Quadrilaterals

Given,

Three angles of quadrilateral = 110^∘, 50^∘, 40^∘

Let fourth angle be = X^∘

As we know sum of all angles of a quadrilateral =360^∘

So,

110^∘+50^∘+40^∘+X^∘ = 360^∘

X^∘= 360^∘ - 200^∘=160^∘

Given,

In quadrilateral ABCD,

A:B:C:D = 1:2:3:4:5

Let angles A, B, C, D = x, 2x, 4x, 5x

So,

x+2x+4x+5x = 360^∘

12x = 360^∘

So,

Given,

In quadrilateral ABCD,

CO is the bisector of ∠C

DO is the bisector of ∠D

In ΔCOD

⇒ ∠COD =

⇒ ∠D+∠C = 360 – (∠A+∠B)

SO,

⇒ ∠COD =

⇒ ∠COD = (∠A+∠B) Proved.

Given,

Ratio of angles of quadrilateral = 3:5:9:13

Let angles are = 3x, 5x, 9x, 13x

So,

3x+5x+9x+13x = 360^∘

30x = 360^∘

Angles would be =

3x = 3×12 = 36

5x = 5×12 = 60^∘

9x = 9×12 = 108^∘

13x = 13×12 = 156^∘

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 = ^∘

Given,

In a parallelogram

Let angle = x

So

We know that opposite angles of parallelogram are equal

So, four angles will be =

And ^∘

^∘

So, x = 108^∘

Adjacent angles will be =

Angles are = 108^∘, 108^∘, 72^∘ and 72^∘

Given,

Let the smallest angle = x^∘

Than the other angle = (2x – 24)°

So,

x^∘+x^∘+2x – 24 + 2x – 24 = 360^∘

6x = 360 + 48 = 408

Other angles = 2x – 24 = 2×68 – 24 = 136-24 = 112^∘

Angles are = 68^∘, 68^∘, 112^∘, 112^∘

Given,

Peremeter of parallelogram = 22cm

Longer side = 6.5cm

Let shorter side be = x cm

So,

2(6.5 + x) = 22cm

13.0 + 2x = 22cm

2x = 22 – 13 = 9cm

Given,

In a parallelogram ABCD

∠D= 135^∘

∠C + ∠D = 180^∘.. (supplementary angles)

∠C = 180^∘ - 135^∘ = 45^∘

∠C = ∠A and ∠D = ∠B.. (Opposite angles of parallelogram)

∠A = 45^∘

∠B = 135^∘

Given,

In a parallelogram ABCD,

∠A =70^∘

∠A + ∠B = 180^∘.. (supplementary angles)

∠B = 180^∘ – 70^∘ = 110^∘

∠A = ∠C and ∠B = ∠D.. (Opposite angles of parallelogram)

∠A = 70, ∠B = 110, ∠C = 70, ∠D = 110^∘

Given,

In a parallelogram ABCD,

∠A = 60^∘

So,

∠B = 180 – 60 = 120 (supplementary angles)

∠ABP = ∠PCB = ^∘

∠DPA = ∠BAP = 30^∘ (AB parallel to DC and AP intersects them)

∠DPA = ∠DAP = 30^∘

AD = DP... (i) (proved)

Similarly,

∠BPC = ∠ABP = 60^∘

In triangle BPC , ∠BPC = ∠PBC = 60^∘

BC = CP = AD... (ii) (proved)

BC = AD

From equation (i) and (ii),

CP = DP

DC = 2AD (proved)

Given,

In a parallelogram ABCD,

∠DAB = 75^∘

∠DBC = 60^∘

∠A + ∠B = 180^∘ (supplementary angles)

∠B = 180^∘ – 75^∘ = 105^∘

∠DBA +∠DBC = 105^∘

∠DBA = 105^∘ – 60^∘ = 45^∘

In a triangle ABD

∠DAB + ∠DBA + ∠ADB = 180^∘

75^∘ + 45^∘ + ∠ADB = 180^∘

∠ADB = 180^∘ -120^∘ = 60^∘

∠A = ∠C = 75^∘ (opposite angles of parallelogram)

∠C + ∠D = 180^∘ (supplementary angles of parallelogram)

