Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.

The figures (i), (iii) and (v) lie on the same base and between the same parallels.

In the given figure ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

We have AE ⊥ DC and AB = 16 cm

∵ AB = CD [Opposite sides of parallelogram]

∴ CD = 16 cm

Now, area of parallelogram ABCD = CD × AE

= 16 × 8 cm² = 128 cm²

Since, CF ⊥ AD

∴ Area of parallelogram ABCD = AD × CF

⇒ AD × CF = 128 ⇒ AD × 10 = 128

[CF = 10 cm]

Thus, the required length of AD is 12.8 cm

If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show

SOLUTION: Join GE and HF, where GE ∥ BC ∥ DA and HF ∥ AB ∥ DC (∵ E, F, G and H are the mid points of a
∥^{gm } ABCD).

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

Now, ∆EFG and parallelogram EBCG are on the same base EG, and between the same parallels EG and BC.

P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar(∆APB) = ar(∆BQC).

SOLUTION: ∵ ABCD is a parallelogram
∴ AB ∥ CD and BC ∥ AD.
Now, ∆APB and parallelogram ABCD are on the same base AB and between the same parallels AB and CD.

Also, ∆BQC and parallelogram ABCD are on the same base BC and between the same parallels BC and AD.

From (1) and (2), we have ar(∆APB) = ar(∆BQC).

In the figure, P is a point in the interior of a parallelogram ABCD. Show that

(i)

(ii) ar(∆APD) + ar(∆PBC) = ar(∆APB) + ar(∆PCD)

[Hint: Through P, draw a line parallel to AB]

We have a parallelogram ABCD, i.e., AB ∥ CD and BC ∥ AD. Let us draw EF ∥ AB and HG ∥ AD through P.

(i) ∆APB and ∥^{gm} AEFB are on the same base AB and between the same parallels AB and EF.

Also, ∆PCD and parallelogram CDEF are on the same base CD and between the same parallels CD and EF.

(ii) ∆APD and ∥^{gm} ADGH are on the same base AD and between the same parallels AD and GH.

In the figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that

(i) ar(PQRS) = ar(ABRS)

(ii)

(i) Parallelogram PQRS and parallelogram ABRS are on the same base RS and between the same parallels RS and PB.

∴ ar(PQRS) = ar(ABRS)

(ii) ∆AXS and ∥^{gm} ABRS are on the same base AS and between the same parallels AS and BR.

But ar(PQRS) = ar(ABRS) [Proved in (i) part] ...(2)

From (1) and (2), we have

A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q.

In how many parts the field is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in

equal portions of the field separately. How should she do it?

The farmer is having the field in the form of parallelogram PQRS and a point A is situated on RS. Join AP and AQ.

Clearly, the field is divided into three parts i.e. in ∆APS, ∆PAQ and ∆QAR.

Since, ∆PAQ and parallelogram PQRS are on the same base PQ and between the same parallels PQ and RS.

From (1) and (2), we have

ar(∆PAQ) = ar[(∆APS) + (∆QAR)]

Thus, the farmer can sow wheat in (∆PAQ) and pulses in[(∆APS) + (∆QAR)]

or wheat in [(∆APS) + (∆QAR)] and pulses in (∆PAQ).

In the figure, E is any point on median AD of a ∆ABC. Show that ar(∆ABE) = ar(∆ACE).

We have a ∆ABC such that AD is a median.

∴ ar(∆ABD) = ar(∆ADC) ...(1)

[∵ A median divides the triangle into two triangles of equal areas]

Similarly, in ∆BEC, we have

ar(∆BED) = ar(∆DEC) ...(2)

Subtracting (2) from (1), we have

ar(∆ABD) – ar(∆BED) = ar(∆ADC) – ar(∆DEC)

⇒ ar(∆ABE) = ar(∆ACE).

In a triangle ABC, E is the mid-point of median AD. Show that ar(∆BED)

SOLUTION: We have a ∆ABC and its median AD.

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

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

SOLUTION:We have a parallelogram ABCD such that its diagonals intersect at O.

∵ Diagonals of a parallelogram bisect each other.

Since, BO = OD

∴ ar(∆BOC) = ar(∆DOC) ...(1)

Similarly, ar(∆AOD) = ar(∆DOC) ...(2)

and ar(∆AOB) = ar(∆BOC) ...(3)

From (1), (2) and (3), we have

ar(∆AOB) = ar(∆BOC) = ar(∆COD) = ar(∆AOD)

Thus, the diagonals of a parallelogram divide it into four triangles of equal area.

In figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar(∆ABC) = ar(∆ABD).

We have ∆ABC and ∆ABD are on the same base AB.

Q CD is bisected at O [Given]

∴ CO = DO

Now, in ