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From the table, it can be observed that the angle sum of a convex polygon of n sides is (n −2) × 180º. Hence, the angle sum of the convex polygons having number of sides as above will be as follows.

(a) (7 − 2) × 180º = 900°

(b) (8 − 2) × 180º = 1080°

(c) (10 − 2) × 180º = 1440°

(d) (n − 2) × 180°

Data: In ∆ABC and ∆DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively 

To Prove: (i) Quadrilateral ABED is a parallelogram. 

(ii) Quadrilateral BEFC is a parallelogram. 

(iii) AD||CF and AD = CF. 

(iv) quadrilateral ACFD is a parallelogram. 

(v) AC = DF 

(vi) ∆ABC ≅ ∆DEF. 

Proof: (i) AB = DE and AB||DE (Data) 

∴ BE = AD and BE || AD 

∴ ABCD is a parallelogram. 

(ii) Similarly, BC = EF and BC || EF. 

∴ BE = CF BE || CF 

∴ BEFC is a parallelogram. 

(iii) ABED is a parallelogram. 

∴ AD = BE AD || BE ………. (i) 

Similarly, BEFC is a parallelogram. 

∴ CF = BE CF || BE ……………. (ii) 

Comparing (i) and (ii), 

AD = CF and AD || CF, 

(iv) In a quadrilateral ACFD, 

AD = CF AD || CF (proved) 

∴ AC = DF AC || DF 

∴ ACFD is a parallelogram. 

(v) ACFD is a parallelogram. 

AC = DF (opposite sides). 

(vi) In ∆ABC and ∆DEF, 

AB = DE (Data) 

BC = EF (opposite sides of a parallelogram) 

AC = DF (Opposite sides of a parallelogram) 

∴ ∆ABC ≅ ∆DEF (SSS postulate).

Given, 

Two opposite angles of a parallelogram (3x - 2) and (50 – x) 

Opposite angles of a parallelogram are equal, 

3x – 2 = 50 – x 

4x = 52 ∘ 

x = 13 ∘ 

So, angles are 

3x – 2 = 3×13-2 = 37^∘ 

50 – x = 50 -13 = 37^∘ 

Sum of other two angles = 360 - 2×37 = 360 – 74 = 286^∘ 

So each angle will be = \(\frac{286}{2}\) = 143^∘

We know that opposite angles of a parallelogram are equal.

Therefore (3x - 2)° = (50 - x)°

3x - 2° = 50° - x

3x° + x = 50° + 2°

4x = 52°

x = \(\frac{52°}{4}\) = 13°

Measure of opposite angles are: 3x - 2 = 3 × 13°-2 = 37°

(50 - x)° = 50 - 13 = 37°

Sum of adjacent angles = 180°

Other two angles are 180° - 37° = 143° each

Correct option is (C) 45

\(\because\) Opposite angles in a parallelogram are equal.

\(\therefore\) 3x - 20 = x+70

\(\Rightarrow\) 3x - x = 70+20 = 90

\(\Rightarrow\) x = \(\frac{90}2\) = 45

Correct option is  C) 45

Given, 

In parallelogram ABCD 

Opposite Angles are equal.

∴ ∠A = ∠C 

⇒ 3x-20 = x + 40 

⇒ 2x = 60 

⇒ x = 30 

Consecutive angles are supplementary (A + B = 180°).

⇒ 3x – 20 + y + 15 = 180

⇒ 3x + y = 185

⇒ 3 × 30 + y = 185

⇒ y = 185 – 90 = 95

Given,

PQRS is an isosceles trapezium 

PS = RQ (Given) 

= ∠P+∠S = 180° 

(sum of adjacent angles are supplementary) 

= 2x + 3x = 180° 

= 5x = 180° 

= x = \(\frac{180° }{5}\) 

= 36° 

∵∠P =∠Q 

∵ PS = RQ 

= 2x = y 

= 2 × 36 = y 

= y = 72°

Data: Diagonal AC of a parallelogram ABCD bisects ∠A. 

To Prove: (i) Diagonal AC also bisects ∠C. 

