R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 10 – Quadrilaterals
Chapter 10 – Quadrilaterals Exercise Ex. 10A
Three angles of a quadrilateral are 75°, 90° and 75°. Find the measure of the fourth angle.
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°.
Prove
that the sum of all the angles of a quadrilateral is 360^{0.}
Given:
ABCD is a quadrilateral.
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The angle of the quadrilateral are in the ratio 2:4:5:7. Find the angles.
In the adjoining figure , ABCD is a trapezium in which AB || DC. If _{}=55^{0} and _{}= 70^{0}, find _{ and }.
Since AB || DC
In the adjoining figure , ABCD is a square and_{}is an equilateral triangle . Prove that
(i)AE=BE, (ii) _{}=15^{0}
Given:_{}
In the adjoining figure , BM_{}AC and DN_{}AC. If BM=DN, prove that AC bisects BD.
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In given figure, ABCD is a quadrilateral in which AB=AD and BC= DC. Prove that (i) AC bisects and , (ii) BE=DE,
(iii)
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In the given figure , ABCD is a square and _{P}QR=90^{0}. If PB=QC= DR, prove that (i) QB=RC, (ii) PQ=QR, (iii) QPR=45^{0}.
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If O is a point with in a quadrilateral ABCD, show that OA+OB+OC+OD> AC+BD.
Given: O is a point within a quadrilateral ABCD
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
Given: ABCD is a quadrilateral and AC is one of its disgonals.
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Chapter 10 – Quadrilaterals Exercise Ex. 10B
In the adjoining figure, ABCD is a parallelogram in which _{}=72^{0}. Calculate _{,and }.
_{In each of the figures given below, ABCD is a rhombus. Find the value of x and y in each case.}
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_{The lengths of the diagonals of a rhombus are 24 cm and 18 cm respectively . Find the length of each side of the rhombus.}
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_{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.}
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_{In each of the figures given below, ABCD is a rectangle . Find the values of x and y in each case.}
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In a rhombus ABCD, the altitude
from D to the side AB bisect AB. Find the angle of the rhombus
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)
⇒ AD = BD (c.p.c.t.)
But,
AD = AB (sides of a rhombus)
⇒ AD = AB = BD
⇒ ΔADB is an
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°.
_{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.}
*Back answer incorrect
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In a rhombus ABCD show that
diagonal AC bisect ∠A as well as ∠C and diagonal BD bisect ∠B
as well as ∠D.
In
ΔABC and ΔADC,
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
ΔBAD and ΔBCD,
AB
= BC (sides of a rhombus are equal)
AD
= 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.
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.
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.
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.
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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.
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In the adjoining figure , ABCD is a parallelogram in which
and . Calculate _{. }
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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.
∠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°.
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.
(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 ΔADC,
AD
= 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.
(ii) In ΔBAD and ΔBCD,
AB = CD
AD = BC
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.
In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE= AB. Prove that ED bisects BC.
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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.
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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.
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.
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.
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.
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
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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 .
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In the adjoining figure, M is the
midpoint of side BC of parallelogram ABCD such that ∠BAM
= ∠DAM.
Prove that AD = 2CD.
ABCD
is a parallelogram.
Hence,
AD || 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)
In a adjoining figure, ABCD is a parallelogram in which _{}=60^{o}. If the parallelogram in which and meet DC at P, prove that (i) PB=90^{o}, (ii) AD=DP and PB=PC=BC, (iii) DC=2AD.
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_{In the adjoining figure, ABCD is a parallelogram in which}
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_{Calculate }
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_{In a ||gm ABCD , if , find the value of x and the measure of each angle of the parallelogram.}
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_{If an angle of a parallelogram is four fifths of its adjacent angle, find the angles of the parallelogram .}
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_{Find the measure of each angle of parallelogram , if one of its angles is less than twice the smallest angle.}
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_{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.}
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Chapter 10 – Quadrilaterals Exercise Ex. 10C
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.
(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.
_{Prove that the line segments joining the middle points of the sides of a triangle divide it into four congruent triangles.}
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_{In the adjoining figure, D,E,F are the midpoints of the sides BC, CA, and AB respectively, of . Show that .}
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Show that the quadrilateral formed by joining the mid points of the pairs of adjacent sides of a rectangle is a rhombus.
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Show that the quadrilateral formed by joining the mid points of the pairs of adjacent sides of a rhombus is a rectangle.
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Show that the quadrilateral formed by joining the mid points of the pairs of adjacent sides of a square is a square.
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Prove
that the line segment joining the midpoints of opposite sides of a
quadrilateral bisect each other.
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The diagonals of a quadrilateral
ABCD are equal. Prove that the quadrilateral formed by joining the midpoints
of its sides is a rhombus.
