# R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 10 – Quadrilaterals

## Chapter 10 – Quadrilaterals Exercise Ex. 10A

Question 1

Three angles of a quadrilateral are 75°, 90° and 75°. Find the measure of the fourth angle.

Solution 1

Let the measure of the fourth angle = x°

For a quadrilateral, sum of four angles = 360°

x° + 75° + 90° + 75° = 360°

x° = 360° – 240°

x° = 120°

Hence, the measure of fourth angle is 120°.

Question 2

Prove
that the sum of all the angles of a quadrilateral is 3600.

Solution 2

Given:

Question 3

The angle of the quadrilateral are in the ratio 2:4:5:7. Find the angles.

Solution 3

Question 4

In the adjoining figure , ABCD is a trapezium in which AB || DC. If =550 and = 700, find and .

Solution 4

Since AB || DC

Question 5

In the adjoining figure , ABCD is a square andis an equilateral triangle . Prove that

(i)AE=BE, (ii) =150

Solution 5

Given:

Question 6

In the adjoining figure , BMAC and DNAC. If BM=DN, prove that AC bisects BD.

Solution 6

Question 7

In given figure, ABCD is a quadrilateral in which AB=AD and BC= DC. Prove that (i) AC bisects  and , (ii) BE=DE,

(iii)

Solution 7

Question 8

In the given figure , ABCD is a square and PQR=900. If PB=QC= DR, prove that (i) QB=RC, (ii) PQ=QR, (iii) QPR=450.

Solution 8

Question 9

If O is a point with in a quadrilateral ABCD, show that OA+OB+OC+OD> AC+BD.

Solution 9

Given: O is a point within a quadrilateral ABCD

Question 10

In the adjoining figure, ABCD is a quadrilateral and AC is one of the diagonals .Prove that:

(i) AB+BC+CD+DA> 2AC

(ii) AB+BC+CD>DA

(iii) AB+BC+CD+DA>AC+BD

Solution 10

Given: ABCD is a quadrilateral and AC is one of its disgonals.

## Chapter 10 – Quadrilaterals Exercise Ex. 10B

Question 1

In the adjoining figure, ABCD is a parallelogram in which =720. Calculate ,and .

Solution 1

Question 2

In each of the figures given below, ABCD is a rhombus. Find the value of x and y in each case.

Solution 2

Question 3

The lengths of the diagonals of a rhombus are 24 cm and 18 cm respectively . Find the length of each side of the rhombus.

Solution 3

Question 4

Each side of a rhombus is 10 cm long and one of its diagonals measures 16 cm. Find the length of the other diagonal and hence find the area of the rhombus.

Solution 4

Question 5

In each of the figures given below, ABCD is a rectangle . Find the values of x and y in each case.

Solution 5

Question 6

In a rhombus ABCD, the altitude
from D to the side AB bisect AB. Find the angle of the rhombus

Solution 6

Let
the altitude from D to the side AB bisect AB at point P.

Join
BD.

In
ΔAMD and ΔBMD,

AM
= BM (M is the mid-point of
AB)

AMD = ∠BMD (Each 90°)

MD
= MD (common)

ΔAMD ≅ ΔBMD (by SAS congruence criterion)

But,
AD = AB (sides of a rhombus)

equilateral triangle.

∠A = 60°

∠C = ∠A = 60° (opposite angles are equal)

∠B = 180° – ∠A = 180° – 60° = 120°

∠D = ∠B = 120°

Hence,
in rhombus ABCD,
∠A = 60°, ∠B = 120°, ∠C = 60° and ∠D = 120°.

Question 7

In the adjoining figure , ABCD is square . A line segment CX cuts AB at X and the diagonal BD at O, such that . Find the value of x.

Solution 7

Question 8

In a rhombus ABCD show that
diagonal AC bisect
∠A as well as ∠C and diagonal BD bisect ∠B
as well as ∠D.

Solution 8

In

AB
= AD (sides of a rhombus are equal)

BC
= CD (sides of a rhombus are equal)

AC
= AC (common)

ΔABC ≅ ΔADC (by SSS congruence criterion)

∠BAC = ∠DAC and ∠BCA = ∠DCA (c.p.c.t.)

AC bisects ∠A as well as ∠C.

Similarly,

In

AB
= BC (sides of a rhombus are equal)

= CD (sides of a rhombus are equal)

BD
= BD (common)

ΔBAD ≅ ΔBCD (by SSS congruence criterion)

∠ABD = ∠CBD and ∠ADB = ∠CDB (c.p.c.t.)

BD bisects ∠B as well as ∠D.

Question 9

In a parallelogram ABCD, points M
and N have been taken on opposite sides AB and CD respectively such that AM =
CN. Show that AC and MN bisect each other.

Solution 9

In
ΔAMO and ΔCNO

MAO = ∠NCO (AB ∥ CD, alternate
angles)

AM
= CN (given)

AOM = ∠CON (vertically opposite angles)

ΔAMO ≅ ΔCNO (by ASA congruence criterion)

AO = CO and MO
= NO (c.p.c.t.)

AC and MN
bisect each other.

Question 10

In the adjoining figure, ABCD is a parallelogram. If P and Q are points on AD and BC respectively such that and , prove that AQCP is a parallelogram.

Solution 10

Question 11

In the adjoining figure, ABCD is a parallelogram whose diagonals intersect each other at O.A line segment EOF is drawn to meet AB at E and DC at F . Prove that OE=OF.

