NCERT Solutions for Class 9 Math Chapter 9 - Area of Parallelograms and Triangles

Answer:

In Fig. (i); $∆$DPC and trap. ABCD are on the same base DC and between same parallels DC and AB.
In Fig. (iii); $∆$RTQ and parallelogram PQRS are on the same base QR and between same parallels QR and PS.
In Fig. (v); parallelogram ABCD and parallelogram APQD are on the same base AD and between the same parallels AD and BQ.

Answer:

ABCD is a parallelogram.
$\therefore$ DC = AB
$\Rightarrow$ DC = 16 cm
AE $\perp$ DC (Given)
Area of parallelogram ABCD
= DC × AE
[$\because$ar (|| gm) = Base × Corresponding
height]
= 16 cm × 8 cm
= 128 cm²
Using base AD and height CF;
Area of parallelogram = AD × CF
$\Rightarrow$ 128 cm² = AD × 10 cm
or AD × 10 cm = 128 cm²

$\Rightarrow$ AD = 12.8 cm

Answer:

Given: ABCD is a parallelogram. Let E, F, G and H are the mid-points of sides AB, BC, CD and DA respectively.

Construction: Join AC and HF.
Proof: In $∆$ABC, E is the mid-point of side AB
and F is the mid-point of side BC.

Answer:

Given: ABCD is a parallelogram. P is a point on DC and Q is a point on AD.
To prove: ar ($∆$APB) = ar ($∆$BQC)
Construction: Draw PM || BC and QN || DC.
Proof: Since QC is the diagonal of || gm QNCD.

From (3) and (6), we get
ar ($∆$BQC) = ar ($∆$APB)
or ar ($∆$APB) = ar ($∆$BQC) (Proved)

Answer:

Draw a line (say l) passing through point P and parallel to AB which intersects AD at Q and BC at R respectively.
Now, $∆$APB and || gm ABRQ are on the same base AB and between same parallels AB and QR.

Also $∆$PCD and || gm DCRQ are on the same base DC and between same parallels DC and QR.

Answer:

Draw a line (say m) through P and parallel to AD which intersects AB at M and DC at N.
Now $∆$APD and ||gm AMND are on the same base AD and between same parallels AD and MN.

From (5) and (6), we get
ar ($∆$APB) + ar ($∆$PCD) = ar ($∆$APD) + ar ($∆$PBC)
or ar ($∆$APD) + ar ($∆$PBC) = ar ($∆$APB)
+ ar ($∆$PCD)
(Proved)

Answer:

Parallelograms PQRS and ABRS are on the same base SR and between the same parallels SR and PB.
$\therefore$ ar (|| gm PQRS) = ar (|| gm ABRS) ...(1)
[$\because$Parallelograms on the same base and between the same parallels are equal in area.]

Answer:

$∆$AXS and ||gm ABRS are on the same base AS and between the same parallels AS and BR.

Answer:

When A is joined with P and Q; the field is divided into three parts viz. $∆$PAS, $∆$APQ and $∆$AQR.
$∆$APQ and parallelogram PQRS are on the same base PQ and between same parallels PQ and SR.

It implies that triangular region APQ covers half portion of parallelogram shaped field PQRS.
So, if farmer sows wheat in triangular shaped field APQ then she will definitely sow pulses in other two triangular parts PAS and AQR.
Or
When she sows pulses in triangular shaped field APQ then she will sow wheat in other two triangular parts PAS and AQR.

Answer:

In $∆$ABC, AD is a median
ar ($∆$ABD) = ar ($∆$ACD) ...(1)

[$\because$ Median divides a $∆$ into two $∆$s of equal area]
Again in $∆$EBC, ED is a median
ar ($∆$EBD) = ar ($∆$ECD) ...(2)
Subtracting (2) from (1), we get
ar ($∆$ABD) – ar ($∆$EBD)
= ar ($∆$ACD) – ar ($∆$ECD)
$\Rightarrow$ ar ($∆$ABE) = ar ($∆$ACE).

