Theorems on Area

Let us consider ABCD be a parallelogram in which E and F are mid-points of AB and CD. Join EF.

Let us construct DG ⊥ AB and let DG = h, where h is the altitude on side AB.

Area of ||gm ABCD = base × height = AB × h

Area of ||gm AEFD = AE × h = $\frac{AB}{2}$ × h .......(i) [Since E is the mid-point of AB]

Area of ||gm EBCF = EB × h = $\frac{AB}{2}$ × h .......(ii) [Since E is the mid-point of AB]

From (i) and (ii)

Area of ||gm AEFD = Area of ||gm EBCF.

Hence proved, that the line segment joining the mid-points of a pair of opposite sides of a parallelogram divides it into two equal parallelograms.

Let us consider a parallelogram ABCD, the diagonals AC and BD cut at point O.

In parallelogram ABCD, the diagonals bisect each other.

AO = OC

In ∆ACD, O is the mid-point of AC.

∴ OD is the median.

Area of ∆AOD = Area of ∆COD ....... (i) [Median of ∆ divides it into two triangles of equal areas.]

Similarly, in ∆ABC

O is the mid-point of AC.

∴ OB is the median.

Area of ∆AOB = Area of ∆COB ....... (ii) [Median of ∆ divides it into two triangles of equal areas.]

In ∆ADB,

O is the mid-point of BD.

∴ OA is the median.

Area of ∆AOD = Area of ∆AOB ....... (iii)

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

Area of ∆AOB = ∆COB = ∆COD = ∆AOD

Hence proved, that the diagonals of a parallelogram divide it into four triangles of equal area.

(i) In ∆ABC,

Area of ∆ABD = Area of ∆ADC [AD is the median] ......(1)

Since, PD is a straight line and base of ∆ABC and ∆PBC,

So PD is median of ∆PBC,

∴ Area of ∆PBD = Area of ∆PDC ........(2)

Hence, proved that Area of ∆PBD = Area of ∆PDC.

(ii) Subtracting eq. 1 from 2 we get,

⇒ Area of ∆ABD - Area of ∆PBD = Area of ∆ADC - Area of ∆PDC

⇒ Area of ∆ABP = Area of ∆ACP.

Hence proved that Area of ∆ABP = Area of ∆ACP.

(i) We know that,

Triangles on the same base and between the same parallel lines are equal in area.

∆BCD and ∆BCE are on the same base BC and between the same || lines DE and BC.

⇒ Area of ∆BCD = Area of ∆BCE

Subtracting area of ∆BCD and ∆BCE from area of ∆ABC

⇒ Area of ∆ABC - Area of ∆BCD = Area of ∆ABC - Area of ∆BCE

⇒ Area of ∆ACD = Area of ∆ABE.

Hence proved, that Area of ∆ACD = Area of ∆ABE.

(ii) We know that,

⇒ Area of ∆BCD = Area of ∆BCE

Subtracting area of ∆OBC from above equation we get,

⇒ Area of ∆BCD - Area of ∆OBC = Area of ∆BCE - Area of ∆OBC

⇒ Area of ∆OBD = Area of ∆OCE.

Hence proved, that Area of ∆OBD = Area of ∆OCE.

We know that,

Area of a triangle is half that of a parallelogram on the same base and between the same parallel lines.

∆APD and || gm ABCD are on the same base AD and between the same || lines AD and BC,

∴ Area of ∆APD = $\frac{1}{2}$ Area of || gm ABCD .......(i)

From figure,

Area of ||gm ABCD = Area of ∆APD + Area of ∆ABP + Area of ∆DPC

Dividing the above equation by 2 we get,

$\frac{\text{Area of ||gm ABCD}}{2} = \frac{\text{Area of ∆APD}}{2} + \frac{\text{Area of ∆ABP}}{2} + \frac{\text{Area of ∆DPC}}{2}$

From (i),

$$\begin{aligned}\text{Area of ∆APD} = \frac{\text{Area of ∆APD}}{2} + \frac{\text{Area of ∆ABP}}{2} + \frac{\text{Area of ∆DPC}}{2} \\[1em] \text{Area of ∆APD} - \frac{\text{Area of ∆APD}}{2} = \frac{\text{Area of ∆ABP}}{2} + \frac{\text{Area of ∆DPC}}{2} \\[1em] \frac{\text{Area of ∆APD}}{2} = \frac{\text{Area of ∆ABP}}{2} + \frac{\text{Area of ∆DPC}}{2} \\[1em] \text{Area of ∆APD} = \text{Area of ∆ABP} + \text{Area of ∆DPC}\end{aligned}$$

Hence, proved that Area of ∆APD = Area of ∆ABP + Area of ∆DPC.

(i) Draw a line PQ || to AB and CD from point O.

AB || PQ and AP || BQ (Since, AD || BC)

ABQP is a || gm

Similarly,

PD || CQ and PQ || DC

PQCD is a || gm

Now, ∆OAB and || gm ABQP are on the same base AB and between same || lines AB and PQ

Area of ∆OAB = $\frac{1}{2}$ Area of ||gm ABQP .....(1)

Similarly, ∆OCD and || gm PQCD are on the same base CD and between same || lines CD and PQ

Area of ∆OCD = $\frac{1}{2}$ Area of || gm PQCD ..... (2)

Now by adding (1) and (2),

Area of ∆OAB + Area of ∆OCD = $\frac{1}{2}$ Area of || gm ABQP + $\frac{1}{2}$ Area of || gm PQCD

= $\frac{1}{2}$ [Area of || gm ABQP + Area of || gm PQCD]

= $\frac{1}{2}$ Area of || gm ABCD

Hence, proved that Area of ∆OAB + Area of ∆OCD = $\frac{1}{2}$ Area of || gm ABCD.

(ii) From figure,

⇒ Area of ∆OAB + Area of ∆OBC + Area of ∆OCD + Area of ∆OAD = Area of || gm ABCD

⇒ Area of ∆OAB + Area of ∆OCD + Area of ∆OBC + Area of ∆OAD = Area of || gm ABCD

⇒ $\frac{1}{2}$ Area of || gm ABCD + Area of ∆OBC + Area of ∆OAD = Area of || gm ABCD

⇒ Area of ∆OBC + Area of ∆OAD = Area of || gm ABCD - $\frac{1}{2}$ Area of || gm ABCD

⇒ Area of ∆OBC + Area of ∆OAD = $\frac{1}{2}$ Area of || gm ABCD.

Hence, proved that Area of ∆OBC + Area of ∆OAD = $\frac{1}{2}$ Area of || gm ABCD.