Solution 11

Question 12

In the adjoining figure , ABCD is a parallelogram in which

and . Calculate .

Solution 12

Question 13

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.

Solution 13

DCM = ∠DCN + ∠MCN

90° = ∠DCN + 60°

∠DCN = 30°

In
ΔDCN,

DNC + ∠DCN + ∠D = 180°

90° + 30° + ∠D = 180°

∠D = 60°

∠B = ∠D = 60° (opposite angles of parallelogram are
equal)

∠A = 180° – ∠B = 180° – 60° = 120°

∠C = ∠A = 120°

Thus,
the angles of a parallelogram are 60
°, 120°, 60° and 120°.

Question 14

ABCD is rectangle in which
diagonal AC bisect
∠A as well as ∠C. Show that (i) ABCD is square,
(ii) diagonal BD bisect ∠B as well as ∠D.

Solution 14

(i) ABCD is a rectangle in which
diagonal AC bisects ∠A as well as ∠C.

∠BAC = ∠DAC ….(i)

And
∠BCA = ∠DCA ….(ii)

Since
every rectangle is a parallelogram, therefore

AB
∥ DC and AC is
the transversal.

∠BAC = ∠DCA (alternate angles)

∠DAC = ∠DCA [From (i)]

Thus,
in

= CD (opposite sides of equal angles
are equal)

But,
AD = BC and CD = AB (ABCD is a
rectangle)

AB = BC = CD = AD

Hence,
ABCD is a square.

AB = CD

BD = BD

ΔBAD ≅ ΔBCD (by SSS congruence criterion)

∠ABD = ∠CBD and ∠ADB = ∠CDB (c.p.c.t.)

Hence, diagonal BD bisects ∠B as well as ∠D.

Question 15

In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE= AB. Prove that ED bisects BC.

Solution 15

Question 16

In the adjoining figure , ABCD is a parallelogram and E is the midpoint of side BC. If DE and AB when produced meet at F, prove that AF=2AB.

Solution 16

Question 17

Two parallel lines l and m are
intersected by a transversal t. Show that the quadrilateral formed by the
bisectors of interior angles is a rectangle.

Solution 17

l ∥ m and t is a
transversal.

∠APR = ∠PRD (alternate angles)

∠SPR = ∠PRQ (PS and RQ are the bisectors of ∠APR and ∠PRD)

Thus,
PR intersects PS and RQ at P and R respectively such that
∠SPR = ∠PRQ i.e.,
alternate angles are equal.

PS ∥ RQ

Similarly,
we have SR
∥ PQ.

Hence,
PQRS is a parallelogram.

Now,
∠BPR + ∠PRD = 180° (interior angles are supplementary)

2∠QPR + 2∠QRP = 180° (PQ and RQ are the bisectors of ∠BPR and ∠PRD)

∠QPR + ∠QRP = 90°

In
ΔPQR, by angle
sum property,

PQR + ∠QPR + ∠QRP = 180°

∠PQR + 90° = 180°

∠PQR = 90°

Since
PQRS is a parallelogram,

PQR = ∠PSR

∠PSR = 90°

Now,
∠SPQ + ∠PQR = 180° (adjacent angles in a parallelogram are
supplementary)

∠SPQ + 90° = 180°

∠SPQ = 90°

∠SRQ = 90°

Thus,
all the interior angles of quadrilateral PQRS are right angles.

Hence,
PQRS is a rectangle.

Question 18

K, L, M and N are point on the
sides AB, BC, CD and DA respectively of a square ABCD such that AK = BL = CM
= DN. Prove that KLMN is a square.

Solution 18

AK
= BL = CM = DN (given)

BK = CL = DM =
AN (i)(since
ABCD is a square)

In
ΔAKN and ΔBLK,

AK
= BL (given)

A = ∠B (Each 90°)

AN
= BK [From (i)]

ΔAKN ≅ ΔBLK (by SAS
congruence criterion)

∠AKN = ∠BLK and ∠ANK = ∠BKL (c.p.c.t.)

But,
∠AKN + ∠ANK = 90° and ∠BLK + ∠BKL = 90°

∠AKN + ∠ANK + ∠BLK + ∠BKL = 90° + 90°

2∠AKN + 2∠BKL = 180°

∠AKN + ∠BKL = 90°

∠NKL = 90°

Similarly,
we have

KLM = ∠LMN = ∠MNK = 90°

Hence,
KLMN is a square.

Question 19

A is given . if lines are drawn through A, B, C parallel respectively to the sides BC, CA and AB forming , as shown in the adjoining figure, show that

Solution 19

Question 20

In the adjoining figure, is a triangle and through A,B, C lines are drawn , parallel respectively to BC, CA and AB, intersecting at P, Q and R. prove that the perimeter of  is double the perimeter of  .

Solution 20

Question 21

In the adjoining figure, M is the
midpoint of side BC of parallelogram ABCD such that
∠BAM
= ∠DAM.

Solution 21

ABCD
is a parallelogram.

Hence,
|| BC.

∠DAM = ∠AMB (alternate angles)

∠BAM = ∠AMB (since ∠BAM = ∠DAM)

BM = AB (sides opposite to equal angles
are equal)

But,
AB = CD (opposite sides of a
parallelogram)

BM = AB =
CD ….(i)

Question 22

In a adjoining figure, ABCD is a parallelogram in which =60o. If the parallelogram in which and meet DC at P, prove that (i) PB=90o, (ii) AD=DP and PB=PC=BC, (iii) DC=2AD.