Answer:

Answer:

Let parallelogram be ABCD and its diagonals
AC and BD intersect each other at O.
In $∆$ABC and $∆$ADC,
AB = DC
(Opposite sides of parallelogram)
BC = AD
(Opposite sides of parallelogram)
AC = AC (Common)
$\therefore$ $∆$ABC $\cong$ $∆$CDA
(SSS criteria of congruency)
As we know that diagonals of a parallelogram bisect each other.
$\therefore$ In fig., O is the point of bisection.
Now in $∆$ADC; DO is the median.
$\therefore$ ar ($∆$AOD) = ar ($∆$COD) ...(1)
($\because$ Median divides a triangle into two triangles of equal area.]
Similarly, in $∆$ABC; OB is the median.
$\therefore$ ar ($∆$AOB) = ar ($∆$BOC) ...(2)
[Same reason as above]
In $∆$AOB and $∆$AOD; AO is the median
$\therefore$ ar ($∆$AOB) = ar ($∆$AOD)
(Same reason as above) ...(3)
From (1), (2) and (3), we get
ar ($∆$AOB) = ar ($∆$AOD) = ar ($∆$BOC)
= ar ($∆$COD)
Thus, diagonals of a parallelogram divide it into four triangles of equal area.

Answer:

Draw CM $\perp$ AB and DN $\perp$ AB
In $∆$CMO and $∆$DNO

∠CMO = ∠DNO
[(Each = 90$°$) construction]
∠COM = ∠DON
(Vertically opp. angles)
OC = OD
(O is the mid-point of CD)
$\therefore$ $∆$COM $\cong$ $∆$DON
[AAS criteria of congruency]
so, CM = DN (c.p.c.t.) ...(1)

From (2) and (4), we get
ar ($∆$ABC) = ar ($∆$ADB)

Answer:

F is the mid-point of AB and E is the midpoint of AC.

[$\because$ Line joining the mid-points of two sides of a triangle is parallel to the third and half of it.]
or FE || BD
[$\because$ BD is the part of BC]
and FE = BD

Now we have
FE || BD and DE || BF
or FE = BD and DE = BF
Therefore BDEF is a parallelogram.

Answer:

BDEF is a parallelogram
$\therefore$ ar ($∆$BDF) = ar ($∆$DEF) ...(1)
[$\because$ Diagonal of a || gm divides it in two
triangles of equal area]
DCEF is also parallelogram. [Using same steps as in part (i)]
$\therefore$ ar ($∆$DEF) = ar ($∆$DEC) ...(2)
Also AEDF is a parallelogram
[Using same steps as in part (i)]
$\therefore$ ar ($∆$AFE) = ar ($∆$DEF) ...(3)
From (1), (2) and (3)
ar ($∆$DEF) = ar ($∆$BDF) = ar ($∆$DEC)
= ar ($∆$AFE) ...(4)
Now ar ($∆$ABC)
= ar ($∆$AFE) + ar ($∆$BDF) + ar ($∆$DEC) + ar ($∆$DEF)
...(5)
$\Rightarrow$ ar ($∆$ABC) = ar ($∆$DEF) + ar ($∆$DEF) +
ar ($∆$DEF)) + ar ($∆$DEF)
[Using (4) in (5)]
$\Rightarrow$ ar ($∆$ABC) = 4 ar (DEF)
or 4 ar ($∆$DEF) = ar ($∆$ABC)

Answer:

ar (|| gm BDEF) = ar ($∆$BDF) + ar ($∆$DEF)
= ar ($∆$DEF) + ar ($∆$DEF)
[Using (4)]
= 2 ar ($∆$DEF)

Answer:

Draw BM $\perp$ AC and DN $\perp$ AC.
In $∆$DON and $∆$BOM
OD = OB (Given)
∠DNO = ∠BMO
[Each = 90$°$ (By construction)]
∠DON = ∠BOM
(Vertically opp. angles)
$\therefore$ $∆$DON $\cong$ $∆$BOM
[AAS criteria of congruency]
So DN = BM (c.p.c.t.)
Also ar ($∆$DON) = ar ($∆$BOM) ...(1)
Now in $∆$DCN and $∆$ABM
∠DNC = ∠BMA
[each angle 90$°$ (by construction)]
CD = AB (Given)
DN = BM (Proved above)
$\therefore$ $∆$DCN $\cong$ $∆$BAM
[RHS criteria of congruency]
$\therefore$ ar ($∆$DCN) = ar ($∆$BAM) ...(2)
Adding (1) and (2), we get
ar ($∆$DON) + ar ($∆$DCN)
= ar ($∆$BOM) + ar ($∆$BAM)
$\Rightarrow$ ar ($∆$DOC) = ar ($∆$AOB)