Parallelogram ABCD with E, F, G and H as mid-points of the sides AB, BC, CD and DA, respectively is shown below:

AH = $\frac{1}{2}$AD (As H is the mid-point of AD)

BF = $\frac{1}{2}$BC (As F is the mid-point of BC)

So, AH = BF and AH || BF (As AD || BC).

So, ABFH is a || gm.

Since, || gm ABFH and △EFH are on same base FH and between same parallel lines AB and FH so,

area of △EFH = $\frac{1}{2}$ area of || gm ABFH .......(i)

Similarly,

HD = $\frac{1}{2}$AD (As H is the mid-point of AD)

FC = $\frac{1}{2}$BC (As F is the mid-point of BC)

So, HD = FC and HD || FC (As AD || BC).

So, HFCD is a || gm.

Since, || gm HFCD and △HFG are on same base HF and between same parallel lines HF and DC so,

area of △HFG = $\frac{1}{2}$ area of || gm HFCD .......(ii)

Adding (i) and (ii) we get,

area of △EFH + area of △HFG = $\frac{1}{2}$ area of || gm ABFH + $\frac{1}{2}$ area of || gm HFCD

area of quad. EFGH = $\frac{1}{2}$ (area of || gm ABFH + area of || gm HFCD)

= $\frac{1}{2}$ (area of || gm ABCD).

Hence, proved that area of the quad. EFGH = $\frac{1}{2}$ area of || gm ABCD.

∆CPD and || gm ABCD are on the same base CD and between the same parallel lines AB and CD.

Area of ∆CPD = $\frac{1}{2}$ Area of ||gm ABCD .......(i)

∆AQD and || gm ABCD are on the same base AD and between the same parallel lines AD and BC.

Area of ∆AQD = $\frac{1}{2}$ Area of ||gm ABCD .......(ii)

From (i) and (ii)

Area of ∆CPD = Area of ∆AQD

Hence, proved that area of ∆CPD = area of ∆AQD.

From figure,

Since, PB is a straight line and PQ || SR so,

PB || SR.

|| gm PQRS and ABRS are on the same base SR and between the same parallel lines PB and SR.

So, Area of ||gm PQRS = Area of ||gm ABRS .........(i)

∆AXS and || gm ABRS are on the same base AS and between the same parallel lines AS and BR.

So, Area of ∆AXS = $\frac{1}{2}$ Area of ||gm ABRS

= $\frac{1}{2}$ area of ||gm PQRS [From (i)]

Hence, proved that Area of ∆AXS = $\frac{1}{2}$ Area of || gm PQRS.

∆ABC with D, E and F as mid-points of the sides BC, CA and AB, respectively is shown below:

(i) F and E are midpoints of AB and AC respectively.

So, by mid-point theorem,

FE || BC and FE = $\frac{1}{2}$ BC .........(1)

Also, D is the mid-point of BC

CD = $\frac{1}{2}$ BC ........ (2)

From (1) and (2),

FE || BC and FE = CD

Since, FE || BC so,

FE || CD and FE = CD ......... (3)

Similarly,

D and F are the midpoints of BC and AB.

So, by mid-point theorem,

DF || AC and DF = $\frac{1}{2}$ AC .........(4)

Also, E is the mid-point of AC

EC = $\frac{1}{2}$ AC ........ (5)

From (4) and (5),

DF || AC and DF = EC

Since, DF || AC so,

DF || EC and DF = EC ......... (6)

From 3 and 6,

FE || CD, FE = CD, DF || EC and DF = EC.

Since, opposite sides of FDCE are parallel and equal.

Hence proved FDCE is a parallelogram.

(ii) Since,

BD = CD (As D is mid-point of BC), FE = CD (Proved above) and FE || CD

So, BD = FE and BD || FE.

Hence, BDEF is a || gm.

Since,

FD = EC (Proved above) and AE = EC (As E is mid-point of AC) and FD || EC.

So, FD = AE and FD || AE.

Hence, AFDE is a || gm.

We know that, FDCE is a parallelogram and DE is a diagonal of || gm FDCE.

So, area of ∆DEF = area of ∆DEC (As diagonal divides || gm into two triangles with equal area) ........(1)

We know that, BDEF is a parallelogram and FD is a diagonal of || gm BDEF.

So, area of ∆DEF = area of ∆FBD (As diagonal divides || gm into two triangles with equal area) ........(2)

We know that, AFDE is a parallelogram and FE is a diagonal of || gm AFDE.

So, area of ∆DEF = area of ∆AFE (As diagonal divides || gm into two triangles with equal area) ........(3)

From 1, 2 and 3 we get,

area of ∆DEF = area of ∆DEC = area of ∆FBD = area of ∆AFE .........(4)

From figure,

area of ∆ABC = area of ∆DEF + area of ∆DEC + area of ∆FBD + area of ∆AFE

= area of ∆DEF + area of ∆DEF + area of ∆DEF + area of ∆DEF (From 4)

= 4 x area of ∆DEF.

∴ area of ∆ABC = 4 x area of ∆DEF

⇒ area of ∆DEF = $\frac{1}{4}$ area of ∆ABC

Hence, proved that area of ∆DEF = $\frac{1}{4}$ area of ∆ABC.

(iii) From figure,

Area of || gm FDCE = Area of ∆DEF + Area of ∆DEC

= Area of ∆DEF + Area of ∆DEF (From part (ii) eqn. 4)

= 2 area of ∆DEF

= $2 \times \frac{1}{4}$ area of ∆ABC [As area of ∆DEF = $\frac{1}{4}$ area of ∆ABC]

= $\frac{1}{2}$ area of ∆ABC.

Hence, proved that area of || gm FDCE = $\frac{1}{2}$ area of ∆ABC.

We know that D and E are the mid-points of BC and CA, respectively.

By mid-point theorem,

DE || AB and DE = $\frac{1}{2}$ AB = BF (As F is mid-point of AB)

From figure,

BF || DE and BF = DE.

Hence, BDEF is a || gm.

Similarly,

F and E are the mid-points of AB and CA.

By mid-point theorem,

EF || BC and EF = $\frac{1}{2}$ BC = DC (As D is mid-point of BC)

From figure,

EF || DC and EF = DC.

Hence, EFDC is a || gm.

Since, FE || BC and FB, EC are not parallel.

Hence, EFBC is a trapezium.

F and D are the mid-points of AB and BC.

By mid-point theorem,

FD || AC and FD = $\frac{1}{2}$ AC = AE (As E is mid-point of AC)

From figure,

FD || AE and FD = AE.

Hence, AFDE is a || gm.

We know that a diagonal divides a || gm in two triangles of equal area.

