Areas Of Parallelograms And Triangles

(i) Yes, It can be observed that trapezium ABCD and triangle PCD have a common base CD and these are lying between the same parallel lines AB and CD

(ii) No, It can be observed that parallelogram PQRS and trapezium MNRS have a common base RS. However, their vertices, (i.e., opposite to the common base) P, Q of parallelogram and M, N of trapezium, are not lying on the same line

(iii) Yes, It can be observed that parallelogram PQRS and triangle TQR have a common base QR and they are lying between the same parallel lines PS and QR

(iv) No, It can be observed that parallelogram ABCD and triangle PQR are lying between same parallel lines AD and BC. However, these do not have any common base

(v) Yes, It can be observed that parallelogram ABCD and parallelogram APQD have a common base AD and these are lying between the same parallel lines AD and BQ

(vi) No, It can be observed that parallelogram PBCS and PQRS are lying on the same base PS. However, these do not lie between the same parallel lines.

We know area of parallelogram = Base × Height

For ABCD,

AE ⊥ DC

Hence AE is the height and DC is the base.

Area of ABCD = DC × AE

As we know sides of parallelogram are equal.

So, DC=AB=16 cm

Hence,

Area of ABCD = 16×8 …. (1)

Also, for ABCD,

CF ⊥ AD

Area of ABCD = CF × AD

= 10 × AD …. (2)

From 1 and 2,

16×8 = 10 × AD

=12.8 cm

Let us join HF

In parallelogram ABCD,

AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)

AB = CD (Opposite sides of a parallelogram are equal)

AD = BC

And AH || BF

AH = BF and AH || BF (H and F are the mid-points of AD and BC)

Therefore, ABFH is a parallelogram

Since,

ΔHEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF

Therefore,

Area of triangle HEF = × Area (ABFH) (i)

Similarly,

It can also be proved that

Area of triangle HGF = × Area (HDCF) (ii)

On adding (i) and (ii), we get

Area of triangle HEF + Area of triangle HGF = Area (ABFH) + Area (HDCF)

Area (EFGH)= [Area (ABFH) + Area (HDCF)]

Area (EFGH) = Area (ABCD)

It can be observed that ΔBQC and parallelogram ABCD lie on the same base BC and these are between the same parallel lines AD and BC

Area (ΔBQC) = Area (ABCD) (i)

Similarly,

ΔAPB and parallelogram ABCD lie on the same base AB
and between the same parallel lines AB and DC

Area (ΔAPB) = Area (ABCD) (ii)

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

Area (ΔBQC) = Area (ΔAPB)

(i) Let us draw a line segment EF, passing through point P and parallel to line segment AB.

In parallelogram ABCD,

AB || EF (By construction) ... (i)

AD || BC (Opposite sides of a parallelogram)

So, AE || BF ... (ii)

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

Quadrilateral ABFE is a parallelogram.

It can be observed that ΔAPB and parallelogram ABFE are lying on the same base AB and between the same parallel lines AB and EF

Therefore,

Area of triangle APB = × Area (ABFE) ... (iii)

Similarly, we show that EFCD is a parallelogram,

For triangle PCD and parallelogram EFCD, lying on same base CD and same parallel lines,

Area of triangle PCD = × Area (EFCD).....(iv)

Adding (iii) and (iv), we get

Area of triangle APB + Area of triangle PCD = [Area (ABFE) + Area (EFCD)]

Area of triangle APB + Area of triangle PCD = Area (ABCD).... (v)

(ii) Let us draw a line segment MN, passing through point P and parallel to line segment AD

In parallelogram ABCD,

MN || AD (By construction) ... (vi)

ABCD is a parallelogram

AB || DC (Opposite sides of a parallelogram)

So, AM || DN (vii)

From equations (vi) and (vii), we get

MN || AD and

AM || DN

Therefore,

Quadrilateral AMND is a parallelogram.