Answer:

In part (i) we have proved that
ar ($∆$DOC) = ar ($∆$AOB)
Adding ar ($∆$BOC) both sides, we get
ar ($∆$DOC) + ar ($∆$BOC) = ar ($∆$AOB) + ar ($∆$BOC)
$\Rightarrow$ ar ($∆$DCB) = ar ($∆$ACB)

Answer:

In part (ii) we have proved that ar ($∆$DCB)
= ar ($∆$ACB)
Therefore these two triangles in addition to be on the same base CB lie between two same parallels CB and DA.
So, DA || CB
Now, we have AB = CD and DA || CB
Therefore ABCD is a parallelogram.

Answer:

Given that,
ar ($∆$DBC) = ar ($∆$EBC)
Two triangles of equal area have common base BC.
Therefore DE || BC

[$\because$ Two triangles having same base (or equal bases) and equal areas lie between the same parallels.]

Answer:

$∆$ABE and || gm BCYE lie on the same base BE
and between the same parallels BE and AC.

But || gm BCYE and || gm BCFX lie on the same
base BC and between the same parallels
BC and EF.
ar (|| gm BCYE) = ar (|| gm BCFX) ...(3)
From (1), (2) and (3), we get
ar ($∆$ABE) = ar ($∆$ACF)

Answer:

Given that ABCD and PBQR are parallelograms.
Also CP || AQ.
We observe that
$∆$ACQ and $∆$APQ lie on the same base AQ and
between the same parallels AQ and CP.
$\therefore$ ar ($∆$ACQ) = ar ($∆$APQ)
Subtracting ar ($∆$ABQ) from both sides, we get
ar ($∆$ACQ) – ar ($∆$ABQ)
= ar ($∆$APQ) – ar ($∆$ABQ)
$\Rightarrow$ ar ($∆$ACB) = ar ($∆$QPB)

$\Rightarrow$ ar (|| gm ABCD) = ar (|| gm PBQR)

Answer:

$∆$ABD and $∆$ABC lie on the same base AB and between the same parallels AB and DC.
$\therefore$ ar ($∆$ABD) = ar ($∆$ABC)
Subtracting ar ($∆$AOB) from both sides, we get
ar ($∆$ABD) – ar ($∆$AOB)
= ar ($∆$ABC) – ar ($∆$AOB)
$\Rightarrow$ ar $∆$(AOD) = ar ($∆$ BOC) (Proved)

Answer:

Given that BF || AC.
$∆$ACB and $∆$ACF lie on the same base AC and between the same parallels AC and BF.
$\therefore$ ar ($∆$ACB) = ar ($∆$ ACF) ...(1)

Answer:

Now
ar ($∆$ABCDE) = ar (quad. AEDC) + ar ($∆$ABC)
...(2)
= ar (quad. AEDC) + ar ($∆$ACF)
[Using (1) in (2)]
= ar (quad. AEDF)
or ar (AEDF) = ar (ABCDE) (Proved)

Answer:

Let Itwaari has land in shape of quadrilateral PQRS.
Draw a line through S parallel to PR, which meets QR produced at M.
Let diagonals PM and RS of new formed quadrilateral intersect each other at point N.
We have PR || SM (By construction)
$\therefore$ ar ($∆$PRS) = ar ($∆$PMR)
[$\because$ Triangles on the same base and same parallel
are equal in area]
Subtracting ar ($∆$PNR) from both sides, we get
ar ($∆$PRS) – ar ($∆$PNR)
= ar ($∆$PMR) – ar ($∆$PNR)
$\Rightarrow$ ar ($∆$PSN) = ar ($∆$MNR)
It implies that Itwaari will give corner triangular shaped plot PSN to the Gram Panchayat for health centre and will take equal amount of land
(denoted by $∆$MNR) adjoining his plot so as to form a triangular plot PQM.