Solution 22

Question 23

In the adjoining figure, ABCD is a parallelogram in which

Calculate

Solution 23

Question 24

In a ||gm ABCD , if , find the value of x and the measure of each angle of the parallelogram.

Solution 24

Question 25

If an angle of a parallelogram is four fifths of its adjacent angle, find the angles of the parallelogram .

Solution 25

Question 26

Find the measure of each angle of parallelogram , if one of its angles is  less than twice the smallest angle.

Solution 26

Question 27

ABCD is a parallelogram in which AB=9.5 cm and its
parameter is 30 cm. Find the length of each side of the parallelogram.

Solution 27

## Chapter 10 – Quadrilaterals Exercise Ex. 10C

Question 1

P, Q, R and S are respectively the
midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD. Show that

(i) PQ ∥ AC and PQ =

(ii) PQ ∥ SR

(iii) PQRS is a parallelogram.

Solution 1

(i) In ΔABC, P and Q
are the mid-points of sides AB and BC respectively.

(ii) In ΔADC, R and S
are the mid-points of sides CD and AD respectively.

From
(i) and (ii), we have

PQ
= SR and PQ
∥ SR

(iii) Thus, in quadrilateral PQRS, one
pair of opposite sides are equal and parallel.

Hence, PQRS is a parallelogram.

Question 2

Prove that the line segments joining the middle points of the sides of a triangle divide it into four congruent triangles.

Solution 2

Question 3

In the adjoining figure, D,E,F are the midpoints of the sides BC, CA, and AB respectively, of . Show that .

Solution 3

Question 4

Show that the quadrilateral formed by joining the mid points of the pairs of adjacent sides of a rectangle is a rhombus.

Solution 4

Question 5

Show that the quadrilateral formed by joining the mid points of the pairs of adjacent sides of a rhombus is a rectangle.

Solution 5

Question 6

Show that the quadrilateral formed by joining the mid points of the pairs of adjacent sides of a square is a square.

Solution 6

Question 7

Prove
that the line segment joining the midpoints of opposite sides of a

Solution 7

Question 8

ABCD are equal. Prove that the quadrilateral formed by joining the midpoints
of its sides is a rhombus.

Solution 8

In
ΔABC, P and Q
are the mid-points of sides AB and BC respectively.

In
ΔBCD, Q and R
are the mid-points of sides BC and CD respectively.

In
are the mid-points of sides AD and CD respectively.

In
ΔABD, P and S
are the mid-points of sides AB and AD respectively.

PQ || RS and QR || SP [From (i), (ii), (iii) and (iv)]

Thus,
PQRS is a parallelogram.

Now,
AC = BD (given)

PQ = QR = RS =
SP  [From (i), (ii), (iii) and (iv)]

Hence,
PQRS is a rhombus.

Question 9

ABCD are perpendicular to each other. Prove that the quadrilateral formed by
joining the midpoints of its sides is a rectangle.

Solution 9

In
ΔABC, P and Q
are the mid-points of sides AB and BC respectively.

In
are the mid-points of sides CD and AD respectively.

From
(i) and (ii),

PQ
|| RS and PQ = RS

Thus,
in quadrilateral PQRS, a pair of opposite sides are equal are parallel.

So,
PQRS is a parallelogram.

Let
the diagonals AC and BD intersect at O.

Now,
in
ΔABD, P and S
are the mid-points of sides AB and AD respectively.

Thus,
|| NO and PN || MO.

PMON is a
parallelogram.

∠MPN = ∠MON (opposite angles of a parallelogram are
equal)

∠MPN = ∠BOA (since ∠BOA = ∠MON)

∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)

∠QPS = 90°

Thus,
PQRS is a parallelogram whose one angle, i.e.
∠QPS = 90°.

Hence,
PQRS is a rectangle.

Question 10

The midpoints of the sides AB, BC,
CD and DA of a quadrilateral ABCD are joined to form a quadrilateral. If AC =
BD and AC
⊥ BD then prove that the quadrilateral formed is a square.

Solution 10

In
ΔABC, P and Q
are the mid-points of sides AB and BC respectively.

In
ΔBCD, Q and R
are the mid-points of sides BC and CD respectively.

In
are the mid-points of sides AD and CD respectively.

In
ΔABD, P and S
are the mid-points of sides AB and AD respectively.

PQ || RS and QR || SP [From (i), (ii), (iii) and (iv)]

Thus,
PQRS is a parallelogram.

Now,
AC = BD (given)

PQ = QR = RS =
SP [From (i), (ii), (iii) and
(iv)]

Let
the diagonals AC and BD intersect at O.

Now,

Thus,
|| NO and PN || MO.

PMON is a
parallelogram.

∠MPN = ∠MON (opposite angles of a parallelogram are
equal)

∠MPN = ∠BOA (since ∠BOA = ∠MON)

∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)

∠QPS = 90°

Thus,
PQRS is a parallelogram such that PQ = QR = RS = SP and
∠QPS = 90°.

Hence,
PQRS is a square.

Question 11

A square is inscribed in an
isosceles right triangle so that the square and the triangle have one angle
common. Show that the vertex of the square opposite the vertex of the common
angle bisect the hypotenuse.