∠D = 180^∘ – 75^∘ = 105^∘

∠ADB + ∠CDB = 105^∘

∠CDB = 105^∘ - ∠ADB

∠CDB = 105^∘ – 60^∘ = 45^∘

Given,

In a parallelgram ABCD,

E = mid point of side BC

AD ⎸⎸BC

AD ⎸⎸BE

E is mid point of BC

So, in ΔDEC and ΔBEF

BE = EC.. (E is the mid point)

∠DEC = ∠BEF

∠DCB = ∠FBE (vertically opposite angles)

So, ΔDEC ≅ ΔBEF

DC = FB

=AB+DC = FB+AB

=2AB =AF (proved)

(i) False

Reason: Imagine a parallelogram and draw its diagonals. Now the areas of the two triangles on one of the bases is equal. But by Heron's formula, the areas are not equal and if the areas are not equal how can be the diagonals because area can only be equal if both the triangles have equal diagonals.

(ii) True

Proof:

Let’s take a parallelogram ABCD,

The diagonals AC and BD intersect each other at O,

AO = OC and BO = OD

In ΔAOB and ΔCOD,

We have

∠BAO = ∠OCD (alternate interior angles)

∠AOB = ∠CDO (alternate interior angles)

AB = DC (opposite sides)

ΔAOB≅ΔCOD (by ASA)

AO = OC and DO = OB

(iii) False

According to the definition a parallelogram is a simple (non-self-intersecting) quadrilateral with two pairs of parallel sides. The opposite or facing sides of aparallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.

It turns out that a parallelogram has its diagonalsmeeting at right angles if and only if the parallelogram is a rhombus (all sides equal).

(iv) True

Reason: Let’s take ABCD is a quadrilateral,

Where AB = CD & AD = BC

We have to prove : ABCD is a parallelogram

Join AC, it’s a diagonal

In ΔABC and ΔCDA

AB = CD

BC = DA

AC = CA

ΔABC≅ΔCDA

Hence,

∠BAC = ∠DCA

For lines AB and CD with transversal AC,

∠BAC & ∠DCA are alternate angles and are equal.

So, AB and CD lines are parallel.

∠BCA = ∠DAC

For lines AD and BC with transversal AC,

∠BCA & ∠DAC are alternate angles and are equal.

So, AD and BC lines are parallel.

Thus in ABCD,

Both pairs of opposite sides are parallel,

So, ABCD is a parallelgram.

(v) False

Reason: If all the angles of a quadrilateral are equal, then it’s a rectangle.

(vi) False

Reason:

As we know that the opposite or facing sides of aparallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.

In case of square and rhombus all side are equal, but if three sides are equal then it doesn’t satisfied the property of parallelogram.

(vii) False

Reason:

Same reason as for the sides, it does not satisfied the property of a parallelogram.

(viii) True

Yes if all side of a parallelogram than it’s a square or rhombus.

Given,

In a parallelogram ABCD,

∠C and ∠D are supplementary angles of parallelogram

So, ∠C + ∠D = 180^∘

Given,

∠B = 135^∘

∠A + ∠B = 180^∘ (supplementary angles of parallelogram)

∠A = 180^∘ – 135^∘ = 45^∘

∠C = ∠A = 45^∘ (opposite angles of parallelogram)

∠D = ∠B = 135^∘

Given,

ABCD is a square

AC and BD intersects at O

∠AOB = 90^∘ (diagonals of square bisects each other at 90^∘)

Given,

ABCD is a rectangle

∠ABD = 40^∘

∠ABD + ∠DBC = 90^∘ (angles of rectangle)

∠DBC = 90^∘ – 40^∘ = 50^∘

Given,

In a parallelogram ABCD

AB & CD bisect at E & F

AB ⎸⎸CD

AB = DC

EB ⎸⎸DF

So, EB = DF

So, EBFD is a parallelogram.

Given,

In a parallelogram ABCD

Since diagonal of parallelogram bisects each other

OA = OC and OB = OD

Since P&Q are the point of trisection of BD

BP = PQ = QD

Now, OB = OD and BP = QD

OB-BP = OD-QD

OP = OQ

Diagonals of quadrilateral bisect each other

Hence, APCQ is a parallelogram.