Answer:

Join CX. $∆$ADX and ACX lie on the same base XA and between the same parallels XA and DC.
$\therefore$ ar ($∆$ADX) = ar ($∆$ACX) ...(1)
Also $∆$ACX and $∆$ACY lie on the same base AC and between same parallels XY and AC.
$\therefore$ ar ($∆$ACX) = ar ($∆$ACY) ...(2)
From (1) and (2), we get
ar ($∆$ADX) = ar ($∆$ACY). (Proved)

Answer:

$∆$ABQ and $∆$BPQ lie on the same base BQ and between same parallels.
AP and BQ.
$\therefore$ ar ($∆$ABQ) = ar ($∆$BPQ) ...(1)
$∆$BQC and $∆$BQR lie on the same base BQ and between same parallels BQ and CR.
$\therefore$ ar ($∆$BQC) = ar ($∆$BQR) ...(2)
Adding (1) and (2), we get
ar ($∆$ABQ) + ar ($∆$BQC)
= ar ($∆$BPQ) + ar ($∆$BQR)
$\Rightarrow$ ar ($∆$AQC) = ar ($∆$PBR) (Proved)

Answer:

Given that,
ar ($∆$AOD) = ar ($∆$BOC)
Adding ar ($∆$AOB) both sides, we get
ar ($∆$AOD) + ar ($∆$AOB) = ar ($∆$BOC) + ar ($∆$AOB)
$\Rightarrow$ ar ($∆$ABD) = ar ($∆$ABC)
As we know that, if two triangles are equal in area, lie on the same base then they lie between same parallels. We have $∆$ABD and $∆$ABC lie on common base AB and are equal in area.
$\therefore$ They lie in same parallels AB and DC.
or AB || DC.
Now in quad. ABCD we have AB || DC.
So, ABCD is a trapezium.
[$\because$ In a trapezium one pair of opposite sides is parallel.]

Answer:

Given that $∆$ DRC and $∆$DPC lie on the same
base DC
also ar ($∆$ DRC) = ar ($∆$DPC) ...(1)
$\therefore$ DC || RP
[$\because$ If two triangles equal in area lie on the
same base then they always lie between same
parallels.]
In quad. DCPR;
DC || RP
So, DCPR is a trapezium.
Also given that
ar ($∆$BDP) = ar ($∆$ARC) ...(2)
Equation (1) can be written as
ar ($∆$DPC) = ar ($∆$DRC) ...(3)
Subtracting (3) from (2), we get
ar ($∆$BDP) – ar ($∆$DPC)
= ar ($∆$ARC) – ar ($∆$DRC)
$\Rightarrow$ ar ($∆$BDC) = ar ($∆$ADC)
We have $∆$BDC and $∆$ADC equal in area; lie on common base DC.
AB || DC
[$\because$ If two triangles equal in area lie on the same base then they always lie between same parallels]
Now in quadrilateral ABCD we have
AB || DC
So, ABCD is a trapezium.

Answer:

Given: Parallelogram ABCD and rectangle ABEF are on same base AB and between the same parallels AB and CF. Therefore ar (||gm ABCD) = ar (rect. ABEF)
To prove: AB + BC + CD + AD >AB + BE + EF + AF.

Proof: AB = CD
[$\because$ Opp. sides of a || gm are always equal]
AB = EF
[$\because$ Opp. sides of rect. are always equal]
CD = EF ...(1)
Adding AB to both sides of eq. (1), we get
AB + CD = AB + EF ...(2)
$\therefore$ If all the segments that can be drawn to a given line from a point not lying on it, the $\perp$ segment is the shortest.
$\therefore$ BE < BC and AF < AD
or BC > BE
and AD > AF
$\therefore$ BC + AD > BE + AF ...(3)
From eq. (2) and (3), we get
AB + BC + CD + AF > AB + BE + EF + AF.
(Proved)

Answer:

In $∆$ABC; points D and E divides BC in three equal parts such that BD = DE = EC

Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide $∆$ABC into n triangles of equal areas.