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
ΔADC, S and R
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.
The diagonals of a quadrilateral
ABCD are perpendicular to each other. Prove that the quadrilateral formed by
joining the midpoints of its sides is a rectangle.
In
ΔABC, P and Q
are the mid-points of sides AB and BC respectively.
In
ΔADC, R and S
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,
in quadrilateral PMON, PM || 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.
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.
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
ΔADC, S and R
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,
in quadrilateral PMON, PM || 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.
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.
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.
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.
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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.
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.
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.
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.
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.
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
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.
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In the adjoining figure, AD is a medium of _{and DE|| BA. Show that BE is also a median of .}
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_{In the adjoining figure , AD and BE are the medians of and DF|| BE. Show that .}
Chapter 10 – Quadrilaterals Exercise MCQ
Three angles of quadrilateral are 80^{°},
95°
and 112°.
Its fourth angle is
(a) 78°
(b) 73°
(c) 85°
(d) 100°
If ABCD is a parallelogram with two adjacent angles ∠A
= ∠B,
then the parallelogram is a
- rhombus
- trapezium
- rectangle
- none of these
In a quadrilateral ABCD, if AO and BO are the bisectors of ∠A
and ∠B
respectively, ∠C
= 70°
and ∠D
= 30°.
Then, ∠AOB
=?
- 40°
- 50°
- 80°
- 100°
The bisectors of any adjacent angles of a parallelogram
intersect at
- 30°
- 45°
- 60°
- 90°
The bisectors of the angles of a parallelogram enclose a
- rhombus
- square
- rectangle
- parallelogram
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
(d) quadrilateral
whose opposite angles are supplementary
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.
The figure formed by joining the
mid-points of the adjacent sides of a quadrilateral is a
- rhombus
- square
- rectangle
- parallelogram
The figure formed by joining the mid-points of the adjacent
sides of a square is a
- rhombus
- square
- rectangle
- parallelogram
The figure formed by joining the mid-points of the adjacent
sides of a parallelogram is a
- rhombus
- square
- rectangle
- parallelogram
The figure formed by joining the mid-points of the adjacent
sides of a rectangle is a
- rhombus
- square
- rectangle
- parallelogram
The figure formed by joining the mid-points of the adjacent
sides of a rhombus is a
- rhombus
- square
- rectangle
- parallelogram
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°
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
Correct
option: (d)
In
ΔABC, P and Q
are the mid-points of sides AB and BC respectively.
In
ΔADC, R and S
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,
in quadrilateral PMON, PM ∥ 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.
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
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
ΔADC, S and R
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.
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
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
ΔADC, S and R
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,
in quadrilateral PMON, PM || 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.
If an angle of a parallelogram is
two-third of its adjacent angle, the smallest angle of the parallelogram is
- 108°
- 54°
- 72°
- 81°
If one angle of a parallelogram is 24°
less than twice the smallest angle, then the largest angles of the
parallelogram is
- 68°
- 102°
- 112°
- 136°
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
- rhombus
- kite
- trapezium
- parallelogram
Which of the following is not true for a parallelogram?
- Opposite sides are
equal. - Opposite angles are
equal. - Opposite angles are
bisected by the diagonals. - Diagonals bisect each
other.
If APB and CQD are two parallel lines, then the bisectors
of ∠APQ,
∠BPQ,
∠CQP
and ∠PQD
enclose a
- square
- rhombus
- rectangle
- kite
In the given figure, ABCD is a parallelogram in which ∠BDC
= 45°
and ∠BAD
= 75°.
Then, ∠CBD
=?
- 45°
- 55°
- 60°
- 75°
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
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°
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,
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
quadrilateral DENM is
(a) a
trapezium
(b) a
parallelogram
(c) a
rectangle
(d) a
rhombus
Correct
option: (b)
In
ΔABC, D and E
are the mid-points of sides AB and AC respectively.
Hence,
DENM is a parallelogram.
The parallel sides of a trapezium are ɑ
and b respectively. The line joining the mid-points of its non-parallel sides
will be
In a trapezium
ABCD, if E and F be the mid-points of the diagonals AC and BD respectively.
Then, EF =?
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=?
- 45°
- 60°
- 90°
- 30°
In the given figures, ABCD is a rhombus. Then
(a) AC^{2}
+ BD^{2} = AB^{2}
(b) AC^{2}
+ BD^{2} = 2AB^{2}
(c) AC^{2}
+ BD^{2} = 4AB^{2}
(d) 2(AC^{2}
+ BD^{2})=3AB^{2}
In a trapezium ABCD, if AB ‖
CD, then (AC^{2} + BD^{2}) =?