Solution 11

Let
ΔABC be an
isosceles right triangle, right-angled at B.

AB = BC

Let
PBSR be a square inscribed in
ΔABC with common ∠B.

PB = BS = SR =
RP

Now,
AB – PB = BC – BS

AP = CS ….(i)

In
ΔAPR and ΔCSR

AP
= CS  [From (i)

APR = ∠CSR (Each 90°)

PR
= SR (sides of a square)

ΔAPR ≅ ΔCSR (by SAS congruence criterion)

AR = CR (c.p.c.t.)

Thus,
point R bisects the hypotenuse AC.

Question 12

In the adjoining figure , ABCD is a ||gm in which E and F are the midpoints of AB and CD respectively . If GH is a line segment that cuts AD, EF, and BC at G, P and H respectively, prove that GP = PH.

Solution 12

Question 13

M and N are points on opposites
sides AD and BC of a parallelogram ABCD such that MN passes through the point
of intersection O of its diagonals AC and BD. show that MN is bisected at O.

Solution 13

In
ΔAOM and ΔCON

MAO = ∠OCN  (Alternate angles)

AO
= OC (Diagonals of a
parallelogram bisect each other)

AOM = ∠CON  (Vertically
opposite angles)

ΔAOM ≅ ΔCON  (by ASA
congruence criterion)

MO = NO (c.p.c.t.)

Thus,
MN is bisected at point O.

Question 14

In the adjoining figure, PQRS is a
trapezium in which PQ
∥ SR and M is the midpoint of PS. A line segment MN ∥
PQ meets QR at N. Show that N is the midpoint of QR.

Solution 14

Construction:
Join diagonal QS. Let QS intersect MN at point O.

PQ
SR and MN ∥ PQ

PQ ∥ MN ∥ SR

By
converse of mid-point theorem a line drawn, through the mid-point of any side
of a triangle and parallel to another side bisects the third side.

Now, in ΔSPQ

MO ∥ PQ and M is
the mid-point of SP

So, this line will
intersect QS at point O and O will be the mid-point of QS.

Also, MN ∥ SR

Thus, in ΔQRS, ON ∥ SR and O is the
midpoint of line QS.

So, by using converse of
mid-point theorem, N is the mid-point of QR.

Question 15

In a parallelogram PQRS, PQ = 12
cm and PS = 9 cm. The bisector of
∠P meets SR in M. PM and QR both when produced meet at T.
Find the length of RT.

Solution 15

PM
is the bisector of
∠P.

∠QPM = ∠SPM ….(i)

PQRS
is a parallelogram.

PQ ∥ SR and PM is the transversal.

∠QPM = ∠MS (ii)(alternate angles)

From
(i) and (ii),

SPM = ∠PMS ….(iii)

MS = PS = 9
cm (sides opposite to equal angles
are equal)

Now,
∠RMT = ∠PMS (iv)(vertically opposite angles)

Also,
PS
∥ QT and PT is
the transversal.

RTM = ∠SPM

∠RTM = ∠RMT

RT = RM (sides opposite to equal angles are
equal)

RM
= SR – MS = 12 – 9 = 3 cm

RT = 3 cm

Question 16

In the adjoining figure , ABCD is a trapezium in which AB|| DC and P, Q are the mid points of AD and BC respectively. DQ and AB when produced meet at E. Also, AC and PQ intersect at R. Prove that (i) DQ=QE, (ii) PR||AB, (iii) AR=RC.

Solution 16

Question 17

In the adjoining figure, AD is a medium of  and DE|| BA. Show that BE is also a median of .

Solution 17

Question 18

In the adjoining figure , AD and BE are the medians of  and DF|| BE. Show that .

Solution 18

## Chapter 10 – Quadrilaterals Exercise MCQ

Question 1

Three angles of quadrilateral are 80°,
95°
and 112°.
Its fourth angle is

(a) 78°

(b) 73°

(c) 85°

(d) 100°

Solution 1

Question 2

If ABCD is a parallelogram with two adjacent angles ∠A
= ∠B,
then the parallelogram is a

1. rhombus
2. trapezium
3. rectangle
4. none of these

Solution 2

Question 3

In a quadrilateral ABCD, if AO and BO are the bisectors of ∠A
and ∠B
respectively, ∠C
= 70°
and ∠D
= 30°.
Then, ∠AOB
=?

1. 40°
2. 50°
3. 80°
4. 100°
Solution 3

Question 4

The bisectors of any adjacent angles of a parallelogram
intersect at

1. 30°
2. 45°
3. 60°
4. 90°

Solution 4

Question 5

The bisectors of the angles of a parallelogram enclose a

1. rhombus
2. square
3. rectangle
4. parallelogram
Solution 5

Question 6

If bisectors of ∠A and ∠B of a quadrilateral ABCD intersect each other at P, of ∠B
and ∠C
at Q, of ∠C and ∠D at R and of ∠D and ∠A at S then PQRS is a

(a) rectangle

(b) parallelogram

(c) rhombus

whose opposite angles are supplementary

Solution 6

Correct
option: (d)

In
ΔAPB, by angle
sum property,

APB + ∠PAB + ∠PBA = 180°

In
ΔCRD, by angle
sum property,

CRD + ∠RDC + ∠RCD = 180°

Now,
∠SPQ + ∠SRQ = ∠APB + ∠CRD

= 360° – 180°

= 180°

Now,
∠PSR + ∠PQR = 360° – (∠SPQ + ∠SRQ)

= 360° – 180°

= 180°

Hence, PQRS is a quadrilateral whose opposite angles are
supplementary.

