Theorems on Area — Exercise 13

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

Answer

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 = $\dfrac{AB}{2}$ × h .......(i) [Since E is the mid-point of AB]

Area of ||gm EBCF = EB × h = $\dfrac{AB}{2}$ × 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.

Prove that the diagonals of a parallelogram divide it into four triangles of equal area.

Answer

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.

In figure (1) given below, AD is the median of ∆ABC and P is any point on AD. Prove that

(i) Area of ∆PBD = area of ∆PDC.

(ii) Area of ∆ABP = area of ∆ACP.

Answer

(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.

In the figure (2) given below, DE || BC. Prove that

(i) area of ∆ACD = area of ∆ABE

(ii) area of ∆OBD = area of ∆OCE.

Answer

(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.

In figure (1) given below, ABCD is a parallelogram and P is any point in BC. Prove that, Area of ∆ABP + area of ∆DPC = Area of ∆APD.

Answer

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 = $\dfrac{1}{2}$ 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,

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

From (i),

$\text{Area of ∆APD} = \dfrac{\text{Area of ∆APD}}{2} + \dfrac{\text{Area of ∆ABP}}{2} + \dfrac{\text{Area of ∆DPC}}{2} \\[1em] \text{Area of ∆APD} - \dfrac{\text{Area of ∆APD}}{2} = \dfrac{\text{Area of ∆ABP}}{2} + \dfrac{\text{Area of ∆DPC}}{2} \\[1em] \dfrac{\text{Area of ∆APD}}{2} = \dfrac{\text{Area of ∆ABP}}{2} + \dfrac{\text{Area of ∆DPC}}{2} \\[1em] \text{Area of ∆APD} = \text{Area of ∆ABP} + \text{Area of ∆DPC}$

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

In the figure (2) given below, O is any point inside a parallelogram ABCD. Prove that

(i) area of ∆OAB + area of ∆OCD = $\dfrac{1}{2}$ area of || gm ABCD.

(ii) area of ∆OBC + area of ∆OAD = $\dfrac{1}{2}$ area of || gm ABCD.

Answer

(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 = $\dfrac{1}{2}$ 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 = $\dfrac{1}{2}$ Area of || gm PQCD ..... (2)

Now by adding (1) and (2),

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

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

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

Hence, proved that Area of ∆OAB + Area of ∆OCD = $\dfrac{1}{2}$ 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

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

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

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

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

If E, F, G and H are mid-points of the sides AB, BC, CD and DA, respectively of a parallelogram ABCD, prove that area of the quad. EFGH = $\dfrac{1}{2}$ area of || gm ABCD.

Answer

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

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

BF = $\dfrac{1}{2}$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 = $\dfrac{1}{2}$ area of || gm ABFH .......(i)

Similarly,

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

FC = $\dfrac{1}{2}$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 = $\dfrac{1}{2}$ area of || gm HFCD .......(ii)

Adding (i) and (ii) we get,

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

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

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

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

In figure (1) given below, ABCD is a parallelogram. P, Q are any two points on the sides AB and BC respectively. Prove that

area of ∆CPD = area of ∆AQD.

Answer

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

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

From (i) and (ii)

Area of ∆CPD = Area of ∆AQD

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

In the figure (2) given below, PQRS and ABRS are parallelograms, and X is any point on the side BR. Show that

area of ∆AXS = $\dfrac{1}{2}$ area of || gm PQRS.

Answer

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 = $\dfrac{1}{2}$ Area of ||gm ABRS

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

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

D, E and F are mid-points of the sides BC, CA and AB respectively of a ∆ABC. Prove that

(i) FDCE is a parallelogram

(ii) area of ∆DEF = $\dfrac{1}{4}$ area of ∆ABC

(iii) area of || gm FDCE = $\dfrac{1}{2}$ area of ∆ABC

Answer

∆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 = $\dfrac{1}{2}$ BC .........(1)

Also, D is the mid-point of BC

CD = $\dfrac{1}{2}$ 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 = $\dfrac{1}{2}$ AC .........(4)

Also, E is the mid-point of AC

EC = $\dfrac{1}{2}$ 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 = $\dfrac{1}{4}$ area of ∆ABC

Hence, proved that area of ∆DEF = $\dfrac{1}{4}$ 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

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

= $\dfrac{1}{2}$ area of ∆ABC.