It can be observed that ΔAPD and parallelogram AMND are lying on the same base AD and between the same parallel lines AD and MN

Area (ΔAPD) = Area (AMND) (viii)

Similarly,

For ΔPCB and parallelogram MNCB,

Area (ΔPCB) = Area (MNCB) (ix)

Adding equations (viii) and (ix), we get

Area (ΔAPD) + Area (ΔPCB) = [Area (AMND) + Area (MNCB)]

Area (ΔAPD) + Area (ΔPCB) = Area (ABCD) (x)

On comparing equations (v) and (x), we get

Area (ΔAPD) + Area (ΔPBC) = Area (ΔAPB) + Area (ΔPCD)

(i) It can be observed that parallelogram PQRS and ABRS lie on the same base SR and also, these lie in between the same parallel lines SR and PB

Area (PQRS) = Area (ABRS) (i)

(ii) Consider ΔAXS and parallelogram ABRS

As these lie on the same base AS and are between the same parallel lines AS and BR,
(ii)

From equations (i) and (ii), we obtain

Hence proved.

From the figure, it can be observed that point A divides the field into three parts. These parts are triangular in shape − ΔPSA, ΔPAQ, and ΔQRA

Area of ΔPSA + Area of ΔPAQ + Area of ΔQRA = Area of Parallelogram PQRS ..............eq(i)

We know that if a parallelogram and a triangle are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram

Therefore,

Area (ΔPAQ) = Area (PQRS) .............eq(ii)

From equations (i) and (ii), we obtain

Area (ΔPSA) + Area (ΔQRA) + 1/2 Area (PQRS) = Area (PQRS)

Area (ΔPSA) + Area (ΔQRA) = 1/2 Area (PQRS) ....(iii)

Clearly, it can be observed that the farmer must sow wheat in triangular part PAQ

and pulses in other two triangular parts PSA and QRA or

wheat in triangular parts PSA and QRA and pulses in triangular parts PAQ.

AD is the median of ΔABC. Therefore, it will divide ΔABC into two triangles of equal areas

Area (ΔABD) = Area (ΔACD) (i)

ED is the median of ΔEBC

Area (ΔEBD) = Area (ΔECD) (ii)

On subtracting equation (ii) from equation (i), we get,

Area (ΔABD) − Area (EBD) = Area (ΔACD) − Area (ΔECD)

Area (ΔABE) = Area (ΔACE)

Given: Δ ABC, with AD as median i.e., BD = CD and E is the mid-point of AD, i.e., AE = DE

To prove: ar (BED) = 1/4 ar (ABC).

Proof: AD is a median of Δ ABC and median divides a triangle into two triangles of equal area

∴ ar(ABD) = ar(ACD)

⇒ ar (ABD) = 1/2 ar (ABC) …(1)

In Δ ABD,

BE is median (As E is mid-point of AD)

median divides a triangle into two triangles of equal area

∴ ar(BED) = ar(BEA)

⇒ ar (BED) = 1/2 ar (ABD)

⇒ ar (BED) = 1/2 × 1/2 ar (ABC) (from (1): ar (ABD) = 1/2 ar (ABC))

⇒ ar (BED) = 1/4 ar (ABC)

Hence proved.

We know that diagonals of a parallelogram bisect each other. Therefore, O is the mid-point of AC and BD

BO is the median in ΔABC. Therefore, it will divide it into two triangles of equal areas

Area (ΔAOB) = Area (ΔBOC) (i) In ΔBCD,

CO is the median

Area (ΔBOC) = Area (ΔCOD) (ii)

Similarly, Area (ΔCOD) = Area (ΔAOD) (iii)

From equations (i), (ii), and (iii), we obtain

Area (ΔAOB) = Area (ΔBOC) = Area (ΔCOD) = Area (ΔAOD)

Therefore, it is evident that the diagonals of a parallelogram divide it into four triangles of equal area

Consider ΔACD.

