In \(\triangle ADM\) and \(\triangle CDM\):
\(AD = CD\) (D is mid-point of AC)
\(\angle ADM = \angle CDM = 90^\circ\) (\(MD \parallel BC\) and \(\angle C = 90^\circ\))
\(DM = DM\) (Common side)
Therefore, \(\triangle ADM \cong \triangle CDM\) by SAS congruence.
Thus, \(AM = CM\) (by CPCT).
Since M is the mid-point of AB, \(AM = MB = \frac{1}{2}AB\).
Hence, \(CM = MA = \frac{1}{2}AB\).
In simple words: Use converse mid-point theorem for part (i) and triangle congruence for part (ii).

Exam Tip: Break multi-part geometry questions into clear steps for each roman numeral.

Question 22. Show that the bisectors of angles of a parallelogram form a rectangle.
Answer:
Let ABCD be a parallelogram. The angle bisectors of \(\angle A\), \(\angle B\), \(\angle C\), and \(\angle D\) form a quadrilateral PQRS.
In \(\triangle ABZ\) (where angle bisectors of A and B meet at Z), the sum of half angles is \(90^\circ\), making the included angle \(90^\circ\).
Similarly, all four interior angles of quadrilateral PQRS are \(90^\circ\).
A quadrilateral with all interior angles equal to \(90^\circ\) is a rectangle.
In simple words: Angle bisectors of consecutive angles in a parallelogram always meet at right angles, forming a rectangle.