# RBSE Class 9 Maths Important Questions Chapter 8 Quadrilaterals

Rajasthan Board RBSE Class 9 Maths Important Questions Chapter 8 Quadrilaterals Important Questions and Answers.

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## RBSE Class 9 Maths Chapter 8 Important Questions Quadrilaterals

I. Multiple Choice Questions :
Choose the correct answer from the given options.

Question 1.
Which of the following is not true for a parallelogram?
(a)    Opposite sides are equal.
(b)    Opposite angles are equal.
(c)    Opposite angles are bisected by the diagonals.
(d)    Diagonals bisect each other.
(c)    Opposite angles are bisected by the diagonals.

Question 2.
Three angles of a quadrilateral are 75°, 90° and 75°. The fourth angle is :
(a)    90°
(b) 95°
(c) 105°
(d) 120°
(d) 120°

Question 3.
ABCD is a rhombus such that ∠ACB = 40°. Then ∠ADB is :
(a)    40°
(b) 45°
(c) 50°
(d) 60°
(c) 50°

Question 4.
A diagonal of a rectangle is inclined to one side of the rectangle at 25°. The acute angle between the diagonal is :
(a)    55°
(b) 50°
(c) 40°
(d) 25°
(b) 50°

Question 5.
The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is square only if:
(a) ABCD is a rhombus.
(b) diagonals of ABCD are equal.
(c) diagonals of ABCD are perpendicular to each other.
(d) diagonals of ABCD are equal and perpendicular to each other.
(d) diagonals of ABCD are equal and perpendicular to each other.

Question 6.
The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle if:
(a) PQRS is a parallelogram.
(b) PQRS is a rectangle.
(c) the diagonals of PQRS are perpendicular to each other.
(d) the diagonals of PQRS are equal.
(c) the diagonals of PQRS are perpendicular to each other.

Question 7.
If APB and CQD are two parallel lines, then the bisectors of ∠APQ, ∠BPQ, ∠CQP and ∠PQD form:
(a) a square
(b) a rhombus
(c) a rectangle
(d) any other parallelogram
(c) a rectangle

Question 8.
D and E are mid-points of the sides AB and AC of ΔABC and O is any point on the side BC. O is joined to A. If P and Q are mid-points of OB and OC respectively, then ΔEQP is:
(a) a square
(b) a rectangle
(c) a rhombus
(d) a parallelogram
(d) a parallelogram

Question 9.
The quadrilateral formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a rhombus if:
(a) ABCD is a parallelogram.
(b) ABCD is a rhombus.
(c) the diagonals of ABCD are equal.
(d) the diagonals of ABCD are perpendicular to each other.
(c) the diagonals of ABCD are equal.

Question 10.
In the given figure, ABCD is a parallelogram and E is the mid-point of BC. Also, DE and AB when produced meet at F. Then

(a)    AF = $$\frac{3}{2}$$AB
(b)    AF = 2AB
(c)    AF = 3AB
(d)    AP2 = 2AB2
(b)    AF = 2AB

Question 11.
In the given figure, ABCD is a parallelogram in which ∠BDC = 45° and ∠BAD = 75°. Then, ∠CBD =?
(a) 45°
(b) 55°
(c) 60°
(d) 75°
(c) 60°

Question 12.
In the given figure, ABCD is a parallelogram, M is the mid-point of BD and BD bisects ∠B as well as ∠D. Then, ∠AMB = ?

(a) 46°
(b)    60°
(c) 90°
(d) 30°
(c) 90°

Question 13.
In the given figure, AD is a median of ΔABC and E is the mid-point of AD. If BE is joined and produced to meet AC in F, then AF = ?

(a)    $$\frac{1}{2}$$AC
(b)    $$\frac{1}{3}$$AC
(c)    $$\frac{2}{3}$$AC
(d)    $$\frac{3}{4}$$AC
(b)    $$\frac{1}{3}$$AC

Question 14.
The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O such that ∠DAC = 30° and ∠AOB = 70°. Then ∠DBC = ?