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

In the adjoining figure, D, E and F are midpoints of the sides BC, CA and AB respectively of ∆ABC. Prove that BCEF is a trapezium and area of the trap. BCEF = $\dfrac{3}{4}$ area of ∆ABC.

Answer

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

By mid-point theorem,

DE || AB and DE = $\dfrac{1}{2}$ 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 = $\dfrac{1}{2}$ 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 = $\dfrac{1}{2}$ 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 = $\dfrac{1}{4}$ 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 = $\dfrac{1}{3}$ x area of trapezium BCEF .......(vi)

From (v) and (vi) we get,

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

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

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

In figure (1) given below, point D divides the side BC of ∆ABC in the ratio m : n. Prove that area of ∆ABD : area of ∆ADC = m : n.

Answer

From fig (1)

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

BD : DC = m : n

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

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

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

Dividing (i) by (ii)

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

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

In the figure (2) given below, P is a point on the side BC of ∆ABC such that PC = 2BP, and Q is a point on AP such that QA = 5PQ, find area of ∆AQC : area of ∆ABC.

Answer

Given, PC = 2BP

From figure,

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

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

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

Let AD be the altitude.

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

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

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

Dividing (ii) by (i) we get,

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

Given,

QA = 5PQ

From figure,

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

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

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

Let CE be the altitude.

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

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

Dividing (iv) by (v) we get,

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

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

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

In the figure (3) given below, AD is a median of △ABC and P is a point in AC such that area of △ADP : area of △ABD = 2 : 3. Find

(i) AP : PC

(ii) area of △PDC : area of △ABC

Answer

(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 = $\dfrac{1}{2}$ Area of ∆ABC .......(1)

It is given that,

⇒ area of ∆ADP : area of ∆ABD = 2 : 3

⇒ area of ∆ADP : area of ∆ADC = 2 : 3

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

Let AP = 2x and AC = 3x.

From figure,

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

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

Hence, AP : PC = 2 : 1

(ii) We know that,

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

So,

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

Since, AD is median of ∆ABC so,

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

Substituting above value in 1 we get,

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

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

In figure (1) given below, area of parallelogram ABCD is 29 cm². Calculate the height of parallelogram ABEF if AB = 5.8 cm.

Answer

||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 = $\dfrac{29}{5.8}$

= 5 cm.

The height of parallelogram ABEF is 5 cm.

In figure (2) given below, area of ∆ABD is 24 sq. units. If AB = 8 units, find the height of △ABC.

Answer

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.

⇒ $\dfrac{1}{2}$ × AB × height = 24

⇒ $\dfrac{1}{2}$ × 8 × height = 24

⇒ 4 × height = 24

⇒ Height = $\dfrac{24}{4}$

⇒ Height = 6 units.

Hence, height of ∆ABC = 6 units.

In figure (3) given below, E and F are midpoints of sides AB and CD, respectively, of parallelogram ABCD. If the area of parallelogram ABCD is 36 cm²,

(i) state the area of ∆APD.

(ii) Name the parallelogram whose area is equal to the area of ∆APD.

Answer

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

= $\dfrac{1}{2}$ × 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 = $\dfrac{1}{2}$ Area of || gm ABCD = $\dfrac{1}{2}$ × 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.

In figure (1) given below, ABCD is a parallelogram. Points P and Q on BC trisect BC into three equal parts. Prove that :

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

Answer

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

From figure,

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

Since, P and Q trisect BC so,

⇒ PQ = $\dfrac{BC}{3}$

⇒ BC = 3PQ.

