# Quadrilaterals Class 9 Mathematics Exam Questions

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.

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.

⇒ ∠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.

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]

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°

[∵ 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.

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.

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.

∴ 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.

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.

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.

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

∴ AC = BC.
Draw, CM ⊥ m and join AE.
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.
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.

∴ ∠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.

∴ 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.

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.

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.

⇒ 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.

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).

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.

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
Therefore, quadrilateral ABCD is a parallelogram.
Also, ∠PAC + ∠CAS = 180° [Linear pair]
So, 1/2 ∠PAC + 1/2 ∠CAS = 1/2 × 180° = 90°
So, ABCD is a parallelogram in which one angle is 90°. Therefore, ABCD is a rectangle.

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.

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.

∴ 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
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

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