Areas Of Parallelograms

(i) and trapezium ABCD are on the same base CD and between the same parallels AB and DC.

(ii) Parallelogram ABCD and APQD are on the same base AD and between the same parallel AD and BQ.

(iii) Parallelogram ABCD and AD and BQ.

(iv) Parallelogram PQRS are on the same base QR and between the same parallels QR and PS.

(v) Parallelogram PQRS and trapezium SMNR are on the same base SR but they are not between the same parallel.

(vi) Parallelograms PQRS, AQRD, BCQR are between the same parallels also parallelograms PQRS, BPSC and APSD are between the same parallels.

Given that,

In a parallelogram ABCD:

CD = AB = 16cm (Opposite sides of parallelogram are equal)

We know that,

Area of parallelogram = Base * Corresponding altitude

Area of parallelogram ABCD:

CD * AE = AD * CF

16 cm * 18 cm = AD * 10 cm

AD =

AD = 12.8 cm

Area of parallelogram ABCD = AD * CF (i)

Again,

Area of parallelogram ABCD = DC * AE (ii)

From (i) and (ii), we get

AD * CF = DC * AF

6 * 10 = CD * 8

CD =

= 7.5 cm

Therefore,

AB = DC = 7.5 cm (Opposite sides of parallelogram are equal)

Given that,

Area of parallelogram ABCD = 124 cm²

Construction: Draw AP perpendicular to DC

Proof: Area of parallelogram AFED = DF * AP (i)

Area of parallelogram EBCF = FC * AP (ii)

And,

DF = FC (iii) (F is the mid-point of DC)

Compare (i), (ii) and (iii), we get

Area of parallelogram AEFD = Area of parallelogram EBCF

Therefore,

Area of parallelogram AEFD =

= = 62 cm²

We know that,

Diagonal of parallelogram divides it into two quadrilaterals.

Since,

AC is the diagonal

Then, Area (ΔABC) = Area (ΔACD)

= Area of parallelogram ABCD (i)

Since,

BD is the diagonal

Then, Area (ΔABD) = Area (ΔBCD)

= Area of parallelogram ABCD (ii)

Compare (i) and (ii), we get

Therefore,

Area (ΔABC) = Area (ΔACD) = Area (ΔABD) = Area (ΔBCD) = Area of parallelogram ABCD

Given that,

DC = 17 cm

AD = 9 cm

And,

BC = 8 cm

In Δ BCD, we have

CD² = BD² + BC²

(17)² = BD² + (8)²

BD² = 289 – 64

=15

In Δ ABD, we have

BD² = AB² + AD²

(15)² = AB² + (9)²

AB² = 225 – 81

= 144

=12

Therefore,

Area of Quadrilateral ABCD = Area (Δ ABD) + Area (Δ BCD)

Area of quadrilateral ABCD = (12 × 9) + (8 × 17)

= 54 + 68

= 112 cm²

From the figure,

T and U are the mid points of PS and QR respectively

Therefore,

TU || PQ

TO||PQ

Thus,

In Δ PQS and T is the mid-point of PS and TO||PQ

Therefore,

TO =* PQ

= 4 cm

Also,

TS =* PS

=4 cm

Therefore,

Area (Δ OTS) = (TO * TS)

=(4 * 4)

=8 cm²

We have,

Area of trapezium PQRS = Area of rectangle PSRT + Area (Δ QRT)

Area of trapezium PQRS = PT * RT + (QT * RT)

= 8 * RT + (8 * RT)

= 12 * RT

In Δ QRT, we have

QR² = QT² + RT²

RT² = QR² - QT²

RT² = (17)² - (8)²

= 225

= 15

Hence,

Area of trapezium PQRS = 12 * 15

= 180 cm²

Since,

The mid-point of the hypotenuse of a right triangle is equidistant from the vertices

Therefore,

CA = CB = OC

CA = CB = 6.5 cm

AB = 13 cm

In right (OAB)

We have,

AB² = OB² - OA²

13² = OB² + 12²

OB = 5 cm

Therefore,

Area (Δ AOB) = (OA * OB)

= (12 * 5)

= 30 cm²

Draw AL perpendicular to DC

And,

BM perpendicular DC

Then,

AL = BM = 4 cm

And,

LM = 7 cm

In Δ ADL, we have

AD² = AL²+ DL²

25 = 16 + DL²

DL = 3 cm

Similarly,

MC =

=

= 3 cm

Therefore,

x = CD = CM + ML + LD

= 3 + 7 + 3

= 13 cm

Area of trapezium ABCD = (AB + CD) * AL

= (7 + 13) * 4

= 40 cm²

We have,

OD = 10 cm

And,

OE = 2 cm

Therefore,

OD² = OE² + DE²

DE =

=

= 4 cm

Therefore,

Area of trapezium OCDE = OE * DE

= 2* 4

= 40 cm²

Given that,

ABCD is a trapezium with AB ‖ DC

To prove: Area ( = Area (

Proof: Since,

ΔABC and ABD are on the same base AB and between the same parallels AB and DC

Therefore,

Area (Δ ABC) = Area (Δ ABD)

Area (Δ ABC) – Area (Δ AOB) = Area (ΔABD) – Area (Δ AOB)

Area (Δ AOD) = Area (Δ BOC)

Hence, proved

Given that,

ABCD is a parallelogram

So,

AD = BC

CDEF is a parallelogram

So,

DE = CF

ABFE is a parallelogram

So,

AE = BF

Thus,

In ΔADE and BCF, we have

AD = BC

DE = CF

And,

AE = BF

So, by SSS congruence rule, we have

Δ ADE ≅ ΔBCF

Therefore,

Area (Δ ADE) = Area (Δ BCF)

Construction: Draw BQ perpendicular to AC

And,

DR perpendicular to AC

Proof: We have,

L.H.S = Area (ΔAPB) * Area (ΔCPD)

= (AP * BQ) * (PC * DR)

= ( * PC * BQ) * ( * AP * DR)

= Area (ΔBPC) * Area (ΔAPD)

= R.H.S

Therefore,

L.H.S = R.H.S

Hence, proved

Given that,

CD bisected AB at O

To prove: Area () = Area ()

Construction: CP perpendicular to AB and DQ perpendicular to AB

Proof: Area ( = (AB * CP) (i)

Area ( = (AB * DQ) (ii)

In , we have

∠CPO = ∠DQO (Each 90^o)

Given that,

CO = DO

∠COP = ∠DOQ (Vertically opposite angle)

Then, by AAS congruence rule

Therefore,

CP = DQ (By c.p.c.t)

Thus,

Area ( = Area ()

Hence, proved

Construction: Draw DN⊥ AB and PM⊥ AB.

Proof: Area of parallelogram ABCD = AB * DN

Area (Δ APB) = (AB * PM)

= AB * PM < AB * DN

= (AB * PM) < (AB * DN)

= Area (Δ APB) < Area of parallelogram ABCD

Construction: Draw AM⊥ BC

Proof: Since,

AD is the median of ΔABC

Therefore,

BD = DC

BD * AM = DC * AM

(BD * AM) = (DC * AM)

Area (Δ ABD) = Area (Δ ACD) (i)

Now, in Δ BGC

GD is the median

Therefore,

Area (BGD) = Area (CGD) (ii)

Also,

In Δ ACD, CG is the median

Therefore, Area (AGC) = Area (CGD) (iii)

From (i), (ii) and (iii) we have

Area (ΔBGD) = Area (ΔAGC)

But,

Area (ΔBGC) = 2 Area (ΔBGD)

Therefore,

Area (BGC) = 2 Area (ΔAGC)

Hence, proved

Given that,

In Δ ABC,

We have

BD = 2DC

To prove: Area () = 2 Area ()

Construction: Take a point E on BD such that, BE = ED

Proof: Since,

BE = ED and,

BD = 2DC

Then,

BE = ED = DC

Median of the triangle divides it into two equal triangles

Since,

AE and AD are the medians of ΔABD and AEC respectively

Therefore,

Area (ΔABD) = 2 Area (ΔAED) (i)

And,

Area (ΔADC) = Area (ΔAED) (ii)

Comparing (i) and (ii), we get

Area (ΔABD) = 2 Area (ΔADC)

Hence, proved

Given that,

ABCD is a parallelogram

To prove: (i) Area () = Area ()

(ii) Area () = Area ()

Proof: We know that,

Diagonals of a parallelogram bisect each other

Therefore,

AO = OC and,

BO = OD

(i) In ΔDAC, DO is a median.

Therefore,

Area (ΔADO) = Area (ΔCDO)

Hence, proved

(ii) In , since BO is a median

Then,

Area (ΔBAO) = Area (ΔBCO) (i)

In a ΔPAC, since PO is the median

Then,

Area (ΔPAO) = Area (ΔPCO) (ii)

Subtract (ii) from (i), we get

Area (ΔBAO) - Area (ΔPAO) = Area (ΔBCO) - Area (ΔPCO)

Area (ΔABP) = Area (ΔCBP)

Hence, proved

In ADF and ECF

We have,

∠ADF = ∠ECF

AD = EC

And,

∠DFA = ∠CFA

So, by AAS congruence rule,

Δ ADF ≅ Δ ECF

Area (ΔADF) = Area (ΔECF)

DF = CF

BF is a median in ΔBCD

Area (ΔBCD) = 2 Area (ΔBDF)

Area (ΔBCD) = 2 * 3

= 6cm²

Hence, Area of parallelogram ABCD = 2 Area (ΔBCD)

= 2 * 6

= 12 cm²

In ΔPOA and QOC, we have

∠AOP = ∠COQ (Vertically opposite angle)

OA = OC (Diagonals of parallelogram bisect each other)

∠PAC = ∠QCA (AB ‖ DC, alternate angles)

So, by ASA congruence rule, we have

ΔPOA ≅ ΔQOC

Area (ΔPOA) = Area (ΔQOC)

Hence, proved

Construction: Draw FG perpendicular to AB

Proof: We have,

BE = 2 EA

And,

DF = 2FC

AB - AE =2 AE

And,

DC - FC = 2 FC

AB = 3 AE

And,

DC = 3 FC

AE = AB and FC = DC (i)

But,

AB = DC

Then,

AE = FC (Opposite sides of a parallelogram)

Thus,

AE || FC such that AE = FC

Then,

AECF is a parallelogram

Now, Area of parallelogram (AECF) = (AB * FG) [From (i)

3 Area of parallelogram AECF = AB * FG (ii)

And,

Area of parallelogram ABCD = AB * FG (iii)

Compare equation (ii) and (iii), we get

3 Area of parallelogram AECF = Area of parallelogram ABCD

Area of parallelogram AECF = Area of parallelogram ABCD

Hence, proved

(i) We know that each median of a triangle divides it into two triangles of equal area.

Since,

CR is a median of ΔCAP

Therefore,

Area (ΔCRA) = Area (ΔCAP) (i)

Also,

CP is a median of ΔCAB

Therefore,

Area (ΔCAP) = Area (ΔCPB) (ii)

From (i) and (ii), we get

Therefore,

Area (ΔARC) = Area (ΔCPB) (iii)

PQ is a median of ΔPBC

Therefore,

Area (ΔCPB) = 2 Area (ΔPQB) (iv)

From (iii) and (iv), we get

Area (ΔARC) = Area (ΔPBQ) (v)

(ii) Since QP and QR medians of ΔQAB and QAP respectively.