Line-segment CD is bisected by AB at O.Therefore, AO is the median of ΔACD

Area (ΔACO) = Area (ΔADO) (i)

Considering ΔBCD, BO is the median

Area (ΔBCO) = Area (ΔBDO) (ii)

Adding equations (i) and (ii), we get

Area (ΔACO) + Area (ΔBCO) = Area (ΔADO) + Area (ΔBDO)

Area (ΔABC) = Area (ΔABD)

(i) In triangle ABC it is given that,

EF is parallel to BC

And,

EF = BC ( By using Mid – point theorem)

Also,

BD = BC (As D is the mid-point)

So,

BD = EF

BF and DE also parallel to each other

Therefore, the pair of opposite sides are equal and parallel in length.
Similarly BF = DE

Therefore,

BDEF is a parallelogram

(ii) Diagonal of a parallelogram divides it into two equal area

Therefore,

⇒ Area of triangle BFD = Area of triangle DEF ......(1)
In parallelogram DCEF

⇒ Area of triangle AFE = Area of triangle DEF .....(2)

⇒ Area of triangle CDE = Area of triangle DEF ......(3)

So from (1), (2) and (3)

Area of triangle BFD = Area of triangle AFE = Area of triangle CDE = Area of triangle DEF

4 (Area of triangle DEF) = Area of triangle ABC

Area of triangle DEF = × Area of triangle

(iii) Area of parallelogram BDEF = Area of triangle DEF + Area of triangle BDE

Area of parallelogram BDEF = Area of triangle DEF + Area of triangle DEF

= 2 × Area of triangle DEF

= 2 × × Area of triangle ABC

= × Area of triangle ABC

Let us draw DN perpendicular to AC and BM perpendicular to AC

(i) In ΔDON and ΔBOM,

∠DNO = ∠BMO ( 90°)

∠DON = ∠BOM (Vertically opposite angles)

OD = OB (Given)

By AAS congruence rule,

ΔDON ΔBOM

DN = BM (BY CPCT) ..... (1)

We know that congruent triangles have equal areas.

Area (ΔDON) = Area (ΔBOM) ..... (2)

In ΔDNC and ΔBMA,

∠DNC = ∠BMA (Angles made by perpendicular)

CD = AB (Given)

DN = BM [Using equation (i)]

ΔDNC ΔBMA (RHS congruence rule)

Area (ΔDNC) = Area (ΔBMA) (iii)

Let us draw DN AC and BM AC

On adding equations (ii) and (iii), we obtain

Area (ΔDON) + Area (ΔDNC) = Area (ΔBOM) + Area (ΔBMA)

Therefore,

Area (ΔDOC) = Area (ΔAOB)

(ii) TO prove :Area (ΔDCB) = Area (ΔACB)
we have proved,
Area (ΔDOC) = Area (ΔAOB)

Area (ΔDOC) + Area (ΔOCB) = Area (ΔAOB) + Area (ΔOCB)

(Adding Area (ΔOCB) to both sides)

Area (ΔDCB) = Area (ΔACB)

(iii)
We have proved above,
Area (ΔDCB) = Area (ΔACB)

If two triangles have the same base and equal areas, then these will lie between the same parallels

So, DA || CB (iv)

In quadrilateral ABCD, one pair of opposite sides is equal (AB = CD) and the other pair of opposite sides is parallel (DA || CB)

Therefore, ABCD is a parallelogram

Since ΔBCE and ΔBCD are lying on a common base BC and also have equal areas,

ΔBCE and ΔBCD will lie between the same parallel lines

Therefore,

DE || BC
Hence proved

It is given that:

In Δ ABC, XY||BC & BE||AC & CF||AB

To prove: Ar(Δ ABE) =Ar(Δ ACF)

Proof:

Let XY intersect AB & BC at points M and N respectively.

The figure of the question is:

Now, XY || BC

And, we know that the parts of the parallel lines are parallel

⇒ EN || BC .... (1)

Also, it is given that BE || AC

⇒ BE || CN .... (2)
From 1 and 2

BCNE is a parallelogram (because both the pair of opposite sides are parallel)

Now, the parallelogram BCNE & ΔABE have a common base i.e. BE & BE is parallel to AC.

⇒ ar(ΔABE) = ½ area of parallelogram BCNE ...(i)

Similarly, we BCFM is also a parallelogram.

And, the ΔACF and the parallelogram BCFM lie on the same base i.e. CF

⇒ ar(ΔACF) = ½ area of parallelogram BCFM ...(ii)

Now,

Parallelograms BCNE and BCFM are on the same base BC and between the same parallels BC and EF
hence there area will be equal.

