Quadrilaterals Class 9 Mathematics Exam Questions

Exam Questions Class 9

Please refer to Quadrilaterals Class 9 Mathematics Exam Questions provided below. These questions and answers for Class 9 Mathematics have been designed based on the past trend of questions and important topics in your class 9 Mathematics books. You should go through all Class 9 Mathematics Important Questions provided by our teachers which will help you to get more marks in upcoming exams.

Class 9 Mathematics Exam Questions Quadrilaterals

Class 9 Mathematics students should read and understand the important questions and answers provided below for Quadrilaterals which will help them to understand all important and difficult topics.

Very Short Answer Type Questions:

Question. In a rhombus ABCD, if ∠A = 60°, then find the sum of ∠A and ∠C.
Ans. Since a rhombus is a parallelogram. 
∴ Its opposite angles are equal.

Quadrilaterals Class 9 Mathematics Exam Questions

⇒ ∠A = ∠C
∴ ∠C = 60° [ ∠A = 60° (Given)]
Now, required sum = ∠A + ∠C
= 60° + 60° = 120°

Question. Can the angles 110°, 80°, 70° and 95° be the angles of a quadrilateral ? Why or why not?
Ans. No.
∵ Sum of the angles = 110° + 80° + 70° + 95°
= 355° ≠ 360°
Thus, the given angles cannot be the angles of a quadrilateral.

Question. In the given figure, ABCD is a parallelogram in which ∠DAB = 60° and ∠DBC = 55° . Compute ∠CDB and ∠ADB.

Quadrilaterals Class 9 Mathematics Exam Questions

Ans. We have, ∠A + ∠B = 180° [Co-interior angles]
⇒ 60° + ∠ABD + 55° = 180° ⇒ ∠ABD = 65°
Also, ∠ABD = ∠CDB [Alternate interior angles are equal]
∴ ∠CDB = ∠ABD = 65°
We have, ∠ADB = ∠DBC [Alternate interior angles are equal]
⇒ ∠ADB = 55°

Question. ABCD is a parallelogram in which ∠A = 78°. Compute ∠B, ∠C and ∠D.
Ans. Since, ∠A + ∠B = 180° [Co-interior angles] 
⇒ ∠B = 180° – 78° = 102°
Now, ∠B = ∠D = 102°
and, ∠A = ∠C = 78°

Quadrilaterals Class 9 Mathematics Exam Questions

[∵ opposite angles of a parallelogram are equal]

Question. In the given figure, AB = AC and CP || BA and AP is the bisector of exterior ∠CAD o f ΔABC. Prove that ∠PAC = ∠BCA and ABCP is a parallelogram.

Quadrilaterals Class 9 Mathematics Exam Questions

Ans. We have, AB = AC ⇒ ∠BCA = ∠B
Now, ∠CAD = ∠B + ∠BCA [Exterior angle property]
⇒ 2∠CAP = 2∠BCA [∵ AP is the bisector of ∠CAD]
⇒ ∠CAP = ∠BCA ⇒ AP || BC
Also, AB || CP [Given]
Hence, ABCP is a parallelogram.

Short Answer Type Questions: 

Question. Diagonals of a quadrilateral ABCD bisect each other. ∠A = 45° and ∠B = 135°. Is it true? Justify your answer.
Ans. True. Given, ABCD is a quadrilateral whose diagonals bisect each other. Then, it should be a parallelogram.
Also, ∠A and ∠B are adjacent angles of parallelogram ABCD. So, their sum should be 180°.
Now, ∠A + ∠B = 45° + 135° = 180°

Question. The side of a rhombus is 10 cm. The smaller diagonal is 1/3 of the greater diagonal. Find the length of the greater diagonal.
Ans. Let ABCD be the rhombus and greater diagonal AC be x cm.
∴ Smaller diagonal, BD = 1/2 AC = x/3 cm
Since diagonals of rhombus are perpendicular bisector of each other.