Area (ΔQAP) = Area (ΔQBP) (vi)

And,

Area (ΔQAP) = 2 Area (ΔQRP) (vii)

From (vi) and (vii), we get

Area (ΔPRQ) = Area (ΔPBQ) (viii)

From (v) and (viii), we get

Area (ΔPRQ) = Area (ΔARC) (ix)

(iii) Since CR is a median of ΔCAP

Therefore,

Area (ΔARC) = Area (ΔCAP)

= * Area (ΔABC) (Therefore, CP is a median of Δ ABC)

= Area (ΔABC) (x)

Since,

RQ is a median of Δ RBC.

Therefore,

Area (ΔRQC) = Area (ΔRBC)

=[Area (ΔABC) – Area (ΔARC)]

= [Area (ΔABC) - Area ()]

= Area (Δ ABC)

Given: ABCD is a parallelogram in which

AG = 2 GB

CE = 2 DE

BF = 2 FC

(i) Since ABCD is a parallelogram, we have AB ‖ CD and AB = CD

Therefore,

BG = AB

And,

DE = CD = AB

Therefore,

BG = DE

ADEH is a parallelogram (Since, AH is parallel to DE and AD is parallel to HE)

Area of parallelogram ADEH = Area of parallelogram BCIG (i)

(Since, DE = BG and AD = BC parallelogram with corresponding sides equal)

Area (ΔHEG) = Area (ΔEGI) (ii)

(Diagonals of a parallelogram divide it into two equal areas)

From (i) and (ii), we get,

Area of parallelogram ADEH + Area (ΔHEG) = Area of parallelogram BCIG + Area (ΔEGI)

Therefore,

Area of parallelogram ADEG = Area of parallelogram GBCE

(ii) Height, h of parallelogram ABCD and ΔEGB is the same

Base of ΔEGB = AB

Area of parallelogram ABCD = h * AB

Area (EGB) = * AB * h

= (h) * AB

= * Area of parallelogram ABCD

(iii) Let the distance between EH and CB = x

Area (EBF) = * BF * x

= * BC * x

= * BC * x

Area (EFC) = * CF * x

= * * BC * x

= * Area (EBF)

Area (EFC) = * Area (EBF)

(iv) As, it has been proved that

Area (EGB) = = * Area of parallelogram ABCD (iii)

Area ( = Area ()

Area ( = * * CE * EP

= * * * CD * EP

= * * Area of parallelogram ABCD

Area ( = * Area ( [By using (iii)]

Area ( = Area (

(i) ΔAYC and Δ BCY are on the same base CY and between the same parallels

CY || AB

Area (ΔAYC) = Area (ΔBCY)

(Triangles on the same base and between the same parallels are equal in area)

Subtracting ΔCXY from both sides we get,

Area (ΔAYC) – Area (ΔCXY) = Area (ΔBCY) – Area (ΔCXY) (Equals subtracted from equals are equals)

Area (ΔCBX) = Area (ΔAXY)

(ii) Since, ΔACC and ΔADE are on the same base AF and between the same parallels

CD || AF

Then,

Area ( = Area ()

Area () + Area ( = Area () + Area (

Area ( = Area () (i)

(iii) Since, ΔCBY and ΔCAY are on the same base CY and between the same parallels

CY || BA

Then,

Area () = Area ()

Adding Area ( on both sides we get

Area ( + Area ( = Area ( + Area ()

Area ( = Area ( (ii)

Compare (i) and (ii), we get

Area ( = Area (

Given that,

PSDA is a parallelogram

Since,

AP ‖ BQ ‖ CR ‖ DS and AD ‖ PS

Therefore,

PQ = CD (i)

In

C is the mid-point of BD and CF ‖ BE

Therefore,

F is the mid-point of ED

EF = PE

Similarly,

PE = FD (ii)

In PQE and , we have

PE = FD

∠EPQ = ∠FDC (Alternate angle)

And,

PQ = CD

So, by SAS theorem, we have

Area (Δ PQE) = Area (Δ CFD)

Hence, proved

(i) Join DY and extend it to meet AB produced at P

∠BYP = ∠CYD (Vertically opposite angles)

∠DCY = ∠PBY (Since DC || AP)

BY = CY (Since Y is the mid-point of BC)

Hence, by A.S.A. congruence rule

ΔBYP ≅ ΔCYD

DY = YP

And,

DC = BP

Also,

X is the mid-point of AD

Therefore,

XY || AP

And,

XY = AP

XY = (AB + BP)

XY = (AB + DC)

XY = (60 + 40)

= × 100

= 50 cm

(ii) We have,

XY || AP

XY || AB and AB || DC

XY || DC

DCYX is a trapezium.

(iii) Since X and Y are the mid-points of AD and BC respectively

Therefore,

Trapezium DCYX and ABYX are of same height and assuming it as 'h' cm

Area (Trapezium DCYX) = (DC + XY) * h

= (40 + 50) h

= 45h cm²

Area (Trapezium ABYX) = (AB + XY) * h

= (60 + 50) * h

= 55h cm²

So,

=

=

Area of trapezium​ DCYX = Area of trapezium ABXY

Given that,

ABC and BDF are two equilateral triangles

Let,

AB = BC = CA = x

Then,

BD = = DE = BF

(i) We have,

Area ( = x²

Area ( = ()²

= * x²

Area ( = Area (

(ii) It is given that triangles ABC and BED are equilateral triangles

∠ACB = ∠DBE = 60^o

BE ‖ AC (Since, alternate angles are equal)

Triangles BAF and BEC re on the same base BE and between the same parallels BE and AC