Answer:

As we know that opposite sides of a parallelogram are always equal.
$\therefore$ In parallelogram ABFE;
AE = BF and AB = EF
In || gm DCFE;
DE = CF and DC = EF
In || gm ABCD;
AD = BC and AB = DC
Now in $∆$ADE and $∆$BCF,
AE = BF
[Opposite sides of || gm ABFE]
DE = CF
[Opp. sides of || gm DCFE]
and AD = BC
[Opp. sides of || gm of ABCD]
$\therefore$ $∆$ADE $\cong$ $∆$BCF
[SSS criteria of congruency]
So, ar ($∆$ADE) = ar ($∆$BCF)
[$\because$Area of two congruent figures is always equal]

Answer:

Join A and C.
$∆$APC and $∆$BPC are on the same base PC and between the same parallels PC and AB.
$\therefore$ ar ($∆$APC) = ar ($∆$BPC) ... (1)

ABCD is a || gm.
$\therefore$ AD = BC
[$\because$ Opposite sides of a parallelogram are equal.]
Also BC = CQ (given)
$\therefore$ AD = CQ
Now AD || CQ
[$\because$ CQ is the extension of BC.]
and AD = CQ
$\therefore$ ADQC is a parallelogram.
[$\because$ If one pair of opposite sides of a quadrilateral are equal and parallel then it is a parallelogram.]
Since diagonals of a || gm bisect each other.
$\therefore$ AP = PQ and CP = DP
Now in $∆$APC and $∆$DPQ
AP = PQ (Proved above)
∠APC = ∠DPQ
(Vert. opp. angles)
PC = PD (Proved above)
$\therefore$ $∆$APC $\cong$ $∆$DPQ
$\Rightarrow$ ar ($∆$APC) = ar ($∆$DPQ) ... (2)
[$\because$ Area of congruent figures is always equal]
From (1) and (2), we get
ar ($∆$BPC) = ar ($∆$DPQ) (Proved)

Answer:

Join EC and AD.
$∆$ABC is an equilateral triangle.
$\therefore$ ∠BAC = ∠ABC = ∠ACB = 60$°$
$∆$BDE is also an equilateral triangle.
$\therefore$ ∠EBD = ∠BED = ∠BDE = 60$°$
Now if we take two lines.
AC and BE and BC as a transversal.
then ∠EBD = ∠ACB
(each angle 60$°$)
$\Rightarrow$ BE || AC
[$\because$ When alternate angles are equal then lines are parallel.]
Similarly, for lines AB and DE and BF as transversal.
∠ABC = ∠BDE
(each = 60$°$) [alternate angles]
$\therefore$ AB || DE

Answer:

Join EC and AD.
$∆$ABC is an equilateral triangle.
$\therefore$ ∠BAC = ∠ABC = ∠ACB = 60$°$
$∆$BDE is also an equilateral triangle.
$\therefore$ ∠EBD = ∠BED = ∠BDE = 60$°$
Now if we take two lines.
AC and BE and BC as a transversal.
then ∠EBD = ∠ACB
(each angle 60$°$)
$\Rightarrow$ BE || AC
[$\because$ When alternate angles are equal then lines are parallel.]
Similarly, for lines AB and DE and BF as transversal.
∠ABC = ∠BDE
(each = 60$°$) [alternate angles]
$\therefore$ AB || DE
In $∆$BEC, ED is the median

[$\because$ Median divides, the triangle in two triangles having equal area.]
We have BE || AC.
Therefore $∆$BEC and $∆$BAE are on the same base BE and between the same parallels BE and AC.
$\therefore$ ar ($∆$BEC) = ar ($∆$BAE) ... (4)
Using (4) in (3), we get

Answer:

Join EC and AD.
$∆$ABC is an equilateral triangle.
$\therefore$ ∠BAC = ∠ABC = ∠ACB = 60$°$
$∆$BDE is also an equilateral triangle.
$\therefore$ ∠EBD = ∠BED = ∠BDE = 60$°$
Now if we take two lines.
AC and BE and BC as a transversal.
then ∠EBD = ∠ACB
(each angle 60$°$)
$\Rightarrow$ BE || AC
[$\because$ When alternate angles are equal then lines are parallel.]
Similarly, for lines AB and DE and BF as transversal.
∠ABC = ∠BDE
(each = 60$°$) [alternate angles]
$\therefore$ AB || DE

Answer:

Join EC and AD.
$∆$ABC is an equilateral triangle.
$\therefore$ ∠BAC = ∠ABC = ∠ACB = 60$°$
$∆$BDE is also an equilateral triangle.
$\therefore$ ∠EBD = ∠BED = ∠BDE = 60$°$
Now if we take two lines.
AC and BE and BC as a transversal.
then ∠EBD = ∠ACB
(each angle 60$°$)
$\Rightarrow$ BE || AC
[$\because$ When alternate angles are equal then lines are parallel.]
Similarly, for lines AB and DE and BF as transversal.
∠ABC = ∠BDE
(each = 60$°$) [alternate angles]
$\therefore$ AB || DE
$∆$BDE and $∆$AED are on the same base DE and between same parallels AB and DE.
$\therefore$ ar ($∆$BDE) = ar ($∆$AED)
Subtracting ar ($∆$FED) from both sides, we get ar($∆$BDE) – ar($∆$FED) = ar($∆$AED) – ar($∆$FED)
$\Rightarrow$ ar ($∆$BFE) = ar ($∆$AFD) ... (8)

Answer:

Join EC and AD.
$∆$ABC is an equilateral triangle.
$\therefore$ ∠BAC = ∠ABC = ∠ACB = 60$°$
$∆$BDE is also an equilateral triangle.
$\therefore$ ∠EBD = ∠BED = ∠BDE = 60$°$
Now if we take two lines.
AC and BE and BC as a transversal.
then ∠EBD = ∠ACB
(each angle 60$°$)
$\Rightarrow$ BE || AC
[$\because$ When alternate angles are equal then lines are parallel.]
Similarly, for lines AB and DE and BF as transversal.
∠ABC = ∠BDE
(each = 60$°$) [alternate angles]
$\therefore$ AB || DE
In an equilateral triangle; median drawn is also perpendicular to the side.
$\therefore$ AD $\perp$ BC
[$\because$ AD is median of $∆$ABC]

Answer:

Join EC and AD.
$∆$ABC is an equilateral triangle.
$\therefore$ ∠BAC = ∠ABC = ∠ACB = 60$°$
$∆$BDE is also an equilateral triangle.
$\therefore$ ∠EBD = ∠BED = ∠BDE = 60$°$
Now if we take two lines.
AC and BE and BC as a transversal.
then ∠EBD = ∠ACB
(each angle 60$°$)
$\Rightarrow$ BE || AC
[$\because$ When alternate angles are equal then lines are parallel.]
Similarly, for lines AB and DE and BF as transversal.
∠ABC = ∠BDE
(each = 60$°$) [alternate angles]
$\therefore$ AB || DE
In an equilateral triangle; median drawn is also perpendicular to the side.
$\therefore$ AD $\perp$ BC
[$\because$ AD is median of $∆$ABC]

Answer:

Join EC and AD.
$∆$ABC is an equilateral triangle.
$\therefore$ ∠BAC = ∠ABC = ∠ACB = 60$°$
$∆$BDE is also an equilateral triangle.
$\therefore$ ∠EBD = ∠BED = ∠BDE = 60$°$
Now if we take two lines.
AC and BE and BC as a transversal.
then ∠EBD = ∠ACB
(each angle 60$°$)
$\Rightarrow$ BE || AC
[$\because$ When alternate angles are equal then lines are parallel.]
Similarly, for lines AB and DE and BF as transversal.
∠ABC = ∠BDE
(each = 60$°$) [alternate angles]
$\therefore$ AB || DE
ar ($∆$AFC) = ar ($∆$AFD) + ar ($∆$ADC)

[Using result of part (i)]
= 2 ar ($∆$FED) + 2 ar ($∆$BDE)
= 2 ar ($∆$FED) + 2 ar($∆$AED)
[$\because$ $∆$BDE and $∆$AED are on the same base ED
and between same parallels AB and DE]
= 2 ar ($∆$FED) + 2 [ar ($∆$AFD) + ar ($∆$FED)]
= 2 ar ($∆$FED) + 2 ar ($∆$AFD) + 2 ar ($∆$FED)
= 4 ar ($∆$FED) + 2(2 ar ($∆$FED)]
[Using result in (11)]
= 4 ar ($∆$FED) + 4 ar ($∆$FED)
$\Rightarrow$ ar ($∆$AFC) = 8 ar ($∆$FED)
or 8 ar ($∆$FED) = ar ($∆$AFC)

Answer:

Given: A quadrilateral ABCD, in which diagonals AC and BD intersect each other at point E.
To prove: ar ($∆$AED) × ar ($∆$BEC) = ar ($∆$ABE) × ar ($∆$CDE)
Construction: From A, draw AM $\perp$ BD and from C, draw CN $\perp$ BD.