Quadrilaterals Class 9 Mathematics Exam Questions

∴ OA = x OB = x/2 cm and OB = x/6 cm
In ΔAOB, we have
AB2 = OA2 + OB2
⇒ 102 = (x/2)2 + (x/6)2
⇒ 100 = x2/4 + x2/36
⇒ 100 = 10/36 x ⇒ x = 6√10 cm

Question. The perimeter of a parallelogram is 30 cm. If longer side is 9.5 cm, then find the length of shorter side.
Ans.  Let ABCD be a parallelogram with AB and DC as longer sides and AD and BC as shorter sides.

Quadrilaterals Class 9 Mathematics Exam Questions

Now, AB = DC = 9.5 cm [Opposite sides of a parallelogram are equal and longer side = 9.5 cm (Given)]
Let AD = BC = x
Now, AB + BC + CD + DA = 30 [Perimeter = 30 cm (Given)]
⇒ 9.5 + x + 9.5 + x = 30
⇒ 2x = 30 – 19 = 11 ⇒ x = 5.5 cm
∴ Length of shorter side = 5.5 cm

Question. In the given figure, K is the mid-point of side SR of a parallelogram PQRS such that ∠SPK = ∠QPK. Prove that PQ = 2QR.

Quadrilaterals Class 9 Mathematics Exam Questions

Ans. We have, ∠SPK = ∠QPK …(i)
Now, PQ || RS and PK is a transversal
∴ ∠SKP = ∠QPK [Alternate angles] …(ii)
From (i) and (ii), ∠SPK = ∠SKP
⇒ PS = SK …(iii) (Sides opposite to equal angles are equal)
But K is the mid-point of SR.
∴ SK = KR …(iv)
PS = QR …(v)(Opposite sides of parallelogram are equal)
From (iii) and (v), SK = PS = QR
Also, PQ = SR = SK + KR = 2SK [From (i)]
= 2QR

Question. In the adjoining figure, points A and B are on the same side of a line m, AD ⊥ m and BE ⊥ m and meet m at D and E, respectively. If C is the mid-point of AB, then prove that CD = CE.

Quadrilaterals Class 9 Mathematics Exam Questions

Ans. We have, C is the mid-point of AB

Quadrilaterals Class 9 Mathematics Exam Questions

∴ AC = BC.
Draw, CM ⊥ m and join AE.
We have, AD ⊥ m,
CM ⊥ m and BE ⊥ m.
∴ AD || CM || BE
In ΔABE, CG || BE [CM || BE]
and C is the mid-point of AB.
Thus, by converse of mid-point theorem, G is the midpoint of AE.
In ΔADE, G is the mid-point of AE and GM || AD. [ CM || AD]
Thus, by converse of mid-point theorem, M is mid-point of DE.
In ΔCMD and ΔCME,
DM = EM (M is the mid-point of DE)
∠CMD = ∠CME = 90° (CM ⊥ m)
CM = CM (Common)
∴ ΔCMD ≅ ΔCME (By SAS congruence rule)
So, CD = CE (By C.P.C.T.)

Question. In a parallelogram ABCD, if ∠A = (3x – 20)°, ∠B = (y + 15)° and ∠C = (x + 40)°, then find x + y (in degrees).
Ans. Since, ABCD is a parallelogram.

Quadrilaterals Class 9 Mathematics Exam Questions

∴ ∠A = ∠C
⇒ (3x – 20)° = (x + 40)°
⇒ 3x – x = 40 + 20
⇒ 2x = 60 ⇒ x = 30
Also, ∠A + ∠B = 180°
⇒ (3x – 20)° + (y + 15)° = 180°
⇒ 3x + y = 185 ⇒ y = 185 – 90 = 95
∴ x + y = 30 + 95 = 125

Question. D and E are the mid-points of sides AB and AC respectively of triangle ABC. If the perimeter of ΔABC = 35 cm, then find the perimeter of ΔADE.
Ans. Since, D and E are the mid-point of sides AB and AC respectively.