Therefore,

Area ( = Area (

Area ( = 2 Area ( (Therefore, ED is the median)

Area ( = Area (BAE)

(iii) Since,

are equilateral triangles

Therefore,

∠ABC = 60^o and,

∠BDE = 60^o

∠ABC = ∠BDE

AB ‖ DE

Triangles BED and AED are on the same base ED and between the same parallels AB and DE

Therefore,

Area ( = Area (

Area ( - Area ( = Area ( - Area (

Area ( = Area (

(iv) Since,

ED is the median of

Therefore,

Area ( = 2 Area (

Area ( = 2 * Area (

Area ( Area (

Area ( = 2 Area (

(v) Let h be the height of vertex E, corresponding to the side BD on

Let H be the vertex A, corresponding to the side BC in

From part (i), we have

Area ( = Area (

* BD * h = ( * BC * H)

= (2 BD * H)

h = H (1)

From part (iii), we have

Area ( = Area (

= * PD * H

= * FD * 2h

= 2 ( * FD * h)

= 2 Area (

(vi) Area ( = Area ( + Area (

= Area ( + Area (

[Using part (iii) and AD is the median of Area

= Area ( + * 4 Area ( [Using part (i)]

Area ( = 2 Area ( (2)

Area ( = Area ( + Area (

2 Area ( + Area (

3 Area ( (3)

From above equations,

Area (= 2 Area ( + 2 * 3 Area ()

= Area ()

Hence,

Area ( = Area (

Join A and D to get AD median

(Median divides the triangle into two triangles of equal area)

Therefore,

Area (ABD) = Area (ABC)

Now,

Join A and E to get AE median

Similarly,

We can prove that,

Area (ABE) = Area (ABD)

Area (ABE) = ABC (Area (ABD) = Area (ABC) (i)

Join B and O and we get BO median

Now,

Area (BOE) = Area (ABE)

Area (BOE) = * Area (ABC)

Area (BOE) = Area (ABC)

In a ΔAXP and ΔCXB,

∠PAX = XCB (Alternative angles AP || BC)

AX = CX (Given)

∠AXP = ∠CXB (Vertically opposite angles)

ΔAXP ≅ ΔCXB (By ASA rule)

AP = BC (By c.p.c.t) (i)

Similarly,

QA = BC (ii)

From (i) and (ii), we get

AP = QA

Now,

AP || BC

And,

AP = QA

Area (APB) = Area (ACQ) (Therefore, Triangles having equal bases and between the same parallels QP and BC)

(i) In and

∠PEA = ∠QFD (Corresponding angle)

∠EPA = ∠FQD (Corresponding angle)

PA = QD (Opposite sides of a parallelogram)

Then,

(By AAS congruence rule)

Therefore,

(c.p.c.t)

Since, and stand on equal bases PE and FQ lies between the same parallel EQ and FQ lies between the same parallel EQ and AD

Therefore,

() = Area (QFD) (1)

Since,

stand on the same base PF and lie between the same parallel PF and AD

Therefore,

Area ( = Area ( (2)

Divide the equation (1) by (2), we get

=

(iii) From (i) part,

Then,

Since,

A diagonal of parallelogram divides it into two triangles of equal area

Therefore,

Area ( = Area ()

Area ( + Area of parallelogram DLOP + Area ()

Area ( + Area of parallelogram DLOP + Area ( (i)

Since,

AO and CO are diagonals of parallelograms AMOP and OQCL respectively

Therefore,

Area ( = Area ( (ii)

Area ( = Area ( (iii)

Subtracting (ii) from (iii), we get

Area of parallelogram DLOP = Area of parallelogram BMOQ.

(i) Clearly, triangles LMB and LMC are on the same base LM and between the same parallels LM and BC.

Therefore,

Area ( = Area () (1)

(ii) We observe that triangles LBC and MBC are on the same base BC and between the same parallels LM and BC.

Therefore,

Area (ΔLBC) = Area (ΔMBC) (2)

(iii) We have,

Area ( = Area ( [From (i)]

Area ( + Area ( = Area ( + Area (

Area ( = Area (

(iv) We have,

Area ( = Area ( [From (ii)]

Area (LBC) - Area ( = Area ( - Area (

Area ( = Area ()

Draw a line through A parallel to BC

Given that,

BD = BE = EC

We observed that the triangles ABD and AEC are on the same base and between the same parallels l and BC

Therefore, their areas are equal

Hence,

Area () = Area () = Area ()

(i) In , we have

MB = AB

BC = BD

And,

∠ MBC = ∠ ABD (Therefore, ∠MBC and ∠ABC are obtained by adding ∠ABC to right angle)

So, by SAS congruence rule, we have

Area () = Area ( (1)

(ii) Clearly, and rectangle BYXD are on the same base BD and between the same parallels AX and BD

Therefore,

Area ( = Area of rectangle BYXD

Area of rectangle BYXD = 2 Area ()

Area of rectangle BYXD = 2 Area () (2)

[Therefore, Area ( = Area (] From (1)

(iii) Since,

and square MBAN are on the same base MB and between the same parallel MB and NC

Therefore,

2 Area () = Area of square MBAN (3)

From (2) and (3), we have

Area of square MBAN = Area of rectangle BXYD

(iv) In , we have

FC = AC

CB = CE

And,

∠FCB = ∠ACE (Therefore, ∠FCB and ∠ACE are obtained by adding ∠ACB to a right angle)

So, by SAS congruence rule, we have

(v) We have,

Area ( Area (

Clearly,

and rectangle CYXE are on the same base CE and between the same parallel CE and AX

Therefore,

2 Area ( = Area of rectangle CYXE

2 Area ( = Area of rectangle CYXE (4)

(vi) Clearly,

and rectangle FCAG are on the same base FC and between the same parallels FC and BG

Therefore,

2 Area ( = Area of rectangle FCAG (5)

From (4) and (5), we get

Area of rectangle CYXE = Area of rectangle ACFG

(vii) Applying Pythagoras theorem in , we have

BC² = AB² + AC²

BC * BD = AB * MB + AC * FC

Area of rectangle BCED = Area of rectangle ABMN + Area of rectangle ACFG

are equilateral triangles

We know that,

Area of equilateral triangle = a²

D is the mid-point of BC then,

Area () = * ()²

= *

Now,

Area (: Area (

* a²: *

1:

4: 1

Hence,

Area (: Area ( is 4: 1

Since, the two opposite line are joined by two another lines connecting the end points.