Given,

ABCD is a square

E,F,G,H are the points of AB, BC, CD, DA

Such that AE = BF = CG = DH

In ΔAEH & ΔBFE

Let,

AE = BF = CG = DH = x

BE = CF = DG = AF = y

In ΔAEH & ΔBFE

AE = BF (given)

∠A = ∠B (each equal)

AH = BE

So, by SAS congruency

ΔAEH ≅ΔBFE

∠1 = ∠2 & ∠3 = ∠4

∠1 + ∠3 = 90^∘

∠2 + ∠4 = 90^∘

∠1+∠2+∠3+∠4 = 180^∘

∠1+∠4+∠1+∠4 = 180^∘

2(∠1+∠4) = 180^∘

∠1+∠4 = = 90^∘

So, ∠HEF = 90^∘

Similarly we have,

∠F =∠G = ∠H = 90^∘

Hence, EFGH is a square.

Given,

ABCD is a rhombus,

EABF is a straight line

EA = AB = BF

OA = OC

OB = OD (diagonals of rhombus are perpendicular bisector of each other)

∠AOD = ∠COD = 90^∘

∠AOB = ∠COB = 90^∘

In ΔBDE,

A and O are mid points of BE and BD

OA ⎸⎸DE

OC ⎸⎸DG

In ΔCFA

B and O are the mid points of AF and AC

OB ⎸⎸CF and

OD ⎸⎸GC

Thus in quadrilateral DOCG

OC ⎸⎸DG and

OD ⎸⎸GC

DOCG is a parallelogram

∠DGC = ∠DOC

∠DGC = 90 (proved)

Given: ABCD is a parallelogram,

In ΔACE

D and O are the mid points of AE and AC

DO ⎸⎸EC

OB ⎸⎸CF and AB = BF → (i)

DC = BF (AB = DC as ABCD is a parallelogram and opposite sides of a parallelogram are equal and parallelP))

In ΔEDC and ΔCBF

DC = BC

∠EDC = ∠CBF

∠ECD = ∠CFB

So, by ASA congruency

ΔEDC≅ΔCBF

DE = BC

DC = BC

AB = BC

BF = BC ( AB = BF from (i) )

Given,

In ΔABC

D, E, F are the mid points of BC, CA and AB

AB = 7cm

BC = 8cm

CA = 9cm

We need to find out the perimeter of DEF

So,

So, perimeter of DEF

DE+EF+DF = 4+3.5+4.5 = 12cm

Given,

In a ΔABC

∠A =50°

∠B =60°

∠C = 70°

Let DEF are mid point of ΔABC

BDEF and CDFE are parallelogram

∠B = ∠E & ∠C = ∠F

∠E = 60°

∠F = 70°

Similarly

∠D = ∠A

∠D = 50°

Given,

In ΔABC

P, Q and R are the mid-points of sides BC, CA and AB

AC = 21cm

BC = 29cm

AB = 30cm

Since ARPQ is a parallelogram

Peremeter of parallelogram ARPQ = 2(AP+AQ)

AB+AC = 30+21 = 51cm

Given,

In ΔABC,

AD is produced to X

AD=DX

In quadrilateral ABXC

AD=DX (Given)

BD = DC (given)

So diagonal AX and BC bisect eachother

Therefore ABXC is a parallelogram.

Given,

In ΔABC,

E & F are the mid-points of AC and AB

EF ⎸⎸BC

SO, FQ ⎸⎸BP

Q is the mid point of AP

AQ = QP (Proved)

Given,

In ΔABC,

BM & CN are perpendiculars from B &C

In ∆BLM and ∆CLN

∠BML =∠CNL= 90°

BL=CL [L is mid point of BC]

∠MLB=∠NLC [vertically opposite angles]

∴ ∆BLM=∆CLN

∴ LM = LN (corresponding sides of congruent triangles)

Given,

ΔABC is right angled at B

AB = 9cm

AC = 15cm

D, E are mid points of side AB and AC

(i) by using pythagores theorem

BC =

BC =

(ii) Area of ΔADE

Given,

M, N, P are mid points AB, AC and BC

MN = 3cm

NP = 3.5cm

MP = 2.5cm

MN =

BC = 2MN = 2×3 = 6cm

MP =

AC = 2MP = 2×2.5 = 5cm

AP =

AB = 2AP = 2×3.5cm

Given,

In ΔABC

ABCQ and ARBC are parallelograms,

BC = AQ &

BC = AR

AQ = AR (A is mid point of QR)

Similarly B and C are the mid point of PR and PQ

AB =

BC =

AC =

PQ = 2AB

QR = 2BC

PR = 2CA

PQ+QR+RP = 2(AB+BC+CA)