(ii) ABCD is a rhombus. 

Proof: i) Diagonal AC also bisects ∠A. 

∴ ∠DAC = ∠BAC …………… (i) 

But, ∠DAC= ∠BCA (alternate angles)(ii) 

∠BAC=∠DCA (alternate angles)(iii) 

From (i), (ii) and (iii), 

∠BCA = ∠DCA 

∴ AC bisects ∠C. 

(ii) ∠DAC = ∠DCA 

∴ AD = DC 

But, AD = BC 

∴ DC = AB 

∴ AB = BC = CD = DA 

∴ ABCD is a rhombus.

(i) ∠A = ∠C

True,

∠C =∠A = 55° [In a parallelogram opposite angles are equal]

(ii) ∠FAB= \(\frac{1}{2}\)∠A

True,

AF is the angle bisector of angle A

(iii) ∠DCE= \(\frac{1}{2}\)∠C

True,

CE is the angle bisector of angle A

(iv) ∠CEB= ∠FAB

True,

∠C = ∠A [In a parallelogram opposite angles are equal]

[AF and CE are angle bisectors]

(v) CE || AF

True,

Since one pair of opposite angles are equal, therefore Quad AEFC is a parallelogram

False

If all angles will be obtuse, then their sum will exceed 360°. This is not possible in case of a quadrilateral.

(c) We know that, the sum of all the angles of a quadrilateral is 360°.

Also, an obtuse angle is more than 90° and less than 180°.

Thus, all the angles of a quadrilateral cannot be obtuse.

Hence, almost 3 angles can be obtuse.

(b) The number of non-overlapping triangles in a n-gon = n – 2, i.e. 2 less than the number of sides.

Correct option is: A) 2

(a) False. Since, squares have all sides are equal.
(b) True. Since, in rhombus, opposite angles are equal and diagonals intersect at mid – point.
(c) True. Since, squares have the same have the same property of rhombus but not a rectangle.
(d) False. Since, all squares have the same property of parallelogram.
(e) False. Since, all kites do not have equal sides.
(f) True. Since, all rhombuses have equal sides and diagonals bisect each other.
(g) True. Since, trapezium has only two parallel sides.
(h) True. Since, all squares have also two parallel lines.

(i) A square is a quadrilateral since it has four sides.

(ii) A square is a parallelogram since its opposite sides are parallel to each other.

(iii) A square is a rhombus since its four sides are of the same length.

(iv) A square is a rectangle since each interior angle measures 90°.

(a) False. A kite does not have all sides of the same length.

(b) True. A rhombus also has two distinct consecutive pairs of sides of equal length.

(c) True. All parallelograms have a pair of parallel sides.

(d) True. All squares have a pair of parallel sides.

(i) AD = BC [In a parallelogram diagonals bisect each other]

(ii) ∠DCB = ∠BAD [alternate interior angles are equal]

(iii) OC = OA [In a parallelogram diagonals bisect each other]

(iv) ∠DAB+ ∠CDA = 180° [Sum of adjacent angles in a parallelogram is 180°]

(i) No, In a parallelogram opposite angles are equal.

(ii) Yes, In a parallelogram opposite sides are equal and parallel.

(iii) No, In a parallelogram diagonals bisect each other.

(i) ∠ABC = ∠Y = 100° [In a parallelogram opposite angles are equal]

∠x + ∠Y = 180° [In a parallelogram sum of the adjacent angles is equal to 180°]

∠x + 100° = 180°

∠x = 180°-100°

∠x = 80°

∠x = ∠z = 80° [In a parallelogram opposite angles are equal]

(ii) ∠PSR + ∠Y = 180° [In a parallelogram sum of the adjacent angles is equal to 180°]

∠Y + 50° = 180°

∠Y = 180°- 50°

∠Y = 130°

∠x = ∠Y = 130° [In a parallelogram opposite angles are equal]

∠PSR = ∠PQR = 50° [In a parallelogram opposite angles are equal]

∠PQR + ∠Z = 180° [Linear pair]