⇒ Area (BCNE) = Area (BCFM)...(iii)

From the equations (i), (ii) and (iii), we can write that,

Ar(Δ ABE) = Ar(Δ ACF)

Let us join AC and PQ

ΔACQ and ΔAQP are on the same base AQ and between the same parallels AQ and CP.

Area (ΔACQ) = Area (ΔAPQ)

Area (ΔACQ) − Area (ΔABQ) = Area (ΔAPQ) − Area (ΔABQ)

Area (ΔABC) = Area (ΔQBP) (i)

Since AC and PQ are diagonals of parallelograms ABCD and PBQR respectively,

Area (ΔABC) = Area (ABCD) (ii)

Area (ΔQBP) = Area (PBQR) (iii)

From equations (i), (ii), and (iii), we obtain

Area (ABCD) = Area (PBQR)

Area (ABCD) = Area (PBQR)

It can be observed that ΔDAC and ΔDBC lie on the same base DC and between the same parallels AB and CD

Area (ΔDAC) = Area (ΔDBC) [Area of triangles with same base and between the same parallels are equal in area.]

Area (ΔDAC) − Area (ΔDOC) = Area (ΔDBC) − Area (ΔDOC)

Area (ΔAOD) = Area (ΔBOC)

(i) ΔACB and ΔACF lie on the same base AC and are between the same parallels AC and BF

Area (ΔACB) = Area (ΔACF)

(ii) It can be observed that:

Area (ΔACB) = Area (ΔACF)

Area (ΔACB) + Area (ACDE) = Area (ACF) + Area (ACDE)

Area (ABCDE) = Area (AEDF)

Let quadrilateral ABCD be the original shape of the field

The proposal may be implemented as follows:

Join diagonal BD and draw a line parallel to BD through point A.

Let it meet the extended side CD of ABCD at point E.

Join BE and AD. Let them intersect each other at O.

Then, portion ΔAOB can be cut from the original field so that the new shape of the field will be ΔBCE

We have to prove that the area of ΔAOB (portion that was cut so as to construct Health Centre)

is equal to the area of ΔDEO (portion added to the field so as to make the area of the new field so formed equal to the area of the original field)

It can be observed that ΔDEB and ΔDAB lie on the same base BD and are between the same parallels BD and AE

So,

Area (ΔDEB) = Area (ΔDAB)

Area (ΔDEB) − Area (ΔDOB) = Area (ΔDAB) − Area (ΔDOB)

Area (ΔDEO) = Area (ΔAOB)

It can be observed that ΔADX and ΔACX lie on the same base AX and are between the same parallels AB and DC.

Area (ΔADX) = Area (ΔACX) (i)

ΔACY and ΔACX lie on the same base AC and are between the same parallels AC and XY

Area (ΔACY) = Area (ACX) (ii)

From equations (i) and (ii), we obtain

Area (ΔADX) = Area (ΔACY)

Since ΔABQ and ΔPBQ lie on the same base BQ and are between the same parallels AP and BQ

Area (ΔABQ) = Area (ΔPBQ) (i)

Again, ΔBCQ and ΔBRQ lie on the same base BQ and are between the same parallels BQ and CR

Area (ΔBCQ) = Area (ΔBRQ) (ii)

On adding equations (i) and (ii), we obtain

Area (ΔABQ) + Area (ΔBCQ) = Area (ΔPBQ) + Area (ΔBRQ)

Area (ΔAQC) = Area (ΔPBR)

It is given that:

Area (ΔAOD) = Area (ΔBOC)

Area (ΔAOD) + Area (ΔAOB) = Area (ΔBOC) + Area (ΔAOB)
⇒Area (ΔADB) = Area (ΔACB)

Therefore, these triangles, ΔADB and ΔACB, are lying between the same parallels i.e.,

AB || CD

Therefore, ABCD is a trapezium.

It is given that:

Area (ΔDRC) = Area (ΔDPC)

As ΔDRC and ΔDPC lie on the same base DC and have equal areas, therefore, they must lie between the same parallel lines

DC || RP

Therefore, DCPR is a trapezium. It is

Also given that:

Area (ΔBDP) = Area (ΔARC)

Area (BDP) − Area (ΔDPC) = Area (ΔARC) − Area (ΔDRC)

Area (ΔBDC) = Area (ΔADC)

Since ΔBDC and ΔADC are on the same base CD and have equal areas, they must lie between the same parallel lines

AB || CD

Therefore,

ABCD is a trapezium.