Question 7

The figure formed by joining the

1. rhombus
2. square
3. rectangle
4. parallelogram
Solution 7

Question 8

The figure formed by joining the mid-points of the adjacent
sides of a square is a

1. rhombus
2. square
3. rectangle
4. parallelogram
Solution 8

Question 9

The figure formed by joining the mid-points of the adjacent
sides of a parallelogram is a

1. rhombus
2. square
3. rectangle
4. parallelogram
Solution 9

Question 10

The figure formed by joining the mid-points of the adjacent
sides of a rectangle is a

1. rhombus
2. square
3. rectangle
4. parallelogram
Solution 10

Question 11

The figure formed by joining the mid-points of the adjacent
sides of a rhombus is a

1. rhombus
2. square
3. rectangle
4. parallelogram
Solution 11

Question 12

The angles of a quadrilateral are in the ratio 3:4:5:6. The
smallest of these angles is

(a) 45°

(b) 60°

(c) 36°

(d) 48°

Solution 12

Question 13

The quadrilateral formed by joining the midpoints of the
sides of a quadrilateral ABCD, taken in order, is a rectangle, if

(a) ABCD
is a parallelogram

(b) ABCD
is a rectangle

(c) diagonals
of ABCD are equal

(d) diagonals
of ABCD are perpendicular to each other

Solution 13

Correct
option: (d)

In
ΔABC, P and Q
are the mid-points of sides AB and BC respectively.

In
are the mid-points of sides CD and AD respectively.

From
(i) and (ii),

PQ
∥ RS and PQ = RS

Thus,
in quadrilateral PQRS, a pair of opposite sides are equal are parallel.

So,
PQRS is a parallelogram.

Let
the diagonals AC and BD intersect at O.

Now,
in
ΔABD, P and S
are the mid-points of sides AB and AD respectively.

Thus,
∥ NO and PN ∥ MO.

PMON is a
parallelogram.

∠MPN = ∠MON (opposite angles of a parallelogram are
equal)

∠MPN = ∠BOA (since ∠BOA = ∠MON)

∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)

∠QPS = 90°

Thus,
PQRS is a parallelogram whose one angle, i.e.
∠QPS = 90°.

Hence,
PQRS is a rectangle if AC
⊥ BD.

Question 14

The quadrilateral formed by joining the midpoints 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) diagonals
of ABCD are equal

(d) diagonals
of ABCD are perpendicular to each other

Solution 14

Correct
option: (c)

In
ΔABC, P and Q
are the mid-points of sides AB and BC respectively.

In
ΔBCD, Q and R
are the mid-points of sides BC and CD respectively.

In
are the mid-points of sides AD and CD respectively.

In
ΔABD, P and S
are the mid-points of sides AB and AD respectively.

PQ ∥ RS and QR ∥ SP [From (i), (ii),
(iii) and (iv)]

Thus,
PQRS is a parallelogram.

Now,
AC = BD (given)

PQ = QR = RS =
SP [From (i), (ii), (iii) and
(iv)]

Hence,
PQRS is a rhombus if diagonals of ABCD are equal.

Question 15

The figure formed by joining the midpoints of the sides of
a quadrilateral ABCD, taken in order, is a square, only if

(a) ABCD
is a rhombus

(b) diagonals
of ABCD are equal

(c) diagonals
of ABCD are perpendicular

(d) diagonals
of ABCD are equal and perpendicular

Solution 15

Correct
option: (d)

In
ΔABC, P and Q
are the mid-points of sides AB and BC respectively.

In
ΔBCD, Q and R
are the mid-points of sides BC and CD respectively.

In
are the mid-points of sides AD and CD respectively.

In
ΔABD, P and S
are the mid-points of sides AB and AD respectively.

PQ || RS and QR || SP [From (i), (ii), (iii) and (iv)]

Thus,
PQRS is a parallelogram.

Now,
AC = BD (given)

PQ = QR = RS =
SP [From (i), (ii), (iii) and
(iv)]

Let
the diagonals AC and BD intersect at O.

Now,

Thus,
|| NO and PN || MO.

PMON is a
parallelogram.

∠MPN = ∠MON (opposite angles of a parallelogram are
equal)

∠MPN = ∠BOA (since ∠BOA = ∠MON)

∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)

∠QPS = 90°

Thus,
PQRS is a parallelogram such that PQ = QR = RS = SP and
∠QPS = 90°.

Hence,
PQRS is a square if
diagonals
of ABCD are equal and perpendicular.

Question 16

If an angle of a parallelogram is
two-third of its adjacent angle, the smallest angle of the parallelogram is

1. 108°
2. 54°
3. 72°
4. 81°
Solution 16

Question 17

If one angle of a parallelogram is 24°
less than twice the smallest angle, then the largest angles of the
parallelogram is

1. 68°
2. 102°
3. 112°
4. 136°

Solution 17

Question 18

If ∠A, ∠B,
∠C
and ∠D
of a quadrilateral ABCD taken in
order, are in the ratio 3:7:6:4, then ABCD is a

1. rhombus
2. kite
3. trapezium
4. parallelogram
Solution 18

Question 19

Which of the following is not true for a parallelogram?