(a) BC^{2}
+ AD^{2} + 2BC. AD
(b) AB^{2}
+CD^{2} + 2AB.CD
(c) AB^{2}
+ CD^{2} + 2AD. BC
(d) BC^{2}
+ AD^{2} + 2AB.CD
Two parallelogram stand on equal bases and between the same
parallels. The ratio of their area is
- 1:2
- 2:1
- 1:3
- 1:1
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 =?
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
=?
- 40°
- 35°
- 45°
- 50°
ABCD is a rhombus such that ∠ACB
= 50°.
Then, ∠ADB
= ?
(a) 40°
(b) 25°
(c) 65°
(d) 130°
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°
⇒ ∠ADO = 40°
⇒ ∠ADB = 40°
Three statement are given below:
- In a ‖gm,
the angle bisectors of two adjacent angles enclose a right angle. - The angle bisectors
of a ‖gm form a rectangle. - The triangle formed
by joining the mid-point of the sides of an isosceles triangle is not
necessarily an isosceles.
Which is true?
- I only
- II only
- I and II
- II and III
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?
- I only
- II and III
- I and III
- I and II
In a quadrilateral PQRS, opposite angles are equal. If SR
= 2 cm and PR = 5 cm then determine PQ.
Since
opposite angles of quadrilateral are equal, PQRS is a parallelogram.
⇒ PQ = SR (opposite sides of parallelogram are
equal)
⇒ PQ = 2 cm
Diagonals of a parallelogram are perpendicular to each
other. Is this statement true? Give reasons for your answer.
The
given statement is false.
Diagonals
of a parallelogram bisect each other.
What special name can be given to a quadrilateral PQRS if ∠P
+ ∠S
= 118°?
In
quadrilateral PQRS, ∠P and ∠S are adjacent
angles.
Since
the sum of adjacent angles ≠ 180°, PQRS is not a
parallelogram.
Hence,
PQRS is a trapezium.
All the angles of a quadrilateral can be acute. Is this
statement true? Give reasons for your answer.
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°.
All the angles of a quadrilateral can be right angles. Is
this statement true? Give reasons for your answer.
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°
All the angles of a quadrilateral can be obtuse. Is this
statement true? Give reasons for your answer.
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°.
Can we form a quadrilateral whose angles are 70°,
115°,
60°
and 120°?
Give reasons for your answer.
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.
What special name can be given to a quadrilateral whose
all angles are equal?
A quadrilateral whose all angles
are equal is a rectangle.
In which of the following figures are the diagonals equal?
- Parallelogram
- Rhombus
- Trapezium
- Rectangle
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.
D and E are respectively the
midpoints of the sides AB and BC of ΔABC.
Thus, by mid-point theorem, we
have
In a quadrilateral PQRS, the diagonals PR and QS bisect
each other. If ∠Q = 56°, determine ∠R.
Since the diagonals PR and QS of
quadrilateral PQRS bisect each, PQRS is a parallelogram.
Now, adjacent angles of
parallelogram are supplementary.
⇒ ∠Q
+ ∠R
= 180°
⇒
56° + ∠R = 180°
⇒ ∠R
= 124°
In the adjoining figure, BDEF and AFDE are parallelograms.
Is AF = FB? Why or why not?
AFDE
is a parallelogram
⇒ AF = ED …(i)
BDEF
is a parallelogram.
⇒ FB = ED …(ii)
From
(i) and (ii),
AF
= FB
In each of the questions are question is followed by two
statements I and II. The answer is
- if the question can
be answered by one of the given statements alone and not by the other; - if the question can
be answered by either statement alone; - if the question can
be answered by both the statements together but not by any one of the
two; - if
the question cannot be answered by using both the statement together.
Is
quadrilateral ABCD a ‖gm?
- Diagonal AC and BD bisect
each other. - Diagonal AC and BD
are equal.
The
correct answer is : (a)/ (b)/ (c)/ (d).
Correct
option: (a)
If the
diagonals of a quad. ABCD bisect each other, then the quad. ABCD is a
parallelogram.
So, I gives the answer.
If the
diagonals are equal, then the quad. ABCD is a parallelogram.
So, II
gives the answer.
In each of the questions are question is followed by two
statements I and II. The answer is
- if the question can
be answered by one of the given statements alone and not by the other; - if the question can
be answered by either statement alone; - if the question can
be answered by both the statements together but not by any one of the
two; - if the question
cannot be answered by using both the statement together
Is
quadrilateral ABCD a rhombus?
- Quad. ABCD is a ‖gm.
- Diagonals AC and BD
are perpendicular to each other.
The
correct answer is: (a) / (b)/ (c)/ (d).
Correct
option: (c)
If the quad.
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.