Perimeter of PQR = 2(Perimetre of ΔABC)

Given,

In ΔABC,

Q, R are the mid points of AB and AC

QR ⎸⎸AC.. (i)

In ΔABH,

Q, P are the mid points of AB and AH respectively,

QP ⎸⎸BH

QP ⎸⎸BE.. (ii)

But AC⌃BE therefore from (i) and (ii)

QP⌃QR

∠PQR = 90^∘

Given,

In figure 14.98

AB = AC

CP ⎸⎸BA

AP is bisector of exterior angle ∠CAD

AB = AC

∠C = ∠B

NOW,

∠CAD = ∠B +∠C

2∠CAP = 2∠C

∠CAP = ∠C

AP ⎸⎸BC

But, AB ⎸⎸CP (given)

Hence ABCP is a parallelogram

Given,

ABCD is a kite , in which

AB=AD and BC =CD

P,Q,R,S are mid points of side AB,BC,CD &DA

In ∆ABC , P&Q are mid points of AB & BC

∴ PQ││AC , PQ =

In ∆ADC , R & S are mid points of CD & AD

∴RS││AC and RS = 1/2 AC………….(ii)

From (i) and(ii) we have

PQ││RS , PQ=RS

Since AB=AD

=

AP=AS……………(iii)

=∠1=∠2…………….(iv)

Now in ∆PBQ and ∆SDR

PB= SD ∵ AD=AB =

BQ = DR ∴ PB=SD

And PQ = SR ∵ PQRS is a parallelogram

So by sss congruency

∆PBQ≅∆SOR

∠3=∠4

Now, ∠3+SPQ+∠2 =180°

∠1+∠PSR+∠4 = 180°

∴ ∠3+∠SPQ+∠2= ∠1+∠PSR+∠4

∠SPQ=∠PSR (∠2=∠1 and ∠3=∠4)

∵∠SPQ+∠PSR = 180°

=2∠SPQ = 180° = ∠SPQ = 90°

Hence , PQRS is a parallelogram.

Given,

ABC is an isosceles triangle , D,E,F are mid points of side BC, CA,AB

∴AB││DF and AC││FD

ABDF is a parallelogram

AF=DE and AE = DF

=

DE=DF (∵AB=AC)

AE=AF=DE=DF

ABDF is a rhombus

= AD and FE bisect each other at right angle.

Given,

Let P and Q be the mid points of AB and AC respectively

Then, PQ ││BC and PQ=…………….(i)

In ∆APQ, D and E are mid points of AP and AQ resp.

∴DE││PQ , DE= ………….(ii)

From (i) and(ii)

=DE=

= DE =

Given,

Join B and D suppose AC and BD cut at D

Then, OC =

Now, CQ =

= CQ = =

In ∆DCO, P & Q are midpoints of DC & OC

∴ PQ

Also in ∆COB , Q is mid point of OC and QR││OB

∴ R is mid point of BC

(i) In ∆ADC , Q is mid point of AC such that

PQ││AD

∴ P is mid point of DC

= DP= DC (converse of mid point theorem)

(ii) Similarly, R is the mid point of BC

= PR=

PR= ∵ diagonals of rectangle are equal

Proved.

Since E and F are midpoints of AB and CD

∴ AE = BE =

And CF = DF =

∵ AB = CD

∴

= BE = CF

∴ BEFC is a parallelogram

= BE││EF and BF= PH---------------(i)

Now, BC││EF

=AD││EF ∵ BC││AD as ABCD is a parallelogram

= AEFD is a parallelogram

= AE = GP

But is the midpoint of AB

∴ AE=BE

= GP=PH proved

draw LS perpendicular to line MN

∴ the line BM , LS,CN being the same perpendiculars , on line MN are parallel to each other

We have , BL=LC (L is mid point of BC)

Using intercept theorem,

MS = SN………………(i)

Now in ∆MLS and ∆LSN

MS=SN

∠LSM=∠LSN = 90° ( LS perpendicular to MN) and SL=LS is common

∴∆MLS ≅∆LSN (SAS congruency )

∴ LM= LN proved

ABCD is a quadrilateral

P,Q,R,S are mid points of sides AB,BC,CD and DA

In ∆ABC , P and PQ are the mid points of AB and AC respectively

So, by using mid point theorem,

PQ││AC and PO ……………….(i)

Similarly in ∆BCD

RS││AC and RS =

From rquation (i) and (ii)

PQ││RS and PQ=RS

Similarly, we have

PS∥QR and PS=QR

Hence , PQRS is a parallelogram.