(a)    40°
(b)    35°
(c)    45°
(d)    50°
(a)    40°

Question 15.
In the given figure, ABCD is a rhombus. Then:

(a)    AC2 + BD2 = AB2
(b)    AC2 + BD2 = 2AB2
(c)    AC2 + BD2 = 4AB2
(d)    2(AC2 + BD2) = 3AB2
(c)    AC2 + BD2 = 4AB2

II. Fill in the Blanks :

Question 1.
If the angles of a quadrilateral are (4x)°, (7x)°, (15x)° and (10x)°. Then the smallest angle of this quadrilateral is _________.
40

Question 2.
In figure, PQRS is a parallelogram in which a pair of opposite angles is given. The value of x is _________.

15

Question 3.
ABCD is rhombus such that ∠ACB = 40°, then ∠ADB is _________.
50

Question 4.
In the figure, ΔPQR is formed by joining the mid-points of sides BC, CA, and AB respectively. If AABC is an equilateral triangle with side 12 cm, then the length of PQ is    _________.

6 cm

Question 5.
In a triangle, the line segment joining the mid-points of any given two sides is to the third side and is _________.
Parallel, half of it.

III. True/False:
State whether the following statements are True or False.

Question 1.
A quadrilateral is a parallelogram if its opposite sides are parallel.
True

Question 2.
In a parallelogram opposite angles are equal.
True

Question 3.
Every square is a rectangle.
True

Question 4.
In a parallelogram, diagonals bisect each other.
True

Question 5.
Every square is a rhombus.
True

Question 6.
The sum of the angles of a quadrilateral is 180°.
True

Question 7.
In a parallelogram, the diagonals interesect at right angles.
False

IV. Match the Columns :

Question 1.
Match the column I with the column II.

 Column I Column II (1) Angle bisectors of a parallelogram form a (i) parallelogram (2) The quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a square is a (ii) rectangle (3) The quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a rectangle is a (iii) square (4) The figure formed by joining the mid-points of the pairs of adjacent sides of a quadrilateral is a (iv) rhombus

 Column I Column II (1) Angle bisectors of a parallelogram form a (ii) rectangle (2) The quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a square is a (iii) square (3) The quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a rectangle is a (iv) rhombus (4) The figure formed by joining the mid-points of the pairs of adjacent sides of a quadrilateral is a (i) parallelogram

V. Very Short Answer Type Questions :

Question 1.
Diagonals of a quadrilateral ABCD bisect each other. If ∠A = 35° then ∠B = 145°. Is it true? Also justify your answer.
It is true that if ∠A = 35° then ∠B = 145°. It is given that ABCD is a quadrilateral and diagonals of quadrilateral ABCD bisect each other. Then, it should be a parallelogram.
Also, we know that ∠A and ∠B are adjacent angles of parallelogram ABCD ∠A + ∠B = 180°
⇒ 35° + 145° = 180°
Hence    ∠B = 145°.

Question 2.
In the following figure, ABCD and AEFG are two parallelograms. If ∠C = 60° then what is the value of ∠GFE?

∠A = ∠C    (Opposite angles of parallelogram ABCD are equal)
Also, AEFG is a parallelogram (Given)
⇒ ∠GFE = ∠A (Opposite angle of parallelogram are equal)
⇒ ∠GFE = 60°

VI. Short Answer Type Questions :

Question 1.
In quadrilateral ABCD, ∠A + ∠C = 140°, ∠A : ∠C = 1 : 3 and ∠B : ∠D = 5:6. Find the ∠A, ∠B, ∠C and ∠D.