As the parallelogram and the rectangle have the same base and equal area, therefore, these will also lie between the same parallels.

Consider the parallelogram ABCD and rectangle ABEF as follows

Here, it can be observed that parallelogram ABCD and rectangle ABEF are between the same parallels AB and CF

We know that opposite sides of a parallelogram or a rectangle are of equal lengths

Therefore,

AB = EF (For rectangle)

AB = CD (For parallelogram)

CD = EF

AB + CD = AB + EF (i)

Of all the line segments that can be drawn to a given line from a point not lying on it, the perpendicular line segment is the shortest

AF < AD

And similarly,

BE < BC

AF + BE < AD + BC (ii)

From equations (i) and (ii), we obtain

AB + EF + AF + BE < AD + BC + AB + CD

Perimeter of rectangle ABEF < Perimeter of parallelogram ABCD

Let us draw a line segment AM perpendicular BC

We know that,

Area of a triangle = * Base * Altitude

Area (ΔADE) = * DE * AM

Area (ΔABD) = * BD * AM

Area (ΔAEC) = * EC * AM

It is given that DE = BD = EC

* DE * AM = * BD * AM = * EC * AM

Area (ΔADE) = Area (ΔABD) = Area (ΔAEC)

It can be observed that Budhia has divided her field into 3 equal parts

It is given that ABCD is a parallelogram. We know that opposite sides of a parallelogram are equal

AD = BC (i)

Similarly, for parallelograms DCEF and ABFE

DE = CF (ii)

And,

EA = FB (iii)

In ΔADE and ΔBCF,

AD = BC [Using equation (i)]

DE = CF [Using equation (ii)]

EA = FB [Using equation (iii)]

ΔADE BCF (SSS congruence rule)

Area (ΔADE) = Area (ΔBCF)

It is given that ABCD is a parallelogram

AD || BC and AB || DC (Opposite sides of a parallelogram are parallel to each other)

Join point A to point C

Consider ΔAPC and ΔBPC

ΔAPC and ΔBPC are lying on the same base PC and between the same parallels PC and AB

Therefore,

Area (ΔAPC) = Area (ΔBPC) (i)

In quadrilateral ACDQ

AD = CQ

Since,

ABCD is a parallelogram

AD || BC (Opposite sides of a parallelogram are parallel)

CQ is a line segment which is obtained when line segment BC is produced

AD || CQ

We have,

AC = DQ and AC || DQ

Hence, ACQD is a parallelogram

Consider ΔDCQ and ΔACQ

These are on the same base CQ and between the same parallels CQ and AD

Therefore,

Area (ΔDCQ) = Area (ΔACQ)

Area (ΔDCQ) − Area (ΔPQC) = Area (ΔACQ) − Area (ΔPQC)

Area (ΔDPQ) = Area (ΔAPC) (ii)

From equations (i) and (ii), we obtain

Area (ΔBPC) = Area (ΔDPQ)

(i) Let G and H be the mid-points of side AB and AC respectively. Line segment GH is joining the mid-points. Therefore, it will be parallel to third side BC and also its length will be half of the length of BC (mid-point theorem)

GH = BC and

GH || BD

GH = BD = DC and GH || BD (D is the mid-point of BC)

Consider quadrilateral GHDB

GH ||BD and GH = BD

Two line segments joining two parallel line segments of equal length will also be equal and parallel to each other

Therefore,

BG = DH and BG || DH

Hence, quadrilateral GHDB is a parallelogram

We know that in a parallelogram, the diagonal bisects it into two triangles of equal area

Hence,

Area (ΔBDG) = Area (ΔHGD)