1. Opposite sides are
equal.
2. Opposite angles are
equal.
3. Opposite angles are
bisected by the diagonals.
4. Diagonals bisect each
other.
Solution 19

Question 20

If APB and CQD are two parallel lines, then the bisectors
of
∠APQ,
∠BPQ,
∠CQP
and ∠PQD
enclose a

1. square
2. rhombus
3. rectangle
4. kite

Solution 20

Question 21

In the given figure, ABCD is a parallelogram in which ∠BDC
= 45°
= 75°.
Then, ∠CBD
=?

1. 45°
2. 55°
3. 60°
4. 75°

Solution 21

Question 22

If area of a ‖gm
with side ɑ
and b is A and that of a rectangle with side ɑ and b is B, then

(a) A
> B

(b) A = B

(c) A
< B

(d) A ≥ B

Solution 22

Question 23

In the given figure, ABCD is a parallelogram in which ∠BAD
= 75°
and ∠CBD
= 60°.
Then, ∠BDC
=?

(a) 60°

(b) 75°

(c) 45°

(d) 50°

Solution 23

Question 24

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

Solution 24

Question 25

P is any point on the side BC of a ΔABC.
P is joined to A. If D and E are the midpoints of the sides AB and AC
respectively and M and N are the midpoints of BP and CP respectively then

(a) a
trapezium

(b) a
parallelogram

(c) a
rectangle

(d) a
rhombus

Solution 25

Correct
option: (b)

In
ΔABC, D and E
are the mid-points of sides AB and AC respectively.

Hence,
DENM is a parallelogram.

Question 26

The parallel sides of a trapezium are ɑ
and b respectively. The line joining the mid-points of its non-parallel sides
will be

Solution 26

Question 27

In a trapezium
ABCD, if E and F be the mid-points of the diagonals AC and BD respectively.
Then, EF =?

Solution 27

Question 28

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=?

1. 45°
2. 60°
3. 90°
4. 30°

Solution 28

Question 29

In the given figures, ABCD is a rhombus. Then

(a) AC2
+ BD2 = AB2

(b) AC2
+ BD2 = 2AB2

(c) AC2
+ BD2 = 4AB2

(d) 2(AC2
+ BD2)=3AB2

Solution 29

Question 30

In a trapezium ABCD, if AB
CD, then (AC2 + BD2) =?

(a) BC2

(b) AB2
+CD2 + 2AB.CD

(c) AB2

(d) BC2

Solution 30

Question 31

Two parallelogram stand on equal bases and between the same
parallels. The ratio of their area is

1. 1:2
2. 2:1
3. 1:3
4. 1:1
Solution 31

Question 32

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 =?

Solution 32

Question 33

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
=?

1. 40°
2. 35°
3. 45°
4. 50°

Solution 33

Question 34

ABCD is a rhombus such that ∠ACB
= 50°.
= ?

(a) 40°

(b) 25°

(c) 65°

(d) 130°

Solution 34

Correct
option: (a)

ABCD
is a rhombus.

AD ∥ BC and AC is
the transversal.

∠DAC = ∠ACB  (alternate angles)

∠DAC = 50°

In
ΔAOD, by angle
sum property,

AOD + ∠DAO + ∠ADO = 180°

90° + ∠50° + ∠ADO = 180°

Question 35

Three statement are given below:

1. In a ‖gm,
the angle bisectors of two adjacent angles enclose a right angle.
2. The angle bisectors
of a ‖gm form a rectangle.
3. The triangle formed
by joining the mid-point of the sides of an isosceles triangle is not
necessarily an isosceles.

Which is true?

1. I only
2. II only
3. I and II
4. II and III
Solution 35

Question 36

Three statements are given below:

I. In a rectangle ABCD, the
diagonal AC bisects ∠A
as well as ∠C.

II. In a square ABCD, the diagonal
AC bisects ∠A
as well as ∠C

III. In a rhombus ABCD, the diagonal
AC bisects ∠A
as well as ∠C

Which is true?

1. I only
2. II and III
3. I and III
4. I and II
Solution 36

Question 37

In a quadrilateral PQRS, opposite angles are equal. If SR
= 2 cm and PR = 5 cm then determine PQ.

Solution 37

Since
opposite angles of quadrilateral are equal, PQRS is a parallelogram.

PQ = SR (opposite sides of parallelogram are
equal)

PQ = 2 cm

Question 38

Diagonals of a parallelogram are perpendicular to each

Solution 38

The
given statement is false.

Diagonals
of a parallelogram bisect each other.

Question 39

What special name can be given to a quadrilateral PQRS if ∠P
+ ∠S
= 118°?

Solution 39

In
angles.

Since
≠ 180°, PQRS is not a
parallelogram.

Hence,
PQRS is a trapezium.

Question 40

All the angles of a quadrilateral can be acute. Is this

Solution 40

The
given statement is false.

We
know that the sum of all the four angles of a quadrilateral is 360
°.

If
all the angles of a quadrilateral are acute, the sum will be less than 360
°.

Question 41

All the angles of a quadrilateral can be right angles. Is

Solution 41

The
given statement is true.

We
know that the sum of all the four angles of a quadrilateral is 360
°.

If
all the angles of a quadrilateral are right angles,

Sum
of all angles of a quadrilateral = 4
× 90° = 360°

Question 42

All the angles of a quadrilateral can be obtuse. Is this

Solution 42

The
given statement is false.

We
know that the sum of all the four angles of a quadrilateral is 360
°.