Since, diagonals of a paralleleogram bisects each other

Hence, PR and QS bisect each other proved

(i) isosceles.

(ii) Right triangle.

(iii) Parallelogram.

no common points

Given,

In parallelogram ABCD

∠A+∠B = 180° (Adjacent angles of a parallelogram are supplementary)

one common point

Given,

In parallelogram ABCD

∠D = 115° Given

∵∠A & ∠D are adjacent angles of parallelogram

∴∠A +∠D = 180°

∠A = 180°-115° = 65°

∠P=100^° , ∠Q=80^° , ∠R=100^°

Given,

In square PQRS

We know that diagonals of a square bisects each other at 90°

Hence, ∠POQ = 90°

Given,

In quadrilateral ABCD

∠x+∠y+75° = 180° (angle sum property of triangle)

= ∠x+∠y = 180°-75° = 105°

=2(∠x+∠y) = 2×105° = 210°

∴ ∠C+∠D = 360°-210° = 150°

one angle between the diagonals is 60^°

square

Given,

ABCD is a rectangle

Diagonals meets at O, ∠BOC = 44°

∵ ∠AOD = ∠BOC (vertically opposite angles)

= ∠AOD = 44°

AO = OD

∠OAD=∠ODA (Angles facing same side)

So, ∠OAD =

rectangle

Given,

PQRS is a square.

We know that each angle of a square is 90° and diagonals of square bisect the angles

Hence , ∠SRP =

rectangle

Given,

ABCD is a rectangle

∠BAC = 32°

∵ we know that diagonals of a rectangle bisects each other

Hence , AO = BO

And, ∠DBA = ∠BAC = 32° (angles infront of same sides)

∵ ∠DBC+∠DBA = 90°

∴ ∠DBC = 90° - 32° = 58°

90°

Given,

ABCD is a rhombus

∠ABC = 56°

Since , AB= BC

∴ ∠BAC = ∠BCA =

∵ AB=BC

∠ACD=∠BAC = 62°

∵ AB││BC

rectangle

Given,

Perimeter of a parallelogram = 22cm

Longer side = 6.5 cm

∵ AB+BC+CD+DA = 22cm

Let length of shorter side= x cm

= 2x + 2×6.5 = 22

= 2x = 22-13 = 9

= X =

parallelogram

Given,

Ratio of angles of quadrilateral = 3:5:9:13

Let sides are = 3x ,5x ,9x, 13x

= 3x+5x+9x+13x = 360° (angle sum property of quadrilateral)

= 30x = 360°

= x =

So, length of smallest angle = 3x = 3×12 =36°

rhombus

Given ,

ABCD is a quadrilateral

∠C+∠D = k ∠AOB

∵∠D+∠C = k

= ∠A+∠B = 360°- k

= ∠OAB+∠OBA =

=

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

rectangle

Given,

Measure of opposite angles of parallelogram = (60-x°) , (3x-4°)

= 60-x° = 3x-4° (opposite angles of parallelogram )

= 4x = 56°

= x =

Hence angles of parallelograms = 60- 14° = 46°

Other two angles = 180-46 = 134°

square

Given,

In parallelogram ABCD ,

Bisector of ∠A bisects BC at X

∵ AD││BC and AX cuts them so

∠DAX = ∠AXB (alternate angles)

∠DAX = ∠XAB (AX is bisector of ∠A)

∴∠AXB = ∠XAB

AB= BX (sides opposite of equal angles)

Now,

=

=

parallelogram

Given,

PQRS is an isosceles trapezium

PS = RQ ( Given)

= ∠P+∠S = 180° (sum of adjacent angles are supplementary)

= 2x + 3x = 180°

= 5x = 180°

= x =

∵∠P =∠Q ∵ PS=RQ

= 2x = y

= 2×36 = y

= y = 72°

Let angles of parallelogram are ∠A, ∠B, ∠C, ∠D

Let smallest angle = ∠A

Let largest angle = ∠B

= ∠B = 2∠A – 24°............(i)

∠A+∠B = 180° [adjacent angle of parallelogram]

So, ∠A+2∠A -24° = 180°

= 3∠A = 180°+24° = 204°

=∠A =

= ∠B = 2×68°-24° = 112°

Given ,

In parallelogram ABCD,

∠DAB = 75°

∠DBC = 60°

∵ ∠DAB= ∠BCD [opposite angles of parallelogram]

∠BCD = 75°

Now in ∆BDC

= ∠DBC+∠BDC+∠BCD = 180°

= 60°+∠BDC+75° = 180°

=∠BDC = 180° - 135°

=∠BDC = 45°.