In || gm BDEF, FD is the diagonal,

∴ area of △DEF = area of △BDF .........(i)

In || gm EFDC, DE is the diagonal,

∴ area of △DEF = area of △EDC .........(ii)

In || gm AFDE, FE is the diagonal,

∴ area of △DEF = area of △AFE .........(iii)

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

area of △DEF = area of △BDF = area of △EDC = area of △AFE .......(iv)

From figure,

area of △ABC = area of △DEF + area of △BDF + area of △EDC + area of △AFE

= area of △DEF + area of △DEF + area of △DEF + area of △DEF

= 4 x area of △DEF

∴ area of △ABC = 4 x area of △DEF

⇒ area of △DEF = $\frac{1}{4}$ area of △ABC ......(v)

From figure,

area of trapezium BCEF = area of △DEF + area of △BDF + area of △EDC

= area of △DEF + area of △DEF + area of △DEF

= 3 x area of △DEF

∴ area of trapezium BCEF = 3 x area of △DEF

⇒ area of △DEF = $\frac{1}{3}$ x area of trapezium BCEF .......(vi)

From (v) and (vi) we get,

⇒ $\frac{1}{4}$ area of △ABC = $\frac{1}{3}$ area of trapezium BCEF

⇒ area of trapezium BCEF = $\frac{3}{4}$ area of △ABC.

Hence, proved that area of trapezium BCEF = $\frac{3}{4}$ area of △ABC.

From fig (1)

In ∆ABC, point D divides the side BC in the ratio m : n.

BD : DC = m : n

area of ∆ABD = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × BD × AE ........ (i)

area of ∆ADC = $\frac{1}{2}$ × DC × AE ...... (ii)

Dividing (i) by (ii)

$$\begin{aligned}\Rightarrow \frac{\text{area of ∆ABD}}{\text{area of ∆ADC}} = \frac{\frac{1}{2} × BD × AE}{\frac{1}{2} × DC × AE} \\[1em] \Rightarrow \frac{\text{area of ∆ABD}}{\text{area of ∆ADC}} = \frac{BD}{DC} \\[1em] \Rightarrow \frac{\text{area of ∆ABD}}{\text{area of ∆ADC}} = \frac{m}{n}.\end{aligned}$$

Hence, proved that area of ∆ABD : area of ∆ADC = m : n.

Given, PC = 2BP

From figure,

BC = BP + PC = BP + 2BP = 3BP.

$\frac{PC}{BC} = \frac{2BP}{3BP} = \frac{2}{3}$

PC = $\frac{2}{3}$ BC

Let AD be the altitude.

Area of △ABC = $\frac{1}{2}$ × BC × AD ......(i)

Area of △APC = $\frac{1}{2}$ × PC × AD

= $\frac{1}{2} \times \frac{2}{3} BC \times AD$ ........(ii)

Dividing (ii) by (i) we get,

$$\begin{aligned}\Rightarrow \frac{\text{Area of △APC}}{\text{Area of △ABC}} = \frac{\frac{1}{2} × \frac{2}{3}BC × AD}{\frac{1}{2} \times BC \times AD} \\[1em] \Rightarrow \frac{\text{Area of △APC}}{\text{Area of △ABC}} = \frac{2}{3} \\[1em] \Rightarrow \text{Area of △APC} = \frac{2}{3}\text{Area of △ABC} ..........(iii)\end{aligned}$$

Given,

QA = 5PQ

From figure,

AP = AQ + QP = 5PQ + PQ = 6PQ.

$\frac{AQ}{AP} = \frac{5PQ}{6PQ} = \frac{5}{6}$.

AQ = $\frac{5}{6}AP$

Let CE be the altitude.

Area of △AQC = $\frac{1}{2}$ × AQ × CE ......(iv)

Area of △APC = $\frac{1}{2}$ × AP × CE .......(v)

Dividing (iv) by (v) we get,

$$\begin{aligned}\Rightarrow \frac{\text{Area of △AQC}}{\text{Area of △APC}} = \frac{\frac{1}{2} \times AQ × CE}{\frac{1}{2} \times AP \times CE} \\[1em] \Rightarrow \frac{\text{Area of △AQC}}{\text{Area of △APC}} = \frac{\frac{1}{2} \times \frac{5}{6}AP × CE}{\frac{1}{2} \times AP \times CE} \\[1em] \Rightarrow \frac{\text{Area of △AQC}}{\text{Area of △APC}} = \frac{5}{6} \\[1em] \Rightarrow \text{Area of △AQC} = \frac{5}{6}\text{Area of △APC} \\[1em] \Rightarrow \text{Area of △AQC} = \frac{5}{6} \times \frac{2}{3} \text{ Area of △ABC (From eq iii)} \\[1em] \Rightarrow \text{Area of △AQC} = \frac{5}{9} \text{ area of △ABC } \\[1em] \Rightarrow \frac{\text{Area of △AQC}}{\text{Area of △ABC}} = \frac{5}{9}\end{aligned}$$

∴ area of △AQC : area of △ABC = 5 : 9.

Hence, area of △AQC : area of △ABC = 5 : 9.

(i) From figure,

Let DE be altitude on base AC.

Median divides a triangle into two triangles of equal area.

AD is the median of ∆ABC,

Area of ∆ABD = Area of ∆ADC = $\frac{1}{2}$ Area of ∆ABC .......(1)

It is given that,

⇒ area of ∆ADP : area of ∆ABD = 2 : 3

⇒ area of ∆ADP : area of ∆ADC = 2 : 3

$$\begin{aligned}\Rightarrow \frac{\text{area of ∆ADP}}{\text{area of ∆ADC}} = \frac{2}{3} \\[1em] \Rightarrow \frac{\frac{1}{2} \times AP \times DE}{\frac{1}{2} \times AC \times DE} = \frac{2}{3} \\[1em] \Rightarrow \frac{AP}{AC} = \frac{2}{3}.\end{aligned}$$

Let AP = 2x and AC = 3x.

From figure,

PC = AC - AP = 3x - 2x = x.

$\frac{AP}{PC} = \frac{2x}{x} = \frac{2}{1}.$

Hence, AP : PC = 2 : 1

(ii) We know that,

PC : AC = x : 3x = 1 : 3

So,

$$\begin{aligned}\Rightarrow \frac{\text{Area of ∆PDC}}{\text{Area of ∆ADC}} = \frac{\frac{1}{2} \times PC \times DE}{\frac{1}{2} \times AC \times DE} \\[1em] \Rightarrow \frac{\text{Area of ∆PDC}}{\text{Area of ∆ADC}} = \frac{PC}{AC} \\[1em] \Rightarrow \frac{\text{Area of ∆PDC}}{\text{Area of ∆ADC}} = \frac{x}{3x} \\[1em] \Rightarrow \frac{\text{Area of ∆PDC}}{\text{Area of ∆ADC}} = \frac{1}{3} .........(1)\end{aligned}$$

Since, AD is median of ∆ABC so,

area of ∆ADC = $\frac{1}{2}$ area of ∆ABC

Substituting above value in 1 we get,

$$\begin{aligned}\Rightarrow \frac{\text{Area of ∆PDC}}{\text{Area of ∆ADC}} = \frac{1}{3} \\[1em] \Rightarrow \frac{\text{Area of ∆PDC}}{\frac{1}{2}\text{Area of ∆ABC}} = \frac{1}{3} \\[1em] \Rightarrow \frac{\text{Area of ∆PDC}}{\text{Area of ∆ABC}} = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}.\end{aligned}$$

Hence, proved that area of △PDC : area of △ABC = 1 : 6.

||gm ABCD and ||gm ABEF are on same base (AB) and between the same parallel lines AB and DE, so their areas are equal.