50° + ∠Z = 180°

∠Z = 180°-50°

∠Z = 130°

(iii) In ΔPMN

∠MPN + ∠PMN + ∠PNM = 180° [Sum of all the angles of a triangle is 180°]

30° + 90° + ∠z = 180°

∠z = 180°-120°

∠z = 60°

∠y = ∠z = 60° [In a parallelogram opposite angles are equal]

∠z = 180°- 120° [In a parallelogram sum of the adjacent angles is equal to 180°]

∠z = 60°

∠z + ∠NML = 180° [In a parallelogram sum of the adjacent angles is equal to 180°]

60° + 90°+ ∠x = 180°

∠x = 180°-150°

∠x = 30°

(iv) ∠x = 90° [vertically opposite angles are equal]

In ΔDOC

∠x + ∠y + 30° = 180° [Sum of all the angles of a triangle is 180°]

90° + 30° + ∠y = 180°

∠y = 180°-120°

∠y = 60°

∠y = ∠z = 60° [alternate interior angles are equal]

(v) ∠x + ∠POR = 180° [In a parallelogram sum of the adjacent angles is equal to 180°]

∠x + 80° = 180°

∠x = 180°- 80°

∠x = 100°

∠y = 80° [In a parallelogram opposite angles are equal]

∠QRS =∠x = 100°

∠QRS + ∠Z = 180° [Linear pair]

100° + ∠Z = 180°

∠Z = 180°-100°

∠Z = 80°

(vi) ∠y = 112° [In a parallelogram opposite angles are equal]

∠y + ∠TUV = 180° [In a parallelogram sum of the adjacent angles is equal to 180°]

∠z + 40° + 112° = 180°

∠z = 180°- 152°

∠z = 28°

∠z =∠x = 28° [alternate interior angles are equal]

(i) x + 100° = 180° (Adjacent angles are supplementary) x = 80°

z = x = 80º(Opposite angles are equal)

y = 100° (Opposite angles are equal)

(ii) 50° + y = 180° (Adjacent angles are supplementary)

y = 130°

x = y = 130° (Opposite angles are equal)

z = x = 130º (Corresponding angles)

(iii) x = 90° (Vertically opposite angles)

x + y + 30° = 180° (Angle sum property of triangles) 120° + y = 180°

y = 60°

z = y = 60° (Alternate interior angles)

(iv) z = 80° (Corresponding angles)

y = 80° (Opposite angles are equal)

x+ y = 180° (Adjacent angles are supplementary)

x = 180° − 80° = 100°

(v) y = 112° (Opposite angles are equal)

x+ y + 40° = 180° (Angle sum property of triangles)

x + 112° + 40° = 180°

x + 152° = 180°

x = 28°

z = x = 28° (Alternate interior angles)

Given: □ABCD is a parallelogram. BE = AB 

To prove: Line ED bisects seg BC at point F i.e. FC = FB

Proof: 

□ABCD is a parallelogram. [Given]

∴ seg AB ≅ seg DC …….(i) [Opposite angles of a parallelogram]

seg AB ≅ seg BE ……..(ii) [Given] 

seg DC ≅ seg BE ……..(iii) [From (i) and (ii)]

 side DC || side AB [Opposite sides of a parallelogram] 

i.e. side DC || seg AE and seg DE is their transversal. 

[A-B-E] 

∴ ∠CDE ≅ ∠AED 

∴ ∠CDF ≅ ∠BEF …..(iv) [D-F-E, A-B-E] 

In ∆DFC and ∆EFB, 

seg DC = seg EB [From (iii)] 

∠CDF ≅ ∠BEF [From (iv)]

 ∠DFC ≅ ∠EFB [Vertically opposite angles] 

∴ ∆DFC ≅ ∆EFB [SAA test] 

∴ FC ≅ FB [c.s.c.t] 

∴ Line ED bisects seg BC at point F.

Correct option is: C) AB = BC = CD = DA, AC ≠ BD

Correct option is: A) 135°, 45°, 135°

ABCD is not a parallelogram, because diagonals of a parallelogram bisect each other. Here OA ≠ OC.

(d) AB = CD.