Similarly, it can be proved that quadrilaterals DCHG, GDHA, and BEDG are parallelograms and their respective diagonals are dividing them into two triangles of equal area

ar (ΔGDH) = ar (ΔCHD) (For parallelogram DCHG) ar (ΔGDH) = ar (ΔHAG) (For parallelogram GDHA) ar (ΔBDE) = ar (ΔDBG) (For parallelogram BEDG)

ar (ΔABC) = ar(ΔBDG) + ar(ΔGDH) + ar(ΔDCH) + ar(ΔAGH)

ar (ΔABC) = 4 × ar(ΔBDE)

Hence,

ar (ΔBDE) = ar (ABC)

(ii)Area (ΔBDE) = Area (ΔAED) (Common base DE and DE||AB)

Area (ΔBDE) − Area (ΔFED) = Area (ΔAED) − Area (ΔFED)

Area (ΔBEF) = Area (ΔAFD) (i)

Area (ΔABD) = Area (ΔABF) + Area (ΔAFD)

Area (ΔABD) = Area (ΔABF) + Area (ΔBEF) [From equation (i)]

Area (ΔABD) = Area (ΔABE) (ii)

AD is the median in ΔABC

Area (ΔABD) = Area (ΔABC)

= Area (ΔBDE)

Area (ΔABD) = 2 Area (ΔBDE) (iii)

From (ii) and (iii), we obtain

2 ar (ΔBDE) = ar (ΔABE)

ar (ΔBDE) = ar (ΔABE)

(iii) ar (ΔABE) = ar (ΔBEC) (Common base BE and BE||AC)

ar (ΔABF) + ar (ΔBEF) = ar (ΔBEC)

Using equation (i), we obtain

ar (ΔABF) + ar (ΔAFD) = ar (ΔBEC) ar (ΔABD) = ar (ΔBEC)

ar (ΔABC) = ar (ΔBEC)

ar (ΔABC) = 2 ar (ΔBEC)

(iv) It is seen that ΔBDE and ar ΔAED lie on the same base (DE) and between the parallels DE and AB

ar (ΔBDE) = ar (ΔAED) ar (ΔBDE) − ar (ΔFED)

= ar (ΔAED) − ar (ΔFED) ar (ΔBFE) = ar (ΔAFD)

(v)

An in equilateral triangle, median drawn is also perpendicular to the side,

Now,

ar Δ AFD = 1/2 × FD × AD ..... (i)

Draw EG ⊥ BC.

Draw EGBC

ar Δ FED = 1/2 × FD × EG ..... (ii)

Dividing eq. (i) by (ii), we get




As Altitude of equilateral triangle =side

As D is the mid-point of BC.

ar (ΔAFD) = 2 ar (ΔFED) ……(iii)

From part (iii)

ar (BFE) = ar (AFD)

So ar (ΔBFE) = 2 ar (ΔFED)

vi)

ar (Δ AFC) = ar (Δ AFD) + ar (Δ ADC) = 2 ar (Δ FED) +ar (Δ ABC) [using (v)

= 2 ar (Δ FED) + 1/2[4 ×ar (Δ BDE)] [Using result of part (i)]

= 2 ar (Δ FED) + 2 ar (Δ BDE) = 2 ar (Δ FED) + 2 ar (Δ AED)

[Δ BDE and Δ AED are on the same base and between same parallels]

= 2 ar (Δ FED) + 2 [ar (Δ AFD) + ar (Δ FED)]

= 2 ar (Δ FED) + 2 ar (Δ AFD) + 2 ar (Δ FED) [Using (viii)]

= 4 ar (Δ FED) + 4 ar (Δ FED)

⇒ ar (Δ AFC) = 8 ar (Δ FED)

⇒ ar (Δ FED) = 1/8[ar (ΔAFC)]

Let us construct AM perpendicular to BD

Now,

Area (APB) x Area (CPD)

=

=

Area (APD) x Area (BPC)

=

=

Hence,

Area (APB) x Area (CPD) = Area (APD) x Area (CPB)

Let ABC is a triangle.



P and Q are the midpoints of AB and BC respectively and R is the mid-point of AP.

Join PQ, QR, AQ, PC, RC as shown in the figure.

Now we know that the median of a triangle divides it into two triangles of equal areas.

In triangle CAP, CR is the median.

So Area(ΔCRA) = (1/2)× Area(ΔCAP) .......(1)

Again in triangle CAB, CP is the median.