Quadrilaterals Class 9 Mathematics Exam Questions

∴ AD = 1/2 AB and AE = 1/2AC
By mid-point theorem, DE = 1/2BC
∴ AD + AE + DE = 1/2(AB + AC + BC)
Perimeter of ΔADE = 1/2 × perimeter of ΔABC
= 1/2 × 35 cm = 17.5 cm
Hence, the perimeter of ΔADE is 17.5 cm.

Question. Let ΔABC be an isosceles triangle with AB = AC and let D, E and F be the mid-points of BC, CA and AB respectively. Show that AD ⊥ FE and AD is bisected by FE. 
Ans. ABC is an isosceles triangle with AB = AC and D, E and F as the mid-points of sides BC, CA and AB
respectively. AD intersects FE at O.

Quadrilaterals Class 9 Mathematics Exam Questions

Join DE and DF.
Since, D, E and F are mid-points of sides BC, AC and AB respectively.
∴ DE || AB and DE = 1/2 AB [By mid-point theorem]
Also, DF || AC and DF = 1/2 AC 
But, AB = AC [Given]
⇒ 1/2 AB = 1/2 AC
⇒ DE = DF …(i)
Now, DE =1/2 AB ⇒ DE = AF …(ii)
and, DF = 1/2 AC ⇒ DF = AE …(iii)
From (i), (ii) and (iii), we have
DE = AE = AF = DF ⇒ DEAF is a rhombus.
Since, diagonals of a rhombus bisect each other at right angles.
∴ AD ⊥ FE and AD is bisected by FE.

Question. ABCD is parallelogram. P is a point on AD such that AP = 1/3 AD and Q is a point on BC such that CQ = 1/3 BC . Prove that AQCP is a parallelogram.
Ans. Q ABCD is parallelogram.

Quadrilaterals Class 9 Mathematics Exam Questions

⇒ AD = BC and AD || BC
⇒ 1/3 AD =1/3 BC and AD|| BC
⇒ AP = CQ and AP || CQ
Thus, APCQ is a quadrilateral such that one pair of opposite sides AP and CQ are parallel and equal.
Hence, APCQ is a parallelogram. 

Question. In the given rectangle ABCD, ∠ABE = 30° and ∠CFE = 144°. Find the measure of ∠BEF.

Quadrilaterals Class 9 Mathematics Exam Questions

Ans. Here, ∠ABE + ∠EBF = 90°
⇒ 30° + ∠EBF = 90°
⇒ ∠EBF = 60° …(i)
and ∠BFE + ∠CFE = 180° [Linear pair]
⇒ ∠BFE + 144° = 180°
⇒ ∠BFE = 180° – 144° = 36° …(ii)
Now, in ΔBEF,
∠EBF + ∠BFE + ∠BEF = 180° (Angle sum property)
⇒ 60° + 36° + ∠BEF = 180° [Using (i) and (ii)]
⇒ ∠BEF = 180° – 96° = 84°

Question. l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure).

Quadrilaterals Class 9 Mathematics Exam Questions

Show that l, m and n cut off equal intercepts DE and EF on q also.
Ans.  We have, AB = BC and have to prove that DE = EF.
Now, trapezium ACFD is divided into two triangles
namely ΔACF and ΔAFD.
In ΔACF, AB = BC ⇒ B is mid-point of AC
and BG || CF [Q m || n]
So, G is the mid-point of AF.
[By converse of mid-point theorem]
Now, in ΔAFD, G is the mid-point of AF.
and GE || AD [Q m || l]
∴ E is the mid-point of FD.
[By converse of mid-point theorem]
⇒ DE = EF
∴ l, m and n cut off equal intercepts on q also. 

Question. Two parallel lines l and m are intersected by l a transversal p (see figure). Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.