If
all the angles of a quadrilateral are obtuse, the sum will be more than 360
°.

Question 43

Can we form a quadrilateral whose angles are 70°,
115°,
60°
and 120°?

Solution 43

We
know that the sum of all the four angles of a quadrilateral is 360
°.

Here,

70° + 115° + 60° + 120° = 365° ≠ 360°

Hence,
we cannot form a quadrilateral with given angles.

Question 44

What special name can be given to a quadrilateral whose
all angles are equal?

Solution 44

are equal is a rectangle.

Question 45

In which of the following figures are the diagonals equal?

1. Parallelogram
2. Rhombus
3. Trapezium
4. Rectangle

Solution 45

Question 46

If D and E are respectively the midpoints of the sides AB
and BC of
ΔABC in which AB = 7.2 cm, BC = 9.8 cm and AC = 3.6 cm then
determine the length of DE.

Solution 46

D and E are respectively the
midpoints of the sides AB and BC of
ΔABC.

Thus, by mid-point theorem, we
have

Question 47

In a quadrilateral PQRS, the diagonals PR and QS bisect
each other. If
∠Q = 56°, determine ∠R.

Solution 47

Since the diagonals PR and QS of
quadrilateral PQRS bisect each, PQRS is a parallelogram.

parallelogram are supplementary.

∠Q
+ ∠R
= 180°

56° + ∠R = 180°

∠R
= 124°

Question 48

In the adjoining figure, BDEF and AFDE are parallelograms.
Is AF = FB? Why or why not?

Solution 48

AFDE
is a parallelogram

AF = ED …(i)

BDEF
is a parallelogram.

FB = ED …(ii)

From
(i) and (ii),

AF
= FB

Question 49

In each of the questions are question is followed by two
statements I and II. The answer is

1. if the question can
be answered by one of the given statements alone and not by the other;
2. if the question can
be answered by either statement alone;
3. if the question can
be answered by both the statements together but not by any one of the
two;
4. if
the question cannot be answered by using both the statement together.

Is
‖gm?

1. Diagonal AC and BD bisect
each other.
2. Diagonal AC and BD
are equal.

The
correct answer is : (a)/ (b)/ (c)/ (d).

Solution 49

Correct
option: (a)

If the
diagonals of a quad. ABCD bisect each other, then the quad. ABCD is a
parallelogram.

If the
diagonals are equal, then the quad. ABCD is a parallelogram.

So, II

Question 50

In each of the questions are question is followed by two
statements I and II. The answer is

1. if the question can
be answered by one of the given statements alone and not by the other;
2. if the question can
be answered by either statement alone;
3. if the question can
be answered by both the statements together but not by any one of the
two;
4. if the question
cannot be answered by using both the statement together

Is

1. Quad. ABCD is a ‖gm.
2. Diagonals AC and BD
are perpendicular to each other.

The
correct answer is: (a) / (b)/ (c)/ (d).

Solution 50

Correct
option: (c)

ABCD is a
‖gm, it could be a rectangle or square
or rhombus.

So,
statement I is not sufficient to answer the
question.

If the
diagonals AC and BD are perpendicular to each other, then the
‖gm
could be a square or rhombus.

So,
statement II is not sufficient to answer the question.

However,
if the statements are combined, then the quad. ABCD is a rhombus.

Question 51

In each of the questions are question is followed by two
statements I and II. The answer is

1. if the question can
be answered by one of the given statements alone and not by the other;
2. if the question can
be answered by either statement alone;
3. if the question can
be answered by both the statements together but not by any one of the
two;
4. if the question
cannot be answered by using both the statement together

Is ‖gm
ABCD a square?

1. Diagonals of ‖gm
ABCD are equal.
2. Diagonals of ‖gm
ABCD intersect at right angles.

The correct
answer is: (a)/ (b)/ (c)/ (d).

Solution 51

Correct
option: (c)

If the
diagonals of a
‖gm ABCD are equal,
then ‖gm
ABCD could either be a rectangle or a square.

If the
diagonals of the
‖gm ABCD intersect at
right angles, then the ‖gm ABCD could be a square or a
rhombus.

However,
if both the statements are combined, then
‖gm
ABCD will be a square.

Question 52

In each of the questions are question is followed by two
statements I and II. The answer is

1. if the question can
be answered by one of the given statements alone and not by the other;
2. if the question can
be answered by either statement alone;
3. if the question can
be answered by both the statements together but not by any one of the
two;
4. if the question
cannot be answered by using both the statement together

1. Its opposite sides
are equal.
2. Its opposite angles
are equal.

The correct
answer is: (a)/ (b)/ (c)/ (d)

Solution 52

Correct
option: (b)

If the
opposite sides of a quad. ABCD are equal, the quadrilateral is a
parallelogram.

If the
opposite angles are equal, then the quad. ABCD is a parallelogram.

Question 53

Each question
consists of two statements, namely, Assertion (A) and Reason (R). For
selecting the correct answer, use the following code:

1. Both Assertion (A)
and Reason (R) are true and Reason (R) is a correct explanation of
Assertion (A).
2. Both Assertion (A)
and Reason (R) are true but Reason (R) is not a correct explanation of
Assertion (A)
3. Assertion (A) is true
and Reason (R) is false.
4. Assertion (A) is
false and Reason (R) is true.