∠A + ∠C = 140°    .................... (1)
∠A : ∠C = 1 : 3 .................. (2)
and    ∠B : ∠D = 5 : 6    ................ (3)
Let ∠A = x and ∠C = 3x.
We know that, sum of angles of a quadrilateral = 360°
∠A + ∠B + ∠C + ∠D = 360°
⇒ ∠B + ∠D = 360° - (∠A + ∠C)
⇒ ∠B + ∠D = 360° - (35° + 105°) = 220°
Since,    ∠B : ∠D = 5 : 6
⇒ ∠B : ∠D = 5 : 6
⇒ 5x + 6x = 11x
⇒ 11x = 220°
x = 20°
∠B = 5x = 5 × 20 = 100°
∠D = 6x = 6 × 20 = 120°
Hence, ∠A = 35°, ∠B = 100°, ∠C = 105° and ∠D = 120°.

Question 2.
In the adjoining figure, PQRS is a rhombus, SQ and PR are the diagonals of the rhombus intersecting at point O. If ∠OPQ = 35°, then find the value of ∠ORS + ∠OQP.

Given : PQRS is a rhombus
⇒ PQRS is a parallelogram
Since,    PQ ∥ SR and PR is a transversal.
∴ ∠ORS = ∠OPQ = 35°    (Alternate interior angles)    ....................(1)
Also, diagonals of a rhombus bisect each other at right angle.
∴ ∠SOR = 90°    .............(2)
Now, in ΔSOR, ∠RSO + ∠SOR + ∠ORS = 180° (Angle sum property of triangles)
⇒ ∠RSO = 180° - ∠SOR - ∠ORS
⇒ ∠RSO = 180° - 90° - 35°
⇒ ∠RSO = 55°
⇒ ∠OQP = 55°
(Alternate interior angles, since PQ ∥ SR and QS is a transversal)

From equations (1) and (3), we get
∠ORS + ∠OQP = 35° + 55° = 90°

Question 3.
ABCD is a kite with AB = AD and CD = CR. Prove that the figure formed by joining the mid-points of the consecutive sides is a rectangle.
We draw the figure as shown below.

Let P, Q, R and S be the mid-points of the sides AB, BC, CD and DA respectively. Join AC and BD. Which intersects each other at O.
Now, in ΔABC, PQ ∥ AC and PQ = $$\frac{1}{2}$$AC ............ (1) (By mid-point theorem)
and in ΔACD, RS ∥ AC and RS = $$\frac{1}{2}$$AC .................(2) (By mid-point theorem)

From equations (1) and (2) we get,
PQ ∥ RS and PQ = RS (Since, a pair of opposite sides equal and parallel)
So, PQRS is a parallelogram
Also, AB = AD (Given)    ...................(1)
So, A lies on the perpendicular bisector of BD further
CB = CD (Given)    ......................(2)
So, C lies on the perpendicular bisector of BD.    .................(3)

From equations (2) and (3), AC is the perpendicular bisector of BD.
i.e.  AC ⊥ BD ⇒ ∠AOD = 90°
Now, clearly HS ∥ OE and SE ∥ OH
So, SEOH is a parallelogram.
Hence, ∠ESH = ∠EOH = 90°
So, parallelogram PQRS is a rectangle.

Question 4.
ABCD is a rhombus such that ∠ACB = 40°, then find ∠ADB.
Given : ABCD is a rhombus.

AB = CD = BC = AD
In ΔABC, AB = BC
∴ ∠ACB = ∠CAB = 40°
(Angles opposite to equal sides are equal)
Now, AB ∥ CD and AC is transversal.
∴ ∠DCA = ∠CAB = 40°    (Alternate angles)
Now,    ∠C = ∠BCD - ∠BCA + ∠DCA
= 40° + 40° = 80°

We know that, ∠D + ∠C = 180° (Co-interior angles of a rhombus)
∠D =180° - 80° = 100°
∴ ∠ADB = $$\frac{1}{2}$$∠D = $$\frac{100^{\circ}}{2}$$ = 50°

Question 5.
In the adjoining figure, P is the mid-point of side BC of a parallelogram ABCD, such that ∠BAP = ∠DAP. Prove that AD = 2 CD.