Quadrilaterals Class 9 Mathematics Exam Questions

Ans. It is given that l || m and transversal p intersects them at points A and C respectively.
The bisectors of ∠PAC and ∠ACQ intersect at B and bisectors of ∠ACR and ∠SAC intersect at D.
Now, ∠PAC = ∠ACR [Alternate angles as l || m and p is a transversal]
So, 1/2 ∠PAC = 1/2 ∠ACR 
⇒ ∠BAC = ∠ACD
These form a pair of alternate angles for lines AB and DC with AC as transversal and they are equal also.
So, AB || DC
Similarly, BC || AD
Therefore, quadrilateral ABCD is a parallelogram.
Also, ∠PAC + ∠CAS = 180° [Linear pair]
So, 1/2 ∠PAC + 1/2 ∠CAS = 1/2 × 180° = 90°
⇒ ∠BAC + ∠CAD = 90° ⇒ ∠BAD = 90°
So, ABCD is a parallelogram in which one angle is 90°. Therefore, ABCD is a rectangle.

Long Answer Type Questions:

Question. In ΔABC, AB = 18 cm, BC = 19 cm and AC = 16 cm. X, Y and Z are mid-points of AO, BO and CO respectively as shown in the figure. Find the perimeter of ΔXYZ.

Quadrilaterals Class 9 Mathematics Exam Questions

Ans. Here, in ΔABC, AB = 18 cm, BC = 19 cm, AC = 16 cm.
In ΔAOB, X and Y are the mid-points of AO and BO.
∴ By mid-point theorem, we have
XY = 1/2 AB = 1/2 × 18 cm = 9 cm
In ΔBOC, Y and Z are the mid-points of BO and CO.
∴ By mid-point theorem, we have
YZ = 1/2 BC = 1/2 × 19 cm = 9.5 cm
And, in ΔCOA, Z and X are the mid-points of CO and AO.
∴ By mid-point theorem, we have
∴ ZX = 1/2 AC = 1/2 × 16 cm = 8 cm
Hence, the perimeter of ΔXYZ = 9 + 9.5 + 8 = 26.5 cm

Question. ABCD is a parallelogram. AB and AD are produced to P and Q respectively such that BP = AB and DQ = AD. Prove that P, C, Q lie on a straight line.
Ans.  CP and CQ are joined.

Quadrilaterals Class 9 Mathematics Exam Questions

∴ ABCD is a parallelogram.
So, BC = AD, AB = DC [Opposite sides of parallelogram]
and ∠ABC = ∠ADC [Opposite angles of parallelogram]
∴ Their supplementary angles are equal
So, ∠PBC = ∠CDQ
In ΔPBC and ΔCDQ, we have
 BC = DQ [BC = AD and AD = DQ (Given)]
 BP = DC [AB = DC and AB = BP (given)]
⇒ ∠PBC = ∠CDQ [Proved above]
∴ ΔPBC ≅ ΔCDQ [By SAS congruency]
⇒ ∠BPC = ∠DCQ and ∠BCP = ∠DQC [By C.P.C.T.]
Again, ∠BCD = ∠PBC [since, AP || DC]
Now, ∠BCP + ∠BCD + ∠DCQ
= ∠BCP + ∠PBC + ∠BPC = 2 right angles
i.e., ∠PCQ is a straight angle.
i.e., P, C, Q lie on a straight line.

Question. In the given figure, ABCD is a parallelogram and E is the mid-point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L. Prove that (i) AF = 2DC (ii) DF = 2DL 

Quadrilaterals Class 9 Mathematics Exam Questions

Ans. (i) As EB || DF ⇒ EB || DL and ED || BL.
Therefore, EBLD is a parallelogram.
∴ BL = ED = 1/2 AD = 1/2 BC = CL …(i) [ABCD is a parallelogram ∴ AD = BC]
Now, in ΔDCL and ΔFBL, we have
CL = BL [from (i)]
∠DLC = ∠FLB (Vertically opposite angles)
∠DCL = ∠FBL (Alternate angles)
∴ ΔDCL ≅ ΔFBL (By ASA congruency criteria)
⇒ CD = BF and DL = FL (By C.P.C.T.)
Now, BF = DC = AB …(ii)
⇒ 2AB = 2DC ⇒ AB + AB = 2DC
⇒ AB + BF = 2DC [Using (ii)]
⇒ AF = 2DC
(ii) ∴ ΔL = FL ⇒ DF = 2DL