Given,

ABCD is a trapezium

= ∠A+∠D = 180°

=2x+10°+x+20° = 180°

= 3x+30°= 180°

= x=

∠B+∠C = 180°

= y + 92° = 180°

= y = 180°-92° = 88°

Given,

ABCD is a parallelogram

E & F are centroids of ∆ABD & ∆BCD respectively

We know that diagonals of parallelogram bisect each other & centroid of a median divides it in 2:1

So, in ∆ABD ,

=

Similarly,

= AO = CO

=

From equations (i) & (ii)

EO = FO

EF = 2 FO…………..(III)

AE = CF……(IV)

From equation (i)

= AE =

=AE =

= AE =

= AE =

=

=

=

= AE= EF

Given,

PQRS is a rhombus

∠SRT = 152°

We know that diagonals of a rhombus bisects each other at 90°

Hence , ∠SOR = 90° = y = 90°

∠SRT+∠SRO= 180° (linear pair of angles)

= 152°+∠SRO = 180°

∠SRO = 180° -152°= 28°

NOW , ∠SRO = z ∵SR=SP

= z = 28°

And , ∠RSO+∠SRO + y = 180° (angle sum property of triangle)

= ∠RSO+28°+90°= 180°

= ∠RSO = 180°-118° = 62°

= ∠X = ∠RSO = 62° (alternate angles)

Given,

ABCD is a rectangle

∠ECD = 146°

= ∠ECD+∠DCO = 180° (Linear pair of angles)

=146°+∠DCO = 180°

= ∠DCO = 180° - 146° = 34°

AND, ∠BOC = 180°- ∠OCB-∠OBC ∵ ∠OCB = ∠OBC

=∠BOC = 180°-34°-34° = 112°

∠BOC+∠AOB = 180° (Linear pair)

= 112°+ ∠AOB = 180°

= ∠AOB = 180°- 112° = 68°

Given,

ABCD is a parallelogram,

M is mid point of BD

BM bisects ∠B

∵ ∠A+∠B = 180° [adjacent angles of parallelogram are supplementary]

In ∆AMB , ∠AMB+

= ∠AMB +

=∠AMB +

= ∠AMB = 180° - 90° = 90°

Given,

ABCD is a parallelogram

E is mid point of BC

DE & AB after producing meet at F

In ∆ECD & ∆BEF ,

∠BEF = ∠CED [vertically opposite angles]

BE = EC

∠EDC = ∠EFB [ alternate angles]

∴ ∆ECD ≅ ∆BEF

So, CD= BF

∵ AB=CD

Thus, AF= AB+ BF

= AF = AB + CD

= AF = AB+AB

= AF = 2AB

Given,

ABCD & AEFG are two parallelograms

∠C = 58°

∵ AB││CD , AE││FG ,BC││AD , EF││AG

GF produced to H so GH││AB

= ∠C = ∠H (corresponding angles)

= ∠H = 58°

AD││BC and GH cuts them

Hence , ∠F = ∠H (corresponding angles)

∠F = 58°.

(i) parallelogram

(ii) trapezium

(iii) parallel toWW

(iv) rectangle

(v) supplementary

(vi) parallelogram

(vii) parallelogram

(viii) rhombus

Given,

In parallelogram ABCD

∠A+∠B = 180° [adjacent angles of parallelogram are supplementary]

∠ A =

So,

=

=

∠A =

Thus smallest angle of parallelogram is 72°

Given,

ABCD is a parallelogram,

Angles of quadrilateral 4x , 7x , 9x , 10x

= 4x+7x+9x+10x = 360° [ angle sum property of quadrilateral]

= 30x = 360°

=

Hence, sum of smallest and largest angles = 4x+10x = 4×12+10×12

= 48°+120° = 168°

Given,

ABCD is a parallelogram

∠A+∠C = 2(∠B+∠D)

∠A = 40°

∵ ∠A+∠B+∠C+∠D = 360° [angle sum property of quadrilateral]