Given : ABCD is a parallelogram. P is the mid-point of BC,
i.e. BC = 2BP ⇒ AD = 2BP
∠BAP = ∠DAP    .
∠BAP = ∠DAP
∠DAP = ∠APB
∠BAP = ∠APB ⇒ BP = AB
(Sides opposite to equal angles of a triangle are equal)
AB = CD (Opposite sides of parallelogram)
BP = CD (∵ BP = AB) ...................(1)
BC = 2BP (Given)
(∵ BC = AD, opposite sides of parallelogram)

From equations (1) and (2), we have :
Hence proved.

Question 6.
In quadrilateral ABCD, AP and BP are bisectors of ∠A and ∠B respectively, then find the value of x.
From the fig.

∠1 = ∠2, ∠3 = ∠4    ...(1)
(AP and BP are bisectors of ∠A and ∠B.)

So,    ∠A + ∠B + ∠C + ∠D = 360°    (Sum of all angles of a quad.)
∴ ∠1 + ∠2 + ∠3 + ∠4 + 130° + 60° = 360°
⇒ 2∠1 + 2∠3 + 190° = 360°    [From (1)]
⇒ 2∠1 + 2∠3 = 360° - 190° = 170°
⇒ ∠1 + ∠3 = $$\frac{170^{\circ}}{2}$$ = 85°    ...(2)

In ΔAPB, ∠1 + x + ∠3 = 180°    (Angle sum property of a A)
⇒ ∠1 + ∠3 + x = 180°
⇒ 85° + x = 180°
∴ x = 180° - 85° = 95°
Hence,    x = 95°

Question 7.
The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°. Find the angles of the parallelogram.

Given : Parallelogram ABCD, in which ∠ADC and ∠ABC are obtuse angles. Now, DE and DF are two altitudes of parallelogram and angle between them is 60°.
Now, BEDF is a quadrilateral, in which
∠BED = ∠BFD = 90°
∴ ∠FBE = 360° - (∠FDE + ∠BED + ∠BFD)
(Angle sum property of a quadrilateral) = 360° - (60° + 90° + 90°) = 360° - 240° = 120°
Since, ABCD is a parallelogram.
∠ADC = 120°    [∵ ∠D = ∠B]
Now,    ∠A + ∠B = 180° (Co-interior angles of a parallelogram)
∴ ∠A = 180° - ∠B = 180° - 120° (∵ ∠FBE = ∠B)
⇒ ∠A = 60°
Also, ∠C = ∠A = 60°
(∵ Opposite angles of a parallelogram are equal.)
Hence, angles of the parallelogram are 60°, 120°, 60° and 120°, respectively.

Question 8.
PQ and RS are two equal and parallel line segments. Any point M not lying on PQ or RS is joined to Q and S and lines through P parallel to QM and through R parallel to SM meet at N. Prove that line segments MN and PQ are equal and parallel to each other.
It is given that PQ = RS and PQ ∥ RS.

Since, a pair of opposite side of a quadrilateral PQSR is equal and parallel.
Therefore, PQSR is a parallelogram.
So,    PR = QS and PR ∥ QS
Now,    PR ∥ QS
Therefore,    ∠RPQ + ∠PQS = 180°
(Interior angles on the same side of the transversal)

i.e.    ∠RPQ + ∠PQM + ∠MQS = 180°
Also,    PN ∥ QM (By Construction)
Therefore, ∠NPQ + ∠PQM = 180°
i.e. ∠NPR + ∠RPQ + ∠PQM = 180° ...........(3)
So, ∠NPR = ∠MQS [From equation (2) and (3)] ...........(4)
Similarly ∠NRP = ∠MSQ, ..........(5)

In ΔPNR and ΔQMR PR = QS [Using equation (1)]
∠NPR = ∠MQS [Using equation (4)]
∠NRP = ∠MSQ [Using equation (5)]
∠PNR = ΔQMS
So, PN = QM and NR = MS
PN = QM and PN ∥ QM, we have PQMN is a parallelogram.
(Since, a pair of opposite side of a quadrilateral equal and parallel)
So,    NM = PQ and NM ∥ PQ.
Hence proved.