If a trapezium is inscribed in a circle, i.e., it is a cyclic trapezium, then it is an isosceles trapezium, i.e., non-parallel sides are equal ⇒ AB = CD.

Given: ABCD is a trapezium in which AB || CD and AD = BC.

To prove:

(i) ∠A = ∠B

(ii) ∠C = ∠D

Construction: Join AC and BD. Extend line AB and draw a line through C parallel to DA which meets AB produced at E.

Proof:

(i) AB || DC

⇒ AE || DC …(i)

and AD || CE …(ii) (by construction)

⇒ ADCE is a parallelogram

⇒ ∠A + ∠E = 180° …(iii) (sum of consecutive interior angles)

⇒ ∠ABC + ∠CBE = 180° …(iv) (linear pair of angles)

AD = CE …(v) (opposite sides of a parallelogram)

and AD = BC (given) …(vi)

⇒ BC = CE from (v) and (vi)

⇒ ∠CBE = ∠CEB …(vii) (angles opposite to equal sides)

⇒ ∠B + ∠E = 180° …(viii) [using (iv) and (vii)]

Now from (iii) and (viii), we have

∠A + ∠E = ∠B + ∠E

⇒ ∠A = ∠B

(ii) ∠A + ∠D = 180°

and ∠B + ∠C = 180°

⇒ ∠A + ∠D = ∠B + ∠C (∵ ∠A = ∠B)

⇒ ∠C = ∠D

Answer is (C) opposite angles are bisected by the diagonals

Answer is (C) DE = EF

Answer is (C) diagonals of ABCD are equal and perpendicular

Since, diagonal of a rectangle are equal and bisect each other and vice versa.

OR = OS

⇒ ∠ORS = ∠OSR …(i)

Also PQ || SR

⇒ ∠OPQ = ∠ORS = 30° …(ii) (alt. angles)

In ΔORS,

∠ORS + ∠OSR + y = 180° (angle sum property)

⇒ 30° + 30° + y = 180° [using (i) and (ii)]

⇒ y = 120°

Also OR = OQ

⇒ ∠ORQ = ∠OQR = x

In ΔORQ,

x + x + ∠ROQ = 180°

⇒ 2x + 60° = 180° (∵ y + ∠ROQ = 180° i.e. linear pair of angles)

⇒ x = \(\frac { 120 }{ 2 }\) = 60°

x + y = 60° + 120° = 180°

Answer is (D) a parallelogram

Since, D and E are mid-points of sides AB and AC respectively. 

Therefore, from mid-point theorem,

DE = \(\frac { 1 }{ 2 }\) BC

⇒ DE = \(\frac { 1 }{ 2 }\) x 6.4 

= 3.2

⇒ DE = 3.2 cm.

Given: ABCD is a parallelogram in which P is the mid-point of side CD.

To prove: DA = AR and CQ = QR

Proof: ABCD is a parallelogram.

BC = AD and BC || AD

Also, DC = AB and DC || AB

Since, P is mid-point of DC

DP = PC = \(\frac { 1 }{ 2 }\) DC

Now, QC || AP and PC || AQ.

Hence, APCQ is a parallelogram.

AQ = PC = \(\frac { 1 }{ 2 }\) DC = \(\frac { 1 }{ 2 }\) AB

= BQ [∵ DC = AB]

Now in ∆AQR and ∆BQC

AQ = BQ

∠AQR = ∠BQC (vertically opposite angles)

and ∠ARQ = ∠BCQ (alternate interior angles)

∆AQR = ∆BQC (by AAS congruence rule)

So, AR = BC (by c.p.c.t)

But, BC = DA

AR = DA

Also, CQ = QR (by c.p.c.t)

Hence proved.

Given: ABCD is a trapezium in which AB || CD and M and N are mid-points of diagonal AC and BD respectively.

Construction: Join AN and produce it to meet CD at E.