Answer:

Answer:

Answer:

Answer:

From fig.,
∠ABM = ∠CBD (Each 90$°$)
Adding ∠ABC both sides, we get
∠ABM + ∠ABC = ∠CBD + ∠ABC
$\Rightarrow$ ∠MBC = ∠ABD ...(1)
Now in $∆$MBC and $∆$ABD;
MB = AB
[Equal sides of square ABMN]
BC = BD [Sides of square BCED]
∠MBC = ∠ABD [Proved above]
$\therefore$ $∆$MBC $\cong$ $∆$ABD
[By SAS criteria of congruency]

Answer:

From (i), we have
($∆$MBC) $\cong$ $∆$ABD
$\Rightarrow$ ar ($∆$MBC) = ar ($∆$ABD)
$\Rightarrow$ ar ($∆$MBC) = ar (trap ABDX) – ar ($∆$ADX)

$\Rightarrow$ 2ar ($∆$MBC) = BD . BY
$\Rightarrow$ 2ar ($∆$MBC) = ar (rect. BYXD)
Hence, ar (BYXD) = 2ar ($∆$MBC) (Proved)

Answer:

Join AM;
ABMN is a square.
$\therefore$ NA || MB
$\Rightarrow$ AC || MB
Now $∆$AMB and $∆$MBC are on the same base
MB and between the same parallels MB and AC.
$\therefore$ ar ($∆$AMB) = ar ($∆$MBC) ...(2)
From result (ii), we have
ar (BYXD) = 2ar ($∆$MBC) ...(3)
Using (2) in (3), we get
ar (BYXD) = 2ar ($∆$AMB)
$\Rightarrow$ ar (BYXD) = ar (square ABMN)
[$\because$ Diagonal AM of square ABMN divides it in two triangles of equal area] ...(4)

Answer:

In $∆$FCB and $∆$ACE
FC = AC
(Sides of square ACFG)
BC = CE
(Sides of square BCED)
∠BCF = ∠ACE
[$\because$ ∠ACF = ∠BCE (Each 90$°$)
∠ACF + ∠ACB = ∠BCE + ∠ACB
$\Rightarrow$ ∠BCF = ∠ACE]
$\therefore$ $∆$FCB $\cong$ $∆$ACE
(SAS criteria of congruency)

Answer:

From (iv) we have
$∆$FCB $\cong$ $∆$ACE
$\Rightarrow$ ar ($∆$FCB) = ar ($∆$ACE)
$\Rightarrow$ ar ($∆$FCB) = ar (trap ACEX) – ($∆$AEX)

$\Rightarrow$ 2ar ($∆$FCB) = CE . CY
$\Rightarrow$ 2ar ($∆$FCB) = ar (rect. CYXE)
Hence, ar (CYXE) = 2ar ($∆$FCB) (Proved)

Answer:

Join AF;
ACFG is a square
$\therefore$ FC || AG
$\Rightarrow$ FC || AB
Now $∆$ACF and $∆$FCB are on the same base FC and between the same parallels FC and AB.
$\therefore$ ar ($∆$ACF) = ar ($∆$FCB) ...(5)
From result (v) we have,
ar (CYXE) = 2ar ($∆$FCB) ...(6)
Using (5) in (6), we get
ar (CYXE) = 2ar ($∆$ACF)
Diagonal AF of square ACFG divides it in two triangles of equal area.
$\therefore$ ar (CYXE) = ar (sq. ACFG) ....(7)

Answer:

Adding (4) and (7), we get
ar (BYXD) + ar (CYXE)
= ar (ABMN) + ar (ACFG)
$\Rightarrow$ ar (BCED)
= ar (ABMN) + ar (ACFG)