Since, quadrilateral is simple closed figure of four line segments.

Given that,

ABCD is a rectangle

CD = 6 cm

AD = 8 cm

We know that,

Area of parallelogram and rectangle on the same base between the same parallels are equal in area

So,

Area of parallelogram CDEF and rectangle ABCD on the same base and between the same parallels, then

We know that,

Area of parallelogram = Base * Height

Area of rectangle ABCD = Area of parallelogram

= AB * AD

= CD * AD (Therefore, AB = CD)

= 6 * 8

= 48 cm²

Hence,

Area of rectangle ABCD is 48 cm²

Since, the quadrilateral with opposite angles equal is a parallelogram.

Given that,

ABCD is rectangle

CD = 6 cm

AD = 8 cm

We know that,

If a rectangle and a parallelogram are on the same base and between the same parallels, then the area of a triangle is equal to the half of parallelogram

So, triangle GEF and parallelogram ABCD are on the same base and between same parallels, then we know

Area of parallelogram = Base * Height

Now,

Area ( = * Area of ABCD

= * AB * AD

= * 6 * 8

= 24 cm²

Hence,

Area ( is 24 cm²

One angle equalling to 60° need not necessarily be a rhombus.

We know that,

If a triangle and a parallelogram are on the same base and between the same parallel then the angle of triangle is equal to the half of the parallelogram

So, triangle EPG and parallelogram ABCD are on the same base and between same parallels. The

We know that,

Area of parallelogram = Base * Height

Now,

Area ( = * Area of ABCD

= * AB * AD

= * 10 * 5

= 25 cm²

Hence,

Area ( is 25 cm²

Each angle measures 45° each after the diagonal bisects them.

Given that,

PQRS is a rectangle

PS = 5 cm

PR = 13 cm

In triangle PSR, by using Pythagoras theorem

SR² = PR² – PS²

SR² = (13)² – (5)²

SR² = 169 – 25

SR² = 114

SR = 12 cm

We have to find the area ,

Area ( = * Base * Height

= * SR * PS

= * 12 * 5

= 30 cm²

Hence, Area ( is 30 cm²

Let ABCD is a rectangle

AC and BD are the diagonals of rectangle

In ΔABC and ΔBCD, we have

AB = CD (Opposite sides of rectangle are equal)

∠ABC = ∠BCD (Each equal to 90^o)

BC = BC (Common)

Therefore,

ABC BCD (By SAS congruence criterion)

AC = BD (c.p.c.t)

Hence, the diagonals of a rectangle are equal.

Given: ABCD is a square

P and Q are the mid points of AB and CD respectively.

AB = 8cm

PQ and BD intersect at O

Now,

AP = BP = AB

AP = BP = * 8

= 4 cm

AB = AD = 8 cm

QP ‖ AD

Then,

AD = QP

So,

OP = AD

OP = * 8

= 4 cm

Now,

Area ( = * BP * PO

= * 4 * 4

= 8 cm²

Hence, Area ( is 8 cm²

Let ABCD is a rectangle such as AB = CD and BC = DA

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

Construction: Join AC and BD

In

P and Q are the mid-points of AB and BC respectively

Therefore,

PQ ‖ AC and PQ = AC (Mid-point theorem) (i)

Similarly,

In

SR ‖ AC and SR = AC (Mid-point theorem) (ii)

Clearly, from (i) and (ii)

PQ ‖ SR and PQ = SR

Since, in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, it is a parallelogram.

Therefore,

PS ‖ QR and PS = QR (Opposite sides of a parallelogram) (iii)

In

Q and R are the mid-points of side BC and CD respectively

Therefore,

QR ‖ BD and QR = BD (Mid-point theorem) (iv)

However, the diagonals of a rectangle are equal

Therefore,

AC = BD (v)

Now, by using equation (i), (ii), (iii), (iv), and (v), we obtain

PQ = QR = SR = PS

Therefore, PQRS is a rhombus.

Given that,

D, E, F are the mid-points of BC, DC, AE respectively

Let, AD is median of triangle ABC

Area ( = Area (

= * 16

= 8 cm²

Now, AE is a median of

Area ( = Area (

= * 8

= 4 cm²

Again,

DE is the median of

Area (F) = Area (

= * 4

= 2 cm²

Let, ABCD is a parallelogram

OA and OD are the bisectors of adjacent angles ∠A and ∠D

As, ABCD is a parallelogram

Therefore,

AB || DC (Opposite sides of the parallelogram are parallel)

AB || DC and AD is the transversal,

Therefore,

∠BAD + ∠CDA = 180^o (Sum of interior angles on the same side of the transversal is 180^o)

∠1 + ∠2 = 90° (AO and DO are angle bisectors ∠A and ∠D) (i)

In ΔAOD,

∠1 + ∠AOD + ∠2 = 180°

∠AOD + 90° = 180° [From (i)]

∠AOD = 180° – 90°

= 90°

Therefore,

In a parallelogram, the bisectors of the adjacent angles intersect at right angle.