In each of the questions are question is followed by two
statements I and II. The answer is
- if the question can
be answered by one of the given statements alone and not by the other; - if the question can
be answered by either statement alone; - if the question can
be answered by both the statements together but not by any one of the
two; - if the question
cannot be answered by using both the statement together
Is ‖gm
ABCD a square?
- Diagonals of ‖gm
ABCD are equal. - Diagonals of ‖gm
ABCD intersect at right angles.
The correct
answer is: (a)/ (b)/ (c)/ (d).
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.
In each of the questions are question is followed by two
statements I and II. The answer is
- if the question can
be answered by one of the given statements alone and not by the other; - if the question can
be answered by either statement alone; - if the question can
be answered by both the statements together but not by any one of the
two; - if the question
cannot be answered by using both the statement together
Is quad. ABCD a parallelogram?
- Its opposite sides
are equal. - Its opposite angles
are equal.
The correct
answer is: (a)/ (b)/ (c)/ (d)
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.
Each question
consists of two statements, namely, Assertion (A) and Reason (R). For
selecting the correct answer, use the following code:
- Both Assertion (A)
and Reason (R) are true and Reason (R) is a correct explanation of
Assertion (A). - Both Assertion (A)
and Reason (R) are true but Reason (R) is not a correct explanation of
Assertion (A) - Assertion (A) is true
and Reason (R) is false. - Assertion (A) is
false and Reason (R) is true.
Assertion (A) | Reason (R) |
If three angles of a quadrilateral are | The sum of all |
The correct
answer is: (a)/ (b)/ (c)/ (d).
Each question consists of two statements, namely, Assertion
(A) and Reason (R). for selecting the correct answer, use the following code:
- Both Assertion (A)
and Reason (R) are true and Reason (R) is a correct explanation of
Assertion (A). - Both Assertion (A)
and Reason (R) are true but Reason (R) is not a correct explanation of
Assertion (A) - Assertion (A) is true
and Reason (R) is false. - Assertion (A) is
false and Reason (R) is true.
Assertion (A) | Reason (R) |
ABCD is a Then, PQRS is | The line |
The correct
answer is: (a)/ (b)/ (c)/ (d).
The
Reason (R) is true and is the correct explanation for the Assertion (A).
Each question consists of two statements, namely, Assertion
(A) and Reason (R). for selecting the correct answer, use the following code:
- Both Assertion (A)
and Reason (R) are true and Reason (R) is a correct explanation of
Assertion (A). - Both Assertion (A)
and Reason (R) are true but Reason (R) is not a correct explanation of
Assertion (A) - Assertion (A) is true
and Reason (R) is false. - Assertion (A) is
false and Reason (R) is true.
Assertion (A) | Reason (R) |
In a rhombus | The diagonals |
The correct
answer is: (a)/ (b)/ (c)/ (d).
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
Each question consists of two statements, namely, Assertion
(A) and Reason (R). for selecting the correct answer, use the following code:
- Both Assertion (A)
and Reason (R) are true and Reason (R) is a correct explanation of
Assertion (A). - Both Assertion (A)
and Reason (R) are true but Reason (R) is not a correct explanation of
Assertion (A) - Assertion (A) is true
and Reason (R) is false. - Assertion (A) is
false and Reason (R) is true.
Assertion (A) | Reason (R) |
Every | The angle |
The correct
answer is: (a)/ (b)/ (c)/ (d).
Each question consists of two statement,
namely, Assertion (A) and Reason (R). for selecting the correct answer, use
the following code:
- Both Assertion (A)
and Reason (R) are true and Reason (R) is a correct explanation of
Assertion (A). - Both Assertion (A)
and Reason (R) are true but Reason (R) is not a correct explanation of
Assertion (A) - Assertion (A) is true
and Reason (R) is false. - Assertion (A) is
false and Reason (R) is true.
Assertion (A) | Reason (R) |
The diagonals | If the |
The correct
answer is: (a)/ (b)/ (c)/ (d).
Column I | Column II |
(a) Angle | (p) parallelogram |
(b) The | (q) rectangle |
(c) The | (r) square |
(d) The | (s) rhombus |
The correct
answer is:
(a) -…….,
(b) -…….,
(c) -…….,
(d)-…….
Column I | Column II |
(a) In | (p) equal |
(b) In | (q) at right angle |
(c) The | (r) 8.5 cm |
(d) The | (s) 6.5 cm |
The correct
answer is:
(a) -…….,
(b) -…….,
(c) -…….,
(d)-…….
The lengths of the diagonals of a rhombus are 16 cm and 12
cm. The length of each side of the rhombus is
- 10 cm
- 12 cm
- 9cm
- 8cm
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
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°
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°