To prove: MN = \(\frac { 1 }{ 2 }\) (CD – AB)

and MN || CD

Proof: In ∆ANB and ∆END

∠ANB = ∠END (vertically opposite angles)

NB = ND (N is the mid-points of BD)

and ∠ABN = ∠EDN (alternate angles)

(∵ AB || CD and BD is a transversal)

∆ANB = ∆END (by ASA congruency rule)

⇒ AN = NE and AB = ED …(i) (by c.p.c.t)

Now in ∆EAC,

N and M are the mid-points of AE and AC respectively.

MN || EC and MN = \(\frac { 1 }{ 2 }\) EC (by mid-point theorem)

⇒ MN || EC

and MN = \(\frac { 1 }{ 2 }\) (CD – ED) = \(\frac { 1 }{ 2 }\) (CD – AB) [using (i)]

Hence, MN || CD (∵ MN || EC)

and MN = \(\frac { 1 }{ 2 }\) (CD – AB).

Given: ABCD is a parallelogram. P and Q are respectively the mid-points of opposite sides AB and DC of a parallelogram ABCD. 

AQ and DP are joined intersecting in S and CP and BQ are joined intersecting in R.

To prove: Quadrilateral PQRS is a parallelogram.

Proof: ∵ ABCD is a parallelogram.

∴ AB = DC and AB || DC

⇒ \(\frac { 1 }{ 2 }\)AB = \(\frac { 1 }{ 2 }\)DC

⇒ AP = QC and AP || QC

⇒ APCQ is a parallelogram.

Similarly. PBQD is also a parallelogram. 

Since in parallelogram APCQ.

AQ || PC

(opposite sides of a parallelogram)

∴ SQ || PR

Now in parallelogram PBQD

PD || BQ

∴ SP || QR

Now in quadrilateral PRQS, SQ || PR and SP || QR.

Therefore, PRQS is a parallelogram.

Hence proved.

Given: ABCD is a parallelogram such that AP = CQ.

To prove: AC and PQ bisect each other.

∠MAP = ∠MCP

(alternate angles)

AP = CQ (given)

∠APM = ∠CQM

(alternate angles)

∴ ∆AMP ≅ ∆CMQ

(by ASA congruency rule)

⇒ AM = CM

and PM = QM (by c.p.c.t)

⇒ AC and PQ bisect each other.

Given: ABCD is a parallelogram in which E and F are the mid-points of sides AB and CD respectively.

To prove: BQ = QP = PD

Proof: ABCD is a parallelogram

⇒ AB || DC

⇒ AE || AF and AE = CF (∵ \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) CD)

⇒ AECF is a parallelogram.

⇒ EC || AF …(i)

If ΔAPB, E is mid-point of AB (given)

and EQ ||AP [using (ii)]

⇒ Q is the mid-point PB

⇒ BQ = PQ …(ii)

Similarly, in ΔDQC, we can show that

DP = PQ

From (ii) and (iii), we have BQ = PQ = PD

Hence, line segment AF and EC trisect the diagonal BD.

Answer is (D) a quadrilateral whose opposite angles are supplementary

Answer is (C) trapezium

Answer is (A) 6 cm

We know that opposite sides of a parallelogram are equal and parallel.

Perimeter = Sum of all sides

Perimeter = 4 + 3 + 4 + 3 = 14 cm

Answer is (B) 75°

It is not be a parallelogram, because we may have ∠A = ∠B = ∠C = 80° and D = 120°. 

Here, ∠B ≠ ∠D.

Let the two consecutive angles be x and 3x.

Since the sum of two consecutive angles of a parallelogram is 180°.

Therefore, x + 3x = 180°

⇒ 4x = 180°

⇒ x = 45°

Smallest angle = 45°.

Correct option is (C) 120°

Let both consecutive angles are x and 2x.

\(\therefore\) x+2x = \(180^\circ\)    \((\because\) Sum of two consecutive angles in a parallelogram is \(180^\circ)\)

\(\Rightarrow\) 3x = \(180^\circ\)

\(\Rightarrow\) x = \(\frac{180^\circ}3=60^\circ\)

\(\therefore\) Greater angle = 2x \(=120^\circ\)

Correct option is  C) 120°

Correct option is (B) Rectangle

The angle bisector of a parallelogram form a rectangle.

Correct option is  B) Rectangle