Given that,

Area ( = 17 cm²

PQRS is a trapezium

PS ‖ QR

A and B are points on PQ and RS respectively

AB ‖ QR

We know that,

If a triangle and a parallelogram are on the same base and between the same parallels, the area of triangle is equal to half area of parallelogram

Here,

Area ( = Area ( (i)

Area ( = Area ( (ii)

We have to find Area (,

Area ( = Area ( + Area (

= Area ( + Area (

= Area (

= 17 cm²

Hence,

Area ( is 17 cm².

Let, ABCD is a parallelogram.

AE bisects ∠BAD and BF bisects ∠ABC

Also,

CG bisects ∠BCD and DH bisects ∠ADC

To prove: LKJI is a rectangle

Proof: ∠BAD + ∠ABC = 180^o (Because adjacent angles of a parallelogram are supplementary)

ΔABJ is a right triangle

Since its acute interior angles are complementary

Similarly,

In CDL, we get

∠DLC = 90°

In ADI, we get

∠AID = 90°

Then,

∠JIL = 90^o because ∠AID and ∠JIL are vertical opposite angles

Since three angles of quadrilateral LKJI are right angles, hence 4th angle is also a right angle

Thus LKJI is a rectangle.

Given that,

CQ: QP = 3: 1

Let,

CQ = 3x

PQ = x

Area ( = 10 cm²

We know that,

Area of triangle = * Base * Height

Area ( = * x * h

10 = * x * h

x * h = 20

Area ( = * 3x * h

= * 3 * 20

= 30 cm²

Now,

Area ( = * PB * H = 30 cm²

PB * H = 60 cm²

We have to find area of parallelogram

We know that,

Area of parallelogram = Base * Height

Area ( = AB * H

Area ( = 2 BP * H

Area (ABCD) = 2 (60)

Area (ABCD) = 120 cm²

Hence,

Area of parallelogram ABCD is 120 cm²

Given that,

ABCD is a quadrilateral and P, Q, R and S are the mid points of the sides AB, BC, CD and DA respectively

To prove: PQRS is a parallelogram

Construction: Join A with C

Proof: In ,

P and Q are the mid-points of AB and BC respectively

Therefore,

PQ ‖ AC and PQ = AC (Mid-point theorem) (i)

Again,

In ,

R and S are mid-points of sides CD and AD respectively

Therefore,

SR ‖ AC and SR = AC (Mid-point theorem) (ii)

From (i) and (ii), we get

PQ ‖ SR and PQ = SR

Hence, PQRS is a parallelogram (One pair of opposite sides is parallel and equal)

Given that,

Area ( = 12 cm²

D is the mid-point of BC

So,

AD is the median of triangle ABC,

Area ( = Area ( = * Area (

Area ( = Area ( = * 12

= 6 cm² (i)

We know that,

Area of triangle between the same parallel and on the same base

Area ( = Area (

Area ( + Area ( = Area ( + Area (

Area ( = Area ( (ii)

ME is the median of triangle ADC,

Area ( = Area ( + Area (

Area ( = Area ( + Area ( [From (ii)]

Area ( = Area (

6 cm² = Area ( [From (i)]

Hence,

Area ( is 6 cm².

Given: ABCD is a rectangle and P, Q, R, S are their midpoints

To Prove: PQRS is a rhombus

Proof: In ABC,

P and Q are the mid points

So, PQ is parallel AC

And,

PQ = AC (The line segment joining the mid points of 2 sides of the triangle is parallel to the third side and half of the third side)

Similarly,

RS is parallel AC

And,

RS = AC

Hence, both PQ and RS are parallel to AC and equal to AC.

Hence, PQRS is a parallelogram

In triangles APS & BPQ,

AP = BP (P is the mid-point of side AB)

∠PAS = ∠PBQ (90^o each)

AS = BQ (S and Q are the mid points of AD and BC respectively and since opposite sides of a rectangle are equal, so their halves will also be equal)

APS BPQ (By SAS congruence rule)

PS=PQ (By c.p.c.t.)

PQRS is a parallelogram in which adjacent sides are equal.

Hence, PQRS is a rhombus.

To prove: That the quadrilateral formed by joining the mid points of sides of a rhombus is a rectangle.

ABCD is a rhombus P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.

Construction: Join AC

Proof: In ΔABC, P and Q are the mid points of AB and BC respectively

Therefore,

PQ || AC and PQ = AC (i) (Mid-point theorem)

Similarly,

RS || AC and RS = AC (ii) (Mid-point theorem)

From (i) and (ii), we get

PQ ‖ RS and PQ = RS

Thus, PQRS is a parallelogram (A quadrilateral is a parallelogram, if one pair of opposite sides is parallel and equal)

AB = BC (Given)

Therefore,

AB = BC

PB = BQ (P and Q are mid points of AB and BC respectively)

In ΔPBQ,

PB = BQ

Therefore,

∠BQP = ∠BPQ (iii) (Equal sides have equal angles opposite to them)

In ΔAPS and ΔCQR,

AP = CQ (AB = BC = AB = BC = AP = CQ)

AS = CR (AD = CD = AD = CD = AS = CR)

PS = RQ (Opposite sides of parallelogram are equal)

Therefore,

APS CQR (By SSS congruence rule)

∠APS = ∠CQR (iv) (By c.p.c.t)

Now,

∠BPQ + ∠SPQ + ∠APS = 180°

∠BQP + ∠PQR + ∠CQR = 180°

Therefore,

∠BPQ + ∠SPQ + ∠APS = ∠BQP + ∠PQR + ∠CQR

∠SPQ = ∠PQR (v) [From (iii) and (iv)]

PS || QR and PQ is the transversal,

Therefore,

∠SPQ + ∠PQR = 180^o (Sum of adjacent interior an angles is 180^o)

∠SPQ + ∠SPQ = 180° [From (v)]

2 ∠SPQ = 180°

∠SPQ = 90°

Thus, PQRS is a parallelogram such that ∠SPQ = 90^o

Hence, PQRS is a rectangle.