So Area(ΔCAP) = (1/2)× Area(ΔCPB) .......... (2)

from equation 1 and 2, we get

Area(ΔCAP) = (1/2)×Area(ΔCPB) .............(3)

Again in triangle PBC, PQ is the median.

So (1/2)×Area(ΔCPB) = Area(ΔPBQ) ............(4)

From equation 3 and 4, we get

Area(ΔARC) = Area(ΔPBQ) ...............(5)

Now QP, And QR are the medians of triangle QAB and QAP respectively

So Area(ΔQAP) = Area(ΔQBP) ............(6)

and Area(ΔQAP) = 2× Area(ΔQRP) ............(7)

from equation 6 and 7, we get

Area(ΔPRQ) = (1/2)× Area(ΔPBQ) ............(8)

from equation 5 and 8, we get

Area(ΔPRQ) = (1/2)× Area(ΔARC)

Now CR is the median of triangle CAP

So Area(ΔARC) = (1/2)× Area(ΔCAP)

= (1/2)× (1/2)× Area(ΔABC)

= (1/4)× Area(ΔABC)

Again RQ is the median of triangle RBC

So Area(ΔRQC) = (1/2)× Area(ΔRBC)

= (1/2)× Area(ΔABC) - (1/2)× Area(ΔARC)

=(1/2)× Area(ΔABC) - (1/2)× (1/2)× (1/2)× Area(ΔABC)

=(1/2)× Area(ΔABC) - (1/8)× Area(ΔABC)

= (3/8)× Area(ΔABC)

So Area(ΔRQC) = = (3/8)× Area(ΔABC)

(i) Since, each angle of a square is 90⁰

Hence, angle ABM = angle DBC = 90⁰

ABM + ABC = DBC + ABC

MBC = ABD (1)

In ΔMBC and ΔABD,

Angle MBC = Angle ABD (From 1)

MB = AB (Sides of square)

BC = BD (Sides of square)

Therefore,

By SAS congruence rule,

(ii) Now, we have:

And we know that adjacent sides of square are perpendicular to each other. Hence,

BD perpendicular DE

And,

AX perpendicular DE

Also,

We know that two lines perpendicular to the same line are parallel to each other.

Therefore,

BD ‖ AX

Now,

= ( and ‖gm BXYD are on same base BD and between the same parallels BD and AX)

Area (BXYD) = 2 area (ΔABD)

Area (BXYD) = 2 area (ΔMBC) (2)

(iii) ΔMBC and parallelogram ABMN are on same base MB and between the same parallels MB and NC.

Therefore,

Area (ΔMBC) = � ar (ABMN)

2 Area (ΔMBC) = Area (ABMN) = Area(BXYD) (From 2) (3)

(iv) Since, each angle of square is 90⁰

Therefore,

Angle FCA = Angle BCE = 90⁰

FCA + ACB = BCE + ACB

Angle FCB = Angle ACE

Now,

In ΔFCB and ΔACE,

Angle FCB = Angle ACE

FC = AC (Sides of square)

CB = CE (Sides of square)

Therefore,

By SAS congruence,

(v) It is given that AX is perpendicular to DE and CE is perpendicular to DE

and,

We know that two lines perpendicular to the same line are parallel to each other.

Hence,

CE ‖ AX

Now,

ΔACE and parallelogram CYXE are on same base CE and between the same parallels CE and AX.

Therefore,

Area (ΔACE) = 1/2 Area (CYXE)

Area (CYXE) = 2 Area (ΔACE) (4)

We had proved that

ΔFCB ΔACE

Hence,

Area (ΔACE) =Area (ΔFCB) (5)

Comparing 4 and 5, we get:

CYXE = 2 Area (ΔFCB) (6)

(vi) ΔFCB and parallelogram ACFG are on same base CF and between the same parallels CF and BG

Hence,

Area (ΔFCB) = � Area (ACFG)

Area (ACFG) = 2 Area (ΔFCB) = Area (CYXE) (Using 6) (7)

(vii) From figure:

Area (BCED) = Area (BXYD) + Area (CYXE)

Area (BCED) = Area (ABMN) + Area (ACFG) (using 3 and 7)