Area of ||gm ABEF = Area of ||gm ABCD = 29 cm².

Area of ||gm ABEF = base × height

⇒ 29 = AB × height

⇒ 29 = 5.8 × height

⇒ Height = $\frac{29}{5.8}$

= 5 cm.

The height of parallelogram ABEF is 5 cm.

Given,

Area of ∆ABD = 24 sq. units

As ∆ABD and ∆ABC lie on same base AB and between same parallel lines AB and CD so,

Area of ∆ABC = Area of ∆ABD = 24 sq. units.

⇒ $\frac{1}{2}$ × AB × height = 24

⇒ $\frac{1}{2}$ × 8 × height = 24

⇒ 4 × height = 24

⇒ Height = $\frac{24}{4}$

⇒ Height = 6 units.

Hence, height of ∆ABC = 6 units.

Join the diagonals AC and BD as shown below:

(i) ∆APD and || gm ABCD are on the same base AD and between the same parallel lines AD and BC.

Area of ∆APD = $\frac{1}{2}$ Area of ||gm ABCD

= $\frac{1}{2}$ × 36

= 18 cm².

Hence, area of ∆APD = 18 cm².

(ii) Let diagonals AC and BD meet at point O.

In ∆ABC,

Since, O is mid-point of AC (as diagonals bisect each other) and E is mid-point of AB so by mid-point theorem,

EO || BC

∴ EF || BC.

Since, BC || AD so,

⇒ EF || AD.

AB || DC (ABCD is a parallelogram)

⇒ AE || DF

Since, EF || AD and AE || DF.

∴ AEFD is a parallelogram.

EF bisects the parallelogram ABCD in two equal halves as E and F are mid-points of AB and CD and EF || BC || AD.

∴ Area of || gm AEFD = $\frac{1}{2}$ Area of || gm ABCD = $\frac{1}{2}$ × 36 = 18 cm².

∴ Area of ∆APD = Area of || gm AEFD.

Hence, AEFD is the required parallelogram which has area equal to the area of ∆APD.

Construct: Through P and Q, draw PR and QS parallel to AB and CD.

Area of ∆APD = Area of ∆AQD [Since ∆APD and ∆AQD lie on the same base AD and between the same parallel lines AD and BC]

Area of ∆APD – Area of ∆AOD = Area of ∆AQD – Area of ∆AOD [On subtracting ar ∆AOD on both sides]

Area of ∆APO = Area of ∆OQD ....... (i)

Area of ∆APO + Area of ∆OPQ = Area of ∆OQD + Area of ∆OPQ [On adding area of ∆OPQ on both sides]

Area of ∆APQ = Area of ∆DPQ ....... (ii)

We know that, ∆APQ and ||gm PQSR are on the same base PQ and between the same parallel lines PQ and AD.

Area of ∆APQ = $\frac{1}{2}$ Area of ||gm PQRS ....... (iii)

From figure,

Height of || gm ABCD = Height of || PQRS = AE

Since, P and Q trisect BC so,

⇒ PQ = $\frac{BC}{3}$

⇒ BC = 3PQ.

$$\begin{aligned}\Rightarrow \frac{\text{Area of || gm ABCD}}{\text{Area of || PQRS}} = \frac{BC \times AE}{PQ \times AE} \\[1em] \Rightarrow \frac{\text{Area of || gm ABCD}}{\text{Area of || PQRS}} = \frac{3PQ \times AE}{PQ \times AE} \\[1em] \Rightarrow \frac{\text{Area of || gm ABCD}}{\text{Area of || PQRS}} = 3 \\[1em] \Rightarrow \text{Area of || PQRS} = \frac{1}{3} \text{Area of || gm ABCD}.\end{aligned}$$

Substituting above value in (iii) we get,

⇒ Area of ∆APQ = $\frac{1}{2} \times \frac{1}{3}$ Area of ||gm ABCD = $\frac{1}{6}$ Area of ||gm ABCD.

⇒ Area of ∆APQ = $\frac{1}{6}$ Area of ||gm ABCD ........(iv)

From (ii) and (iv) we get,

area of ∆APQ = area of ∆DPQ = $\frac{1}{6}$ (area of ||gm ABCD).

Hence, proved that area of ∆APQ = area of ∆DPQ = $\frac{1}{6}$ (area of ||gm ABCD).

We know that, ∆ACE and ∆ADC are on the same base AC and between the same parallel lines AC and DE.

Area of ∆ACE = Area of ∆ADC

Now, adding ar (∆ABC) on both sides, we get

⇒ Area of ∆ACE + Area of ∆ABC = Area of ∆ADC + Area of ∆ABC

⇒ Area of ∆ABE = Area of quad. ABCD

Hence, proved that area of quad. ABCD = area of ∆ABE.

Join BD, which meets AC at P.

In ∆ABD, AP is the median (As P is mid-point of BD because diagonals of || gm bisect each other).

Since, median of triangle divides it into two triangles of equal area.

∴ Area of ∆ABP = Area of ∆ADP .......(i)

Similarly,

PO is median of ∆BOD,

∴ Area of ∆BOP = Area of ∆POD .......(ii)

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

⇒ Area of ∆ABP + Area of ∆BOP = Area of ∆ADP + Area of ∆POD

⇒ Area of ∆AOB = Area of ∆AOD.

Hence, proved that area of ∆AOB = area of ∆AOD.

(a) Join BG.

We know that,

Since, ∆ABG and || gm ABCD lie on same base AB and between same parallel lines AB and CD.

Area of ∆ABG = $\frac{1}{2}$ Area of ||gm ABCD ......(i)

Since, ∆ABG and || gm AEFG lie on same base AG and between same parallel lines AG and EF.

Area of ∆ABG = $\frac{1}{2}$ Area of ||gm AEFG ......(ii)

From (i) and (ii) we get,

$\frac{1}{2}$ Area of || gm ABCD = $\frac{1}{2}$ Area of || gm AEFG

⇒ Area of || gm ABCD = Area of || gm AEFG

Hence, proved that area of || gm ABCD = area of || gm AEFG.