 Assertion (A) Reason (R) If three angles of a quadrilateral are 130°, 70° and 60°, then the fourth angle is 100°. The sum of all the angle of a quadrilateral is 360°.

The correct
answer is: (a)/ (b)/ (c)/ (d).

Solution 53

Question 54

Each question consists of two statements, namely, Assertion
(A) and Reason (R). for selecting the correct answer, use the following code:

1. Both Assertion (A)
and Reason (R) are true and Reason (R) is a correct explanation of
Assertion (A).
2. Both Assertion (A)
and Reason (R) are true but Reason (R) is not a correct explanation of
Assertion (A)
3. Assertion (A) is true
and Reason (R) is false.
4. Assertion (A) is
false and Reason (R) is true.

 Assertion (A) Reason (R) ABCD is a quadrilateral in which P, Q, R and S are the mid-points of AB, BC, CD and DA respectively.Then, PQRS is a parallelogram. The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

The correct
answer is: (a)/ (b)/ (c)/ (d).

Solution 54

The
Reason (R) is true and is the correct explanation for the Assertion (A).

Question 55

Each question consists of two statements, namely, Assertion
(A) and Reason (R). for selecting the correct answer, use the following code:

1. Both Assertion (A)
and Reason (R) are true and Reason (R) is a correct explanation of
Assertion (A).
2. Both Assertion (A)
and Reason (R) are true but Reason (R) is not a correct explanation of
Assertion (A)
3. Assertion (A) is true
and Reason (R) is false.
4. Assertion (A) is
false and Reason (R) is true.

 Assertion (A) Reason (R) In a rhombus ABCD, the diagonal AC bisects ∠A as well as ∠C. The diagonals of a rhombus bisect each other at right angles.

The correct
answer is: (a)/ (b)/ (c)/ (d).

Solution 55

Question 56

If the diagonals of a quadrilateral bisect each other at
right angles, then the figure is a

a. trapezium

b. parallelogram

c. rectangle

d. rhombus

Solution 56

Question 57

Each question consists of two statements, namely, Assertion
(A) and Reason (R). for selecting the correct answer, use the following code:

1. Both Assertion (A)
and Reason (R) are true and Reason (R) is a correct explanation of
Assertion (A).
2. Both Assertion (A)
and Reason (R) are true but Reason (R) is not a correct explanation of
Assertion (A)
3. Assertion (A) is true
and Reason (R) is false.
4. Assertion (A) is
false and Reason (R) is true.

 Assertion (A) Reason (R) Every parallelogram is a rectangle. The angle bisectors of a parallelogram form a rectangle.

The correct
answer is: (a)/ (b)/ (c)/ (d).

Solution 57

Question 58

Each question consists of two statement,
namely, Assertion (A) and Reason (R). for selecting the correct answer, use
the following code:

1. Both Assertion (A)
and Reason (R) are true and Reason (R) is a correct explanation of
Assertion (A).
2. Both Assertion (A)
and Reason (R) are true but Reason (R) is not a correct explanation of
Assertion (A)
3. Assertion (A) is true
and Reason (R) is false.
4. Assertion (A) is
false and Reason (R) is true.

 Assertion (A) Reason (R) The diagonals of a ‖gm bisect each other. If the diagonals of a ‖gm are equal and intersect at right angles, then the parallelogram is a square.

The correct
answer is: (a)/ (b)/ (c)/ (d).

Solution 58

Question 59

 Column I Column II (a) Angle bisectors of a parallelogram form a (p) parallelogram (b) The quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a square is a (q) rectangle (c) The quadrilateral formed by joining the mid-points of the pairs of adjacent side of a rectangle is a (r) square (d) The figure formed by joining the mid-points of the pairs of adjacent sides of a quadrilateral is (s) rhombus

The correct

(a) -…….,

(b) -…….,

(c) -…….,

(d)-…….

Solution 59

Question 60

 Column I Column II (a) In the given figure, ABCD is a trapezium in which AB = 10 cm and CD = 7cm. If P and Q are the mid-points of AD and BC respectively, then PQ = (p) equal (b) In the given figure, PQRS is a ‖gm whose diagonal intersect at O. If PR = 13 cm, then OR= (q) at right angle (c) The diagonals of a square are (r) 8.5 cm (d) The diagonals of a rhombus bisect each other (s) 6.5 cm

The correct

(a) -…….,

(b) -…….,

(c) -…….,

(d)-…….

Solution 60

Question 61

The lengths of the diagonals of a rhombus are 16 cm and 12
cm. The length of each side of the rhombus is

1. 10 cm
2. 12 cm
3. 9cm
4. 8cm
Solution 61

Question 62

The length of each side of a rhombus is 10 cm and one of
its diagonals is of length 16 cm. The length of the other diagonal is

Solution 62

Question 63

A diagonal of a rectangle is inclined to one side of the
rectangle at 35
°. The acute angle between the diagonals is

(a) 55°

(b) 70°

(c) 45°

(d) 50°

Solution 63

Correct
option: (b)

DAO + ∠OAB = ∠DAB

∠DAO + 35° = 90°

∠DAO = 55°

ABCD
is a rectangle and diagonals of a rectangle are equal and bisect each other.

OA = OD

∠ODA = ∠DAO (angles opposte
to equal sides are equal)

∠ODA = 55°

In DODA, by angle
sum property,

ODA + ∠DAO + ∠AOD = 180°

55° + ∠55° + ∠AOD = 180°

∠AOD = 70°

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