Question 9.
In the given figure, ABCD is a square and EF ∥ BD. M is the mid-point of EF. Prove that AM bisects ∠BAD.

Given : ABCD is a square and BD is a diagonal.
∠CBD = ∠CDB = x 90° = 45°
(∵ Diagonal of a square bisect each angle at the vertex)
Also, EF ∥ BD    (Given)
So, ∠CEF = ∠CBD = 45°    (Corresponding angles)
and ∠CFE = ∠CDB = 45°    .    (Corresponding angles)
⇒ CE = CF (∵ Sides opposite to equal angles are equal.)
⇒ BC - CE = CD - CF    (∵ BC = CD)
⇒ BE = DF     ................(1)
BE = DF [From equation (1)]

So, ΔABE ≅ ΔADF (By SAS congruence rule)
Then, AE = AF (By CPCT)    .........(2)
and ∠BAE = ∠DAF .............(3)
Now, in ΔAEM and ΔAFM,
AE = AF [From equation (2)]
ME = MF (M is mid-point of EF)
AM = AM (Common side)
ΔAEM ≅ ΔAFM (By SSS congruence rule)
So, ∠EAM = ∠FAM    (By CPCT)    ...........(4)

On adding equations (3) and (4), we get
∠BAE + ∠EAM = ∠DAF + ∠FAM
⇒ ∠BAM = ∠DAM
Hence proved.

Question 10.
ABCD is a parallelogram in which P and Q are mid-points of opposite sides AB and DC (given in figure). If AQ intersects DP at S and BQ intersects CP at R, show that:

(i) APCQ is a parallelogram.
(ii) DPBQ is a parallelogram.
(iii) PSQR is a parallelogram.
Given : ABCD is a parallelogram.
∴    AB ∥ DC, AB = DC
P and Q are mid-points of sides AB and CD.
AP ∥ QC [∵ AB ∥ DC]    ...................(1)

Now,    AP = $$\frac{1}{2}$$AB, QC = $$\frac{1}{2}$$DC ..................    (2)
(∵ P and Q are mid-points of AB and CD, respectively.)
Since,    AB = DC    (Given)
$$\frac{1}{2}$$AB = $$\frac{1}{2}$$CD
⇒ AP = QC    [Using equation (2)]
Thus, we have AP ∥ QC and AP = QC
Hence, APCQ is a parallelogram.

BP ∥ QD and BP = QD
DPBQ is a parallelogram.

(iii) Since, DPBQ is a parallelogram,
∴ DP ∥ QB
⇒ SP ∥ QR .................(3)
and APCQ is also a parallelogram.
∴ AQ ∥ PC
SQ ∥ PR ..............(4)
From equations (3) and (4), we have :
SP ∥ QR and SQ ∥ PR
Hence, PSQR is also a parallelogram.
Hence proved.

Question 11.
In the following figure, ABCD is a rhombus. If ∠ABC = 68°, then determine ∠ACD.

Given : ABCD is a rhombus.
⇒ ABCD is a parallelogram
⇒ ∠ADC = ∠ABC = 68° (v Opposite angles of a parallelogram are equal.)
∠ODC = $$\frac{1}{2}$$∠ADC ( Diagonal DB of rhombus ABCD bisects its ∠ABC and ∠ADC)
∠ODO = $$\frac{1}{2}$$ × 68°= 34°

Now, in ∠OCD
∠OCD + ∠ODC + ∠COD = 180° (Angle sum property of a triangle)
∠OCD + 340 + 90° = 180° (∵ ∠COD = 90°, diagonals of a rhombus bisect each other at right angles)
⇒ ∠OCD + 124° = 180°
⇒ ∠OCD = 180° - 124°
⇒ ∠OCD = 56° or ∠ACD = 56°

Question 12.
In the given figure, ABCD is a square, if ∠PQR = 90° and PB - QC = DR, then prove that QB = RC, PQ = QR and ∠QPR = 45°.