Join AC and EF.

Since, ∆AFC and ∆AFE lie on same base AF and between same parallel lines AF and CE so,

area of ∆AFC = area of ∆AFE

Now, subtract area of ∆ABF on both sides,

area of (∆AFC - ∆ABF) = area of (∆AFE – ∆ABF)

area of ∆ABC = area of ∆BEF

2 area of ∆ABC = 2 area of ∆BEF .......(i)

From figure,

As || gm ABCD and ∆ABC lie on same base AB and between same parallel lines AB and DC.

Area of ∆ABC = $\frac{1}{2}$ Area of || gm ABCD

Area of || ABCD = 2 Area of ∆ABC .........(ii)

As || gm BFGE and ∆BEF lie on same base BE and between same parallel lines FG and BE.

Area of ∆BEF = $\frac{1}{2}$ Area of || gm BFGE

Area of || BFGE = 2 Area of ∆BEF .........(iii)

Substituting value from (ii) and (iii) in (i) we get,

Area of || ABCD = Area of || BFGE.

Hence, proved that area of || ABCD = area of || BFGE.

We know that,

AD || BE ⇒ AD || EG

ED || FA ⇒ ED || GA

Since, opposite sides are parallel.

Hence, ADEG is a parallelogram.

Since || gm ABCD and || gm ADEG lie on same base AD and between same parallel lines AD and EB,

area of || gm ABCD = area of ||gm ADEG ....... (i)

We know that,

ED || FA ⇒ DE || FH

DH || EF

Since, opposite sides are parallel.

Hence, DEFH is a parallelogram.

Since || gm DEFH and || gm ADEG lie on same base DE and between same parallel lines DE and FA,

area of ||gm DEFH = area of ||gm ADEG ....... (ii)

From (i) and (ii) we get,

⇒ area of ||gm ABCD = area of ||gm DEFH

Hence, proved that area of ||gm ABCD = area of ||gm DEFH.

∆ABC with point D on the side BC and AD produced to E such that AD = DE is shown below:

In ∆ABE, it is given that AD = DE.

∴ BD is the median of ∆ABE

⇒ area of ∆ABD = area of ∆BED ........ (i)

Similarly,

In ∆ACE, it is given that AD = AE

∴ CD is the median of ∆ACE

⇒ area of ∆ACD = area of ∆CED ........ (ii)

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

⇒ area of ∆ABD + area of ∆ACD = area of ∆BED + area of ∆CED

⇒ area of ∆ABC = area of ∆BCE.

Hence, proved that area of ∆ABC = area of ∆BCE.

In ∆APD and ∆PQB,

AP = BP [Since P is the mid-point of AB]

∠DAP = ∠QBP [each angle is 90° as ABCD is a rectangle]

∠APD = ∠BPQ [Vertically opposite angles are equal]

So, ∆APD ≅ ∆PQB [By using ASA axiom]

So, area of ∆APD = area of ∆PQB .......(i) (As both triangles are congruent.)

From figure,

area of rectangle ABCD = area of ∆APD + area of quad. PBCD

= area of ∆PQB + area of quad. PBCD ......(From i)

= area of ∆DQC.

Hence, proved that area of rectangle ABCD = area of ∆DQC..

Let AB = p

Since, opposite sides of || gm are equal.

Perimeter of || gm ABCD = 2(AB + BC)

⇒ 42 = 2(p + BC)

⇒ p + BC = $\frac{42}{2}$

⇒ p + BC = 21

⇒ BC = 21 – P

area of ||gm ABCD = base × height = AB × DM = p × 6 = 6p .......(i)

Also, area of ||gm ABCD = BC × DN

= (21 – p) × 8

= 8(21 – p) ............ (ii)

From (i) and (ii), we get

⇒ 6p = 8(21 – p)

⇒ 6p = 168 – 8p

⇒ 6p + 8p = 168

⇒ 14p = 168

⇒ p = $\frac{168}{14}$ = 12 cm.

⇒ 21 - p = 21 - 12 = 9 cm.

Hence, the sides of ||gm are AB = 12 cm and BC = 9 cm.

Let us consider BC = p, CA = q.

From figure,

Perimeter of ∆ABC = AB + BC + CA

⇒ 37 = AB + p + q

⇒ AB = 37 – (p + q)

Area of ∆ABC = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × BC × AM

= $\frac{1}{2}$ × p × $5x$

∴ Area of ∆ABC = $\frac{1}{2}$ × p × $5x$ .........(i)

Also,

Area of ∆ABC = $\frac{1}{2}$ × CA × BN

= $\frac{1}{2}$ × q × $6x$

∴ Area of ∆ABC = $\frac{1}{2}$ × q × $6x$ .........(ii)

Also,

Area of ∆ABC = $\frac{1}{2}$ × AB × CL

= $\frac{1}{2}$ × (37 - p - q) × $4x$

∴ Area of ∆ABC = $\frac{1}{2}$ × (37 - p - q) × $4x$ .........(iii)

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

$$\begin{aligned}\Rightarrow \frac{1}{2} \times p \times 5x = \frac{1}{2} \times q \times 6x \\[1em] \Rightarrow p = q \times \frac{6x}{5x} \\[1em] \Rightarrow p = \frac{6q}{5}.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{1}{2} \times q \times 6x = \frac{1}{2} \times (37 - p - q) \times 4x \\[1em] \Rightarrow q = (37 - p - q) \times \frac{4x}{6x} \\[1em] \Rightarrow q = \Big(37 - \frac{6q}{5} - q\Big) \times \frac{2}{3} .....\text{ (From (i))} \\[1em] \Rightarrow \frac{3q}{2} = \frac{185 - 6q - 5q}{5} \\[1em] \Rightarrow 3q \times 5 = 2(185 - 11q) \\[1em] \Rightarrow 15q = 370 - 22q \\[1em] \Rightarrow 37q = 370 \\[1em] \Rightarrow q = \frac{370}{37} \\[1em] \Rightarrow q = 10.\end{aligned}$$

⇒ CA = q = 10 cm.

⇒ BC = p = $\frac{6q}{5} = \frac{6}{5} \times 10 = 12$ cm.

⇒ AB = (37 - p - q) = (37 - 12 - 10) = 15 cm.

Hence, AB = 15 cm, BC = 12 cm and CA = 10 cm.

(i) Since,

∆APB and || gm ABCD have same base AB and are between same parallel lines AB and DC. So,

area of ∆APB = $\frac{1}{2}$ area of ||gm ABCD

From figure,

⇒ $\frac{1}{2}$ area of ||gm ABCD = area of (∆DAP + ∆BCP)

⇒ $\frac{1}{2}$ area of ||gm ABCD = 25 + 15 = 40

⇒ area of ||gm ABCD = 2 × 40 = 80 cm².