= ∠A+∠C+∠B+∠D = 360°

= 2(∠B+∠D)+ ∠B+∠D = 360°

= 3(∠B+∠D) = 360°

= ∠B+∠D =

∵∠D= 60° [given]

∴ ∠B = 120°-60° = 60°

Given,

ABCD is a rhombus

AC=24 cm, BD = 18 cm

= AB=BC=CD=DA [side of rhombus]

We know that diagonals of rhombus bisect each other at 90°

In right ∆ AOB

AB² = BO²+AO²

AB²= 12²+9² = 144+81 = 225

AB =

Side of rhombus = 15 cm

Given,

ABCD is a rectangle

Diagonals AC & BD intersect each other at P

∠ABD = 50°

∵ diagonals of rectangle bisect each other and are equal in length

=∠ABD= ∠PDC [alternate angles]

∠PDC=∠PCD = 50°

In ∆DPC

=∠DPC+∠PCD+∠PDC = 180°

= ∠DPC+50°+50°= 180°

= ∠DPC = 180°-100° = 80°

Given,

ABCD is a parallelogram

Diagonal AC bisects ∠BAD

∠BAC = 35°

∵∠A+∠B = 180° …………(i) [ angle sum property of quadrilateral]

∠A = 2∠BAC = 2×35° = 70°

Putting value of ∠A in equation (i)

= 70°+∠B = 180°

= ∠B = 180°-70° = 110°

∠ABC = 110°

Given,

ABCD is a rhombus

∠ACB = 40°

∵ ∠DAC =∠ACB = 40° [alternate angles]

∠AOD = 90° [ diagonals of rhombus are perpendicular to each other]

In ∆AOD

∠AOD+∠ADO+∠DAO = 180°

∠ADO+90°+40° = 180°

= ∠ADO = 180°- 130° = 50°

=∠ADO=∠ADB = 50°

Given,

In ∆ABC

∠A=30°,∠B=40°,∠C=110°

In figure, BDEF,DCEF,DEAF are parallelograms so,

∠B = ∠E = 40° [∵∠B&∠E are opposite angles of parallelogram BDEF]

∠C=∠F = 110° [ ∵ ∠C &∠F are opposite angles of parallelogram DCEF]

∠A=∠D =30° [ ∵∠A &∠D are opposite angles of parallelogram DEAF]

Hence, ∠D = 30°

∠E = 40°

∠F = 110°

Given,

ABCD is a parallelogram

Diagonals of parallelogram intersects at O

∠BOC = 90° , ∠BDC = 50°

∵ ∠BOC+∠AOB = 180° [Linear pair of angles]

90°+∠AOB = 180°

=∠AOB = 180°-90° = 90°

∠BDC = ∠OBA = 50° [Alternate angles]

In ∆AOB

∠AOB+∠OBA+∠OAB = 180°

= 90°+50°+∠OAB = 180°

= ∠OAB = 180°-140° = 40°

Given,

ABCD is a trapezium

AB││DC

M , N are mid points of AD & BC

AB = 12cm , MN = 14 cm

∵ AB││MN││CD [ M, N are mid points of AD &BC]

MP = NP

By mid point theorem,

MP =

∴MN =

= 14 =

Given,

ABCD is a quadrilateral

∠A = 45° ,

∵ diagonals of quadrilateral bisects each other hence ABCD is a parallelogram,

= ∠A +∠B = 180°

=45°+∠B = 180°

=∠B = 180°- 45° = 135°

Given,

ABCD is a parallelogram,

P is mid point of side BC

∠BAP = ∠DAP

AD = 10cm

∵ AD││BC

∠DAP = ∠APB [alternate angles]

∠DAP = ∠BAP

AB = BP (side opposite to equal angles)

= BP =

∴ AB =

AB = CD = 5cm (sides of parallelogram)

Hence , CD = 5 cm

Given,

In ∆ABC

E is mid point of median AD

AC = 10.5 cm

Draw DM││EF

∵ E is mid point of AD so F is mid point of AM

AF= FM ……………………(i)

In ∆BFC

EF││DM

SO, FM = MC ………………….(ii)

From(i) &(ii)

AF= MC……………..(iii)

AC = AF+MC+FM

= AC= AF+AF+AF From (i) (ii) &(iii)

AC = 3AF

AF =

AF =