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

Substituting above value in (iii) we get,

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

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

From (ii) and (iv) we get,

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

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

In figure (2) given below, DE is drawn parallel to the diagonal AC of the quadrilateral ABCD to meet BC produced at point E. Prove that area of quad. ABCD = area of ∆ABE.

Answer

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.

In the figure (3) given below, ABCD is a parallelogram. O is any point on the diagonal AC of the parallelogram. Show that the area of ∆AOB is equal to the area of ∆AOD.

Answer

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.

In the figure (1) given, ABCD and AEFG are two parallelograms. Prove that area of || gm ABCD = area of || gm AEFG.

Answer

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

From (i) and (ii) we get,

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

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

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

In figure (2) given below, the side AB of the parallelogram ABCD is produced to E. A straight line through A is drawn parallel to CE to meet CB produced at F and parallelogram BFGE is completed. Prove that

area of || gm BFGE = area of || gm ABCD.

Answer

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 = $\dfrac{1}{2}$ 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 = $\dfrac{1}{2}$ 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.

In figure (3) given below, AB || DC || EF, AD || BE and DE || AF. Prove that the area of DEFH is equal to the area of ABCD.

Answer

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.

Any point D is taken on the side BC of a ∆ABC and AD is produced to E such that AD = DE, prove that area of ∆BCE = area of ∆ABC.

Answer

∆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.

ABCD is a rectangle and P is the mid-point of AB. DP is produced to meet CB at Q. Prove that the area of rectangle ABCD = area of ∆DQC.

Answer

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..

In figure (1) given below, the perimeter of the parallelogram is 42 cm. Calculate the lengths of the sides of the parallelogram.

Answer

Let AB = p

Since, opposite sides of || gm are equal.

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

⇒ 42 = 2(p + BC)

⇒ p + BC = $\dfrac{42}{2}$

⇒ 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 = $\dfrac{168}{14}$ = 12 cm.

⇒ 21 - p = 21 - 12 = 9 cm.

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

In the figure (2) given below, the perimeter of ∆ABC is 37 cm. If the lengths of the altitudes AM, BN and CL are 5x, 6x, and 4x respectively, calculate the lengths of the sides of ∆ABC.

Answer

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 = $\dfrac{1}{2}$ × base × height

= $\dfrac{1}{2}$ × BC × AM

= $\dfrac{1}{2}$ × p × $5x$

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

Also,

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

= $\dfrac{1}{2}$ × q × $6x$

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

Also,

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

= $\dfrac{1}{2}$ × (37 - p - q) × $4x$

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

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

$\Rightarrow \dfrac{1}{2} \times p \times 5x = \dfrac{1}{2} \times q \times 6x \\[1em] \Rightarrow p = q \times \dfrac{6x}{5x} \\[1em] \Rightarrow p = \dfrac{6q}{5}.$

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

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

⇒ CA = q = 10 cm.

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

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

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

In the figure(3) given below, ABCD is a parallelogram. P is a point on DC such that area of ∆DAP = 25 cm² and the area of ∆BCP = 15 cm². Find

(i) area of || gm ABCD

(ii) DP : PC.

Answer

(i) Since,

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

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

From figure,

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

⇒ $\dfrac{1}{2}$ 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.

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

Hence, DP : PC = 5 : 3.

In the adjoining figure, E is the midpoint of the side AB of a triangle ABC and EBCF is a parallelogram. If the area of ∆ ABC is 25 sq. units, find the area of || gm EBCF.

Answer

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.

In the figure (1) given below, BC || AE and CD || BE. Prove that

area of ∆ABC= area of ∆EBD.

Answer

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.

In figure (2) given below, ABC is a right-angled triangle at A. AGFB is a square on the side AB and BCDE is a square on the hypotenuse BC. If AN ⊥ ED, prove that

(i) ∆BCF ≅ ∆ABE.

(ii) area of square ABFG = area of rectangle BENM.

Answer

(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 = $\dfrac{1}{2}$ 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 = $\dfrac{1}{2}$ area of rectangle BENM ....... (6)

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

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

⇒ area of square AGFB = area of rectangle BENM

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