Let ABCD is a square such that AB = BC = CD = DA, AC = BD and P, Q, R and S are the mid points of the sides AB, BC, CD and DA respectively.

In

P and Q are the mid-points of AB and BC respectively.

Therefore,

PQ ‖ AC and PQ = AC (Mid-point theorem) (i)

Similarly,

In

SR ‖ AC and SR = AC (Mid-point theorem) (ii)

Clearly,

PQ ‖ SR and PQ = SR

Since, in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other. Hence, it is a parallelogram.

Therefore,

PS ‖ QR and PS = QR (Opposite sides of a parallelogram) (iii)

In

Q and R are the mid-points of sides BC and CD respectively

Therefore,

QR ‖ BD and QR = BD (Mid-point theorem) (iv)

However, the diagonals of a square are equal

Therefore,

AC = BD (v)

By using equation (i), (ii), (iii), (iv) and (v), we obtain

PQ = QR = SR = PS

We know that, diagonals of a square are perpendicular bisector of each other

Therefore,

∠AOD = ∠AOB = ∠COD = ∠BOC = 90^o

Now, in quadrilateral EHOS, we have

SE || OH

Therefore,

∠AOD + ∠AES = 180^o (Corresponding angle)

∠AES = 180° - 90°

= 90°

Again,

∠AES + ∠SEO = 180^o (Linear pair)

∠SEO = 180° - 90°

= 90°

Similarly,

SH || EO

Therefore,

∠AOD + ∠DHS = 180^o (Corresponding angle)

∠DHS = 180° - 90° = 90°

Again,

∠DHS + ∠SHO = 180° (Linear pair)

∠SHO = 180° - 90°

= 90°

Again,

In quadrilateral EHOS, we have

∠SEO = ∠SHO = ∠EOH = 90°

Therefore, by angle sum property of quadrilateral in EHOS, we get

∠SEO + ∠SHO + ∠EOH + ∠ESH = 360°

90^o + 90^o + 90^o + ∠ESH = 360°

∠ESH = 90°

In the same manner, in quadrilateral EFOP, FGOQ, GHOR, we get

∠HRG = ∠FQG = ∠EPF = 90°

Therefore, in quadrilateral PQRS, we have

PQ = QR = SR = PS and ∠ESH = ∠HRG = ∠FQG = ∠EPF = 90°

Hence, PQRS is a square.

Let ABCD be a quadrilateral, with points

E, F, G and H the midpoints of

AB, BC, CD, DA respectively.

(I suggest you draw this, and add segments EF, FG, GH, and HE, along with diagonals AC and BD)

EF = AB (Definition of midpoint)

Similarly,

BF = BC

Thus, triangle BEF is similar to triangle BAC (SAS similarity)

Therefore EF is half the length of diagonal AC, since that's the proportion of the similar triangles.

Similarly, we can show that triangle DHG is similar to triangle DAC,

Therefore,

HG is half the length of diagonal AC

So,

EF = HG

Similarly,

We can use similar triangles to prove that EH and FG are both half the length of diagonal BD, and therefore equal

This means that both pairs of opposite sides of quadrilateral EFGH are equal, so it is a parallelogram.

Let the small angle be = x

Then the second angle = 2x - 24°

Since,

Opposite angles are equal the 4 angles will be x, 2x - 24° , x , 2x - 24°

So now by angle sum property:

x + 2x - 24° + x + 2x - 24° = 360°

6x - 48° = 360°

6x = 360° + 48°

6x = 408°

x =

x = 68°

Thus, the smallest angle is 68°

The second angle = 2 (68°) - 24°

= 112°

We know that,

The opposite angles of a parallelogram are equal

Therefore,

∠BCD = ∠BAD = 75°

Now, in ∆ BCD, we have

∠CDB + ∠DBC + ∠BCD = 180° (Since, sum of the angles of a triangle is 180^o)

∠CDB + 60° + 75° = 180°

∠CDB + 135° = 180°

∠CDB = (180° - 135°) = 45°

Given: ABCD is a parallelogram

E and F are the centroids of triangle ABD and BCD

Since, the diagonals of parallelogram bisect each other

AO is the median of triangle ABD

And,

CO is the median of triangle CBD

EO= AO (Since, centroid divides the median in the ratio 2:1)

Similarly,

FO = CO

EO + FO = AO + CO

= (AO + CO)

EF = AC

AE = AO

= * AC

= AC

Therefore,

EF = AE

ABCD is a parallelogram. BD is the diagonal and M is the mid-point of BD.

BD is a bisector of ∠B

We know that,

Diagonals of the parallelogram bisect each other

Therefore,

M is the mid-point of AC

AB || CD and BD is the transversal,

Therefore,

∠ ABD = ∠ BDC (i) (Alternate interior angle)

∠ ABD = ∠ DBC (ii) (Given)

From (i) and (ii), we get

∠ BDC = ∠ DBC

In Δ BCD,

∠ BDC = ∠ DBC

BC = CD (iii) (In a triangle, equal angles have equal sides opposite to them)

AB = CD and BC = AD (iv) (Opposite sides of the parallelogram are equal)

From (iii) and (iv), we get

AB = BC = CD = DA

Therefore,

ABCD is a rhombus

∠AMB = 90° (Diagonals of rhombus are perpendicular to each other)

ABCD is a parallelogram. E is the midpoint of BC. So, BE = CE

DE produced meets the AB produced at F

Consider the triangles CDE and BFE

BE = CE (Given)

∠CED = ∠BEF (Vertically opposite angles)

∠DCE = ∠FBE (Alternate angles)

Therefore,

ΔCDE ≅ ΔBFE

So,

CD = BF (c.p.c.t)

But,

CD = AB

Therefore,

AB = BF

AF = AB + BF

AF = AB + AB

AF = 2 AB

Since the adjacent angle of a parallelogram are supplementary.