Hence, area of ||gm ABCD = 80 cm².

(ii) From figure,

∆DAP and ∆BCP are on the same base CD and between the same parallel lines CD and AB.

$$\begin{aligned}\Rightarrow \frac{\text{Area of ∆DAP}}{\text{Area of ∆BCP}} = \frac{DP}{PC} \\[1em] \Rightarrow \frac{25}{15} = \frac{DP}{PC} \\[1em] \Rightarrow \frac{DP}{PC} = \frac{5}{3}.\end{aligned}$$

Hence, DP : PC = 5 : 3.

In △ABC,

As E is mid-point of AB and EF || BC so,

G is mid-point of AC (By mid-point theorem)

∴ AG = GC.

In ∆AEG and ∆CFG,

∠EAG = ∠GCF (Alternate angles are equal)

∠EGA = ∠CGF (Vertically opposite angles are equal)

AG = GC (Proved above)

Hence, ∆AEG ≅ ∆CFG (By ASA axiom)

∴ area of ∆AEG = area of ∆CFG ..........(i)

From figure,

area of || gm EBCF = area of quad. BCGE + area of ∆CFG

= area of quad. BCGE + area of ∆AEG ........(from i)

= area of ∆ABC = 25 sq. units.

Hence, area of ||gm EBCF = 25 sq. units.

Join CE and AC.

From figure,

∆ABC and ∆EBC lie on same base BC and between same parallel lines BC and AE.

∴ area of ∆ABC = area of ∆EBC ....... (i)

From figure,

∆EBC and ∆EBD lie on same base BE and between same parallel lines CD and BE.

∴ area of ∆EBC = area of ∆EBD ....... (ii)

From (i) and (ii), we get

area of ∆ABC = area of ∆EBD.

Hence, proved that area of ∆ABC = area of ∆EBD.

(i) From figure,

⇒ ∠FBC = ∠FBA + ∠ABC

∠FBA = 90° (As each angle of a square = 90°)

⇒ ∠FBC = 90° + ∠ABC ....... (1)

Also,

⇒ ∠ABE = ∠CBE + ∠ABC

∠CBE = 90° (As each angle of a square = 90°)

⇒ ∠ABE = 90° + ∠ABC .......(2)

From (1) and (2), we get

∠FBC = ∠ABE ........ (3)

Now, in ∆BCF and ∆ABE

BF = AB (As FBAG is a square)

∠FBC = ∠ABE (Proved above)

BC = BE (As BCDE is a square)

By using the SAS axiom rule of congruency,

∴ ∆BCF ≅ ∆ ABE.

Hence, proved that ∆BCF ≅ ∆ ABE.

(ii) We know that,

∆BCF ≅ ∆ABE

So, area of ∆BCF = area of ∆ABE ...... (4)

Since, ∆BCF and square AGFB have same base FB and are between same parallel lines FB and GC.

area of ∆BCF = $\frac{1}{2}$ area of square AGFB ........(5)

∆ABE and rectangle BENM are on same base BE and are between same parallel lines BE and AN.

area of ∆ABE = $\frac{1}{2}$ area of rectangle BENM ....... (6)

From (4), (5) and (6)

⇒ $\frac{1}{2}$ area of square AGFB = $\frac{1}{2}$ area of rectangle BENM

⇒ area of square AGFB = area of rectangle BENM

Hence, proved that area of square AGFB = area of rectangle BENM.

We know that,

A parallelogram and a rectangle on the same base and between the same parallel lines are equal in area.

Since, || ABEF and rectangle CDEF are on same base EF and between same parallel lines l and m.

∴ area of || ABEF = area of rect. CDEF.

Hence, Option 1 is the correct option.

We know that,

Two parallelograms on equal bases and between the same parallel lines have equal areas.

So, ratio of their areas = 1 : 1.

Hence, Option 2 is the correct option.

We know that,

Area of a triangle is half that of a parallelogram on the same base and between the same parallel lines.

Let area of parallelogram be x so area of triangle = $\frac{x}{2}$.

Ratio of area of the triangle to the area of the parallelogram = $\frac{\frac{x}{2}}{x} = \frac{1}{2} = 1 : 2.$

Hence, Option 2 is the correct option.

A median of a triangle divides it into two triangles of equal area.

Hence, Option 1 is the correct option.

Area of || gm = base × height.

From figure,

Base = DC and Height = DL.

∴ Area of || gm = DC × DL.

Hence, Option 3 is the correct option.

Let CDEF be a parallelogram.

So, diagonal DF divides it into two triangles of equal area.

∴ area of △CDF = area of △EDF ........(i)

E and F are mid-points of side AB and AC respectively.

By mid-point theorem,

EF = $\frac{1}{2}$ BC = BD (As D is mid-point of BC) and EF || BC.

Since, FE || BC so from figure,

EF || BD and EF = BD.

So, EBDF is a paralleogram with diagonal ED dividing it into two triangles of equal area.

∴ area of △EBD = area of △EDF ........(ii)

E and D are mid-points of side AB and BC respectively.

By mid-point theorem,

ED = $\frac{1}{2}$ AC = AF (As F is mid-point of AC) and ED || AC.

Since, ED || AC so from figure,

ED || AF.

So, AEDF is a paralleogram with diagonal EF dividing it into two triangles of equal area.

∴ area of △AEF = area of △EDF ........(iii)

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

area of △EDF = area of △CDF = area of △EBD = area of △AEF = x

From figure,

area of △ABC = area of △EDF + area of △CDF + area of △EBD + area of △AEF = 4x.

area of || gm EDCF = area of △EDF + area of △CDF = 2x.

So,

$$\begin{aligned}\Rightarrow \frac{\text{area of || gm EDCF}}{\text{area of △ABC}} = \frac{2x}{4x} \\[1em] \Rightarrow \frac{\text{area of || gm EDCF}}{\text{area of △ABC}} = \frac{1}{2} \\[1em]\end{aligned}$$

area of || gm EDCF = $\frac{1}{2}$ area of △ABC.

Hence, Option 1 is the correct option.

It is given that

AB = a cm

DC = b cm

AB || DC

E and F are the mid-points of AD and BC

Consider h as the distance between AB, CD and EF

Now join BD which intersects EF at M

In ∆ABD,

E is the midpoint of AD and EM || AB

By midpoint theorem,

M is the midpoint of BD

and

EM = $\frac{1}{2}$ AB ........ (1)

In ∆CBD,

F is mid-point of BC and M is mid-point of BD so by mid-point theorem,

MF = $\frac{1}{2}$ CD ......... (2)

Adding equations (1) and (2)

EM + MF = $\frac{1}{2}$ AB + $\frac{1}{2}$ CD

EF = $\frac{1}{2}$ (AB + CD)

EF = $\frac{1}{2}$ (a + b)

Here,

Area of trapezium ABFE = $\frac{1}{2}$ [sum of parallel sides] × [distance between parallel sides]

Substituting the values,

$$\begin{aligned}\text{Area of trap. ABFE} = \frac{1}{2} \Big[a + \frac{1}{2} (a + b)\Big] × h \\[1em] = \frac{1}{2}\Big[\frac{2a + a + b}{2}\Big]h \\[1em] = \frac{h}{4}[3a + b].\end{aligned}$$

Similarly,

$$\begin{aligned}\text{Area of trap. EFCD } = \frac{1}{2}\Big[b + \frac{1}{2}(a + b)\Big] × h \\[1em] = \frac{1}{2}\Big[\frac{2b + a + b}{2}\Big]h \\[1em] = \frac{h}{4}[3b + a].\end{aligned}$$

Required ratio = Area of trapezium ABFE / Area of trapezium EFCD

By substituting the values,

$$\begin{aligned}\text{Ratio } = \frac{\frac{h}{4}[3a + b]}{\frac{h}{4}[3b + a]} \\[1em] = \frac{3a + b}{3b + a} \\[1em] = (3a + b) : (3b + a).\end{aligned}$$

Hence, Option 2 is the correct option.

We know that,

Triangles on the same base and between same parallel lines are equal in area.

Hence,

⇒ area of △ABC = area of △ABD ........ (1)

⇒ area of △ACD = area of △BCD

From figure and eqn. (1),

area of (△AOB + △OAD) = area of (△AOB + △OBC)

⇒ area of △OAD = area of △OBC.

Hence, Option 3 is the correct option.

Let ABCD be a parallelogram in which E and F are mid-points of AB and CD respectively. Join EF.

Let us construct DG ⊥ AB and let DG = h, where h is the altitude on side AB.

Area of || gm ABCD = base × height = AB × h

Area of ||gm AEFD = AE × h = $\frac{AB}{2}$ × h ...................(1) [Since E is the mid-point of AB]

Area of ||gm EBCF = EB × h = $\frac{AB}{2}$ × h ...................(2) [Since E is the mid-point of AB]

From (1) and (2),

Area of || gm AEFD = Area of || gm EBCF.

Thus, the line segment joining the mid-points of a pair of opposite sides of a parallelogram divides it into two equal parallelograms.

∴ Statement 1 is true.

From figure,

The diagonals AC and BD cut at point O.

In parallelogram, the diagonals bisect each other.

∴ AO = OC

In ∆ACD, O is the mid-point of AC.

∴ OD is the median.

Area of ∆AOD = Area of ∆COD ................. (3) [Median of ∆ divides it into two triangles of equal areas.]

Similarly, in ∆ABC

O is the mid-point of AC.

∴ OB is the median.

Area of ∆AOB = Area of ∆COB ................. (4) [Median of ∆ divides it into two triangles of equal areas.]

In ∆ADB,

O is the mid-point of BD.

∴ OA is the median.

Area of ∆AOD = Area of ∆AOB ................. (5)

From (3), (4) and (5) we get,

Area of ∆AOB = Area of ∆COB = Area of ∆COD = Area of ∆AOD

So proved, that the diagonals of a parallelogram divide it into four triangles of equal area.

∴ Statement 2 is true.

∴ Both the statements are true.

Hence, option 1 is the correct option.

By formula,

Area of trapezium = $\frac{1}{2}$ (sum of parallel sides) x height.

∴ Assertion (A) is true.

We know that,

Parallelograms on the same base and between the same parallel lines are equal in area.

A rectangle is a special type of parallelogram.

Thus, a parallelogram and a rectangle on the same base and between the same parallel lines have equal areas.

∴ Reason (R) is true.

∴ Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Hence, option 4 is the correct option.

When two figures are referred to as "equal figures," it specifically means that they have the same area. It does not imply that they are congruent.

∴ Reason (R) is true.

As we know that area of a triangle = $\frac{1}{2}$ x base x height

Area of first triangle = $\frac{1}{2}$ x 8 x 6

= $\frac{1}{2}$ x 48

= 24 cm².

Area of second triangle = $\frac{1}{2}$ x 12 x 4

= $\frac{1}{2}$ x 48

= 24 cm².

So, area of first triangle = area of second triangle.

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Having equal areas does not necessarily mean two triangles are congruent.Congruent triangles must have the same shape and the same size.

Triangles with the same area can have different shapes.

∴ Assertion (A) is false.

Congruent triangles are identical in every aspect; they have the same shape and the same size.

If two figures are identical in size, they must occupy the same amount of space, and therefore, they must have the same area.

∴ Reason (R) is true.

∴ Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

(i) We know that,

A parallelogram and a rectangle on the same base and between the same parallel lines are equal in area.

From figure,

rectangle ABCD and parallelogram DEFC are on the same base DC and between same parallel lines DG and AF.

Hence, area of || gm DEFC = area of rectangle ABCD

= AB × AD

= 4 × 6

= 24 cm².

Hence, area of || gm DEFC = 24 cm².

(ii) We know that,

Area of a triangle is half that of a parallelogram on the same base and between the same parallel lines.

Since, triangle GEF and || gm DEFC are on same base EF and between same parallel lines DG and AF so,

area of △EFG = $\frac{1}{2}$ area of || gm DEFC

= $\frac{1}{2} \times 24 = 12$ cm².

Hence, area of △EFG = 12 cm².

From figure,

Area of ∆ACD = $\frac{1}{2}$ Area of || gm ACRS [As both are on same base AC and between the same parallel lines AC and SR]

⇒ Area of || gm ACRS = 2 Area of ∆ACD .......(i)

Similarly,

Area of ∆ABC = $\frac{1}{2}$ Area of || gm ∆APQC [As both are on same base AC and between the same parallel lines AC and PQ]

⇒ Area of || gm APQC = 2 Area of ∆ABC .......(ii)

Adding (i) and (ii),

⇒ Area of || gm ACRS + Area of || gm APQC = 2 Area of ∆ACD + 2 Area of ∆ABC

⇒ Area of || gm PQRS = 2[Area of ∆ACD + Area of ∆ ABC]

⇒ Area of || gm PQRS = 2(Area of quad. ABCD)

Hence, proved that area of || gm PQRS = 2 x area of quad. ABCD.

Parallelogram ABCD with point P on the side AB and Q on the side BC is shown below:

∆CPD and || gm ABCD are on the same base CD and between the same parallels lines AB and CD.

∴ Area of ∆CPD = $\frac{1}{2}$ Area of || gm ABCD ......(1)

∆AQD and || gm ABCD are on the same base AD and between the same parallel lines AD and BC,

∴ Area of ∆AQD = $\frac{1}{2}$ Area of || gm ABCD .......(2)

From (1) and (2),

Area of ∆CPD = Area of ∆AQD.

Hence, proved that area of ∆CPD = area of ∆AQD.

(ii) From part (i) we get,

Area of ∆CPD = $\frac{1}{2}$ Area of || gm ABCD

∴ Area of || gm ABCD - Area of ∆CPD = $\frac{1}{2}$ Area of || gm ABCD ........(3)

From figure,

Area of || gm ABCD - Area of ∆CPD = Area of ∆APD + Area of ∆CPB .........(4)

From (3) and (4) we get,

⇒ Area of ∆APD + Area of ∆CPB = $\frac{1}{2}$ Area of || gm ABCD

Since,

Area of ∆ADQ = $\frac{1}{2}$ Area of || gm ABCD (From eq 2)

⇒ Area of ∆APD + Area of ∆CPB = Area of ∆ADQ.

Hence, proved that area of ∆ADQ = area of ∆APD + area of ∆CPB.

From figure,

∆LZX and ∆MZX are on the same base XZ and between the same parallel lines LM and XZ.

∴ Area of ∆LZX = Area of ∆MZX

Adding area ∆XZY to both sides of the above equation we get,

⇒ area of ∆LZX + area ∆XZY = area ∆MZX + area ∆XZY

From figure,

∆LZX + ∆XZY = ∆LZY and ∆MZX + ∆XZY = MZYX.

∴ Area of ∆LZY = Area of quadrilateral MZYX.

Hence, proved that area of ∆LZY = area of quadrilateral MZYX.

Since, ABC is an equilateral triangle. Let each side be x cm.

PN, PM, and PL are perpendicular on side AB, AC and BC respectively. AD is any altitude from point A on side BC.

Join PA, PB and PC.

Area of ∆ABC = $\frac{1}{2}$ × Base × Altitude

= $\frac{1}{2} × BC × AD = \frac{x}{2}AD$........(1)

Area of ∆APB = $\frac{1}{2}$ × AB × NP ........(2)

Area of ∆APC = $\frac{1}{2}$ × AC × MP ........(3)

Area of ∆BPC = $\frac{1}{2}$ × BC × LP ........(4)

Adding (2), (3) and (4)

⇒ Area of (∆APB + ∆APC + ∆BPC) = $\frac{1}{2}$ × (AB × NP + AC × MP + BC × LP)

From figure,

$$\begin{aligned}⇒ \text{Area of ∆ABC} = \frac{1}{2} [\text{AB} \times \text{NP} + \text{AC} \times \text{MP} + \text{BC} \times \text{LP}] \\[1em] = \frac{1}{2} [x \times \text{NP} + x \times \text{MP} + x \times \text{LP}] \\[1em] = \frac{x}{2} [\text{NP} + \text{MP} + \text{LP}]\end{aligned}$$

∴ Area of ∆ABC = $\frac{x}{2}$ [NP + MP + LP] .......(5)

From (1) and (5),

$\frac{x}{2}$ × AD = $\frac{x}{2}$ × (NP + LP + MP)

⇒ AD = NP + LP + MP.

Hence, proved that the sum of three perpendiculars is equal to the altitude of the triangle.

Let ABCD be a quadrilateral such that each diagonal divides it into triangles of equal areas, then

area of △ABC = $\frac{1}{2}$ Area of ABCD, .......(1)

area of △ABD = $\frac{1}{2}$ Area of ABCD, ......(2)

area of △BCD = $\frac{1}{2}$ Area of ABCD, .......(3)

We know that,

Triangles on the same base and having equal areas lie between the same parallel lines.

From (1) and (2) we get,

Area of △ABC = Area of △ABD.

Since, △ABC and △ABD lie on same base AB and have equal area.

So, AB || CD.

From (1) and (3) we get,

∴ Area of △ABC = Area of △BCD.

Since, △ABC and △BCD lie on same base BC and have equal area.

So, BC || AD.

Since, AB || CD and BC || AD.

Hence, proved ABCD is a parallelogram.

From figure,

BE is a straight line.

Since, BC || AD,

∴ CE || AD.

Hence, ACED is a parallelogram.

Diagonals AE and DC of || ACED bisect each other, so F is mid-point of DC.

So, BF is median of △BDC.

Since, median divides triangle into two triangles with equal areas.

∴ area of △BFC = area of △DFB = 3 cm².

area of △BDC = area of △BFC + area of △DFB = 6 cm².

From figure,

⇒ area of △BDC = $\frac{1}{2}$ area of || gm ABCD (As || gm ABCD and △BDC lie on same base CD and between same parallel lines AB and CD.)

⇒ 6 = $\frac{1}{2}$ area of || gm ABCD

⇒ area of || gm ABCD = 12 cm².

Hence, area of || gm ABCD = 12 cm².

In ∆ABE and ∆ADF,

AB = AD (Sides of a square)

∠B = ∠D (Each angle of a square = 90°)

BE = DF (E is mid-point of BC and F is mid-point of DC)

∴ ∆ABE ≅ ∆ADF (SAS axiom)

∴ AE = AF (c.p.c.t.)

Again in ∆AER and ∆AFR

AE = AF (Proved above)

AR = AR (Common)

ER = FR (R is mid-point of EF)

∴ ∆AER ≅ ∆AFR (SSS axiom)

∴ Area of ∆AER = Area of ∆AFR

Hence, proved that area of ∆AER = area of ∆AFR.

X and Y are the mid-points of sides AC and AB respectively.

Since, X and Y are midpoints of AC and AB respectively.

In ∆ABC,

XY || BC (By midpoint theorem).

Given, QP || BC

∴ QP || BC || XY

In ∆BAP, Y is mid of AB and XY || AP

∴ X is mid-point of BP (Converse of mid-point theorem)

∴ XY = $\frac{1}{2}$AP .......(1)

Similarly we can prove in ∆AQC

X is mid-point of AC and XY is parallel to QA

∴ Y is mid-point of QC (Converse of mid-point theorem)

XY = $\frac{1}{2}$ QA .......(2)

From (1) and (2),

⇒ $\frac{1}{2}QA = \frac{1}{2}AP$

⇒ QA = AP.

Thus, ∆ABP and ∆ACQ are on the equal bases (QA = AP) and between the same parallel lines BC and QP

∴ Area of ∆ABP = Area of ∆ACQ.

Hence, proved that Area of ∆ABP = Area of ∆ACQ.