Given : ABCD is a square, ∠PQR = 90° and PB = QC = DR.
To prove : QB = RC, PQ = QR and ∠QPR = 450
Proof: Since, ABCD is a square.
BC = CD
But QC = DR (Given)
⇒ BC - QC = CD - DR
⇒ QB = RC ................. (1)

Now, in ΔPBQ and ΔQCR, PB = QC (Given)
BQ = CR [From equation (1)]
∠PBQ = ∠QCR = 90°
∴ ΔPBQ ≅ ΔQCR (By SAS congruence rule)
So, PQ = QR (By CPCT)
Also, ∠BPQ = ∠CQR and ∠BQP = ∠CRQ

Now, since APQR is an isosceles right angled triangle.
So, ∠PRQ + ∠QPR + ∠PQR = 180° (Angle sum property of triangle)
⇒ ∠QPR + ∠QPR + 90° = 180° (Since, ∠PQR is isosceles)
⇒ ∠QPR = 90°
∴ ∠QPR = 45°
Hence proved.

Question 1.
Prove that the opposite angles of an isosceles trapezium are supplementary.
Let ABCD be an isosceles trapezium AB ∥ DC and AD = BC.

Through C, draw CE ∥ DA, AB ∥ DC  i.e., AE ∥ DC and CE ∥ DA    (By construction)
AECD is a parallelogram AD = EC (Opp. sides of a ∥gm)
BC = EC
In ΔCEB, BC = EC
∠3 = ∠2 (Angles opp. equal sides are equal.) ...(1)

As DA ∥ CE and AE is a transversal.
∠1 = ∠2 (Corresponding ∠s) ...(2)
Also AE ∥ DC and CE is a transversal
∠5  = ∠2 (Alternate ∠s)
Now ∠A + ∠C  = ∠1 + (∠5 + ∠4)
⇒ ∠A + ∠C = ∠2 + ∠2 + ∠4 [Using (2) and (3)]
∠A + ∠C = ∠2 + ∠3 + ∠4 [Using (1)]
∠A + ∠C = 180° (Sum of ∠s of a Δ = 180°)
Now    ∠A + ∠B + ∠C + ∠D = 360°
∠B + ∠D + 180    = 360° (Sum of ∠s of quadrilateral)
⇒ ∠B + ∠D = 180° (∵ ∠A + ∠C= 180°)
Hence, the opposite angles of an isosceles trapezium are supplementary.
Hence proved.

Question 2.
P, Q,R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD, such that AC ⊥ BD. Prove that PQRS is a rectangle.
In the figure:

AC ⊥ BD
∠COD = ∠AOD = ∠AOB = ∠COB = 90°
SR ∥ AC and SR = $$\frac{1}{2}$$AC
Similarly in ΔABC, PQ ∥ AC and PQ = $$\frac{1}{2}$$AC
From equations (1) and (2) PQ ∥ SR and PQ = SR = $$\frac{1}{2}$$AC
Similarly, SP ∥ RQ and SP = RQ = $$\frac{1}{2}$$BD
In quadrilateral EDFR, DE ∥ FR, OF ∥ ER
∠EOF = ∠ERF = 90°
Hence, PQRS is a rectangle.

Question 3.
E is mid-point of the median AD of AABC and BE is produced to meet AC at F. Show that AF = $$\frac{1}{2}$$AC.

Through D, draw a line parallel to BF to meet AC at G.
In ΔADG, E is mid-point of side AD and DG ∥ BF i.e.
DG ∥ EF, therefore by converse of mid-point theorem,
F is mid-point of AG i. e. AF = FG.    ..........(i)
In ABCF, D is mid-point of side BC and DG ∥ BF, therefore, by converse of mid-point theorem, G is mid-point of CF,
i.e. FG = GC .......(ii)
From (i) and (ii), we get:
AF = FG = GC ........(iii)
Now    AC = AF + FG + GC = AF + AF + AF [Using (iii)]
⇒ AC = 3AF ⇒ AF = $$\frac{1}{3}$$AC
Hence Proved.

Last Updated on April 30, 2022, 12:10 p.m.
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Published April 30, 2022