Hence,

x + x = 180°

x = 180°

x = 108°

Now,

x = * 108°

= 72°

The total must be equal to 360° (Sum of quadrilaterals)

So,

4x + 7x + 9x + 10x = 360°

30x = 360°

x = 12°

Now,

Substitute x = 12 into 4x (Smallest) + 10x (Biggest)

So,

(4 * 12) + (10 * 12)

= 48 + 120

= 168 so the answer is C

Given that,

∠A = 140°

∠D = 60°

According to question,

∠A + ∠C = 2 (∠B + ∠D)

140 + ∠C = 2 (∠B + 60°)

∠B = (∠C) + 10° (i)

We know,

∠A + ∠B + ∠C + ∠D= 360°

140° + (∠C) + 10° + ∠C + 60° = 360°

∠C = 150°

∠C = 100°

∠B = (100°) + 10°

= 60°

ABCD is a rhombus

AC = 18cm

BD = 24cm

We have to find the sides of the rhombus

In triangle AOB,

AO = 9cm (Diagonals of a parallelogram bisect each other)

BO = 12cm

AOB is a right - triangle right angled at O (Diagonals of a rhombus are perpendicular to each other)

So,

AB² = AO² + BO² (By Pythagoras theorem)

AB² = 9² + 12²

AB² = 81 + 144

AB² = 225

AB = 15cm

In a rhombus, all sides are equal

Thus, each side of the rhombus is 15cm.

Given that,

∠ABD = ∠ABP = 50°

∠PBC +∠ABP = 90° (Each angle of a rectangle is a right angle)

∠PBC = 40°

Now,

PB = PC (Diagonals of a rectangle are equal and bisect each other)

Therefore,

∠BCP = 40° (Equal sides has equal angle)

In triangle BPC,

∠BPC + ∠PBC + ∠BCP = 180° (Angle sum property of a triangle)

∠BPC = 100°

∠BPC + ∠DPC = 180° (Angles in a straight line)

∠DPC = 180^o – 100^o

= 80°

Given: ABCD is a parallelogram

AC is a bisector of angle BAD

∠BAC = 35°

∠A = 2 ∠BAC

∠A = 2 (35°)

∠A = 70°

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

70 + ∠B = 180°

∠B = 110^o

The diagonals in a rhombus are perpendicular,

So,

∠BPC = 90^o

From triangle BPC,

The sum of angles is 180°

So,

∠CBP = 180^o – 40^o – 90^o

= 50°

Since, triangle ABC is isosceles

We have,

AB = BC

So,

∠ACB = ∠CAB = 40^o

Again from triangle APB,

∠PBA = 180^o – 40^o – 90^o

= 50^o

Again, triangle ADB is isosceles,

So,

∠ADB = ∠DBA = 50^o

∠ADB = 50^o

Since, the triangle formed by joining the mid points of the triangle would be similar to it hence the angles would be equal to the outer triangle's angles.

∠BOC is 90°

So,

∠COD and ∠AOB all should be 90° by linear pair

∠BDC is 50°,

So,

Now as in a parallelogram the opposite​ sides are equal

We say,

AB parallel to CD

∠DCA = 50°

So,

In triangle COA

∠C = 50° (Stated above)

∠COA = 90° (Proved above)

Therefore,

90° + 50° + x° = 180°

x = 40°

Construction: Join A to C

Mark the intersection point of AC and MN as O

Now, M and N are mid points of the non-parallel of a trapezium

Therefore,

MN || AB || DC

So,

MO || BC

And,

M is a mid-point of AD

Therefore,

MO= BC

Similarly,

NO = AB

Therefore,

MN = MO + NO

= (AB + CD)

But,

MN = 14 cm

Hence,

(AB + CD) = 14 cm

12 + CD = 28

CD = 16 cm

Since, diagonals of quadrilateral bisect each other

Hence, it's a parallelogram

We know,

The adjacent angles of parallelogram are supplementary

Therefore,

∠A + ∠B = 180°

45° + ∠B = 180°

∠B = 135°

Given that,

ABCD is a parallelogram

P is the mid-point of BC

∠DAP =∠PAB =x

AD=10 cm

To find: The length of CD

∠ABP = 180 - 2x (Co interior angle of parallelogram)

∠APB = 180^o - (180^o - 2x + x) = x

Therefore,

In triangle ABP,

∠APB =∠PAB = x

Therefore,

AB = PB (In a triangle sides opposite to equal angles are equal in length)

CD = AB = PB = = = = 5 cm

Complete the parallelogram ADCP

So the diagonals DP & AC bisect each other at O

Thus O is the midpoint of AC as well as DP (i)

Since ADCP is a parallelogram,

AP = DC

And,

AP parallel DC

But,

D is mid-point of BC (Given)

AP = BD

And,

AP parallel BD

Hence,

BDPA is also a parallelogram.

So, diagonals AD & BP bisect each other at E (E being given mid-point of AD)

So, BEP is a single straight line intersecting AC at F

In triangle ADP,

E is the mid-point of AD and

O is the midpoint of PD.

Thus, these two medians of triangle ADP intersect at F, which is centroid of triangle ADP

By property of centroid of triangles,

It lies at of the median from vertex

So,

AF = AO (ii)

So,

From (i) and (ii),

AF = * * AC

= AC

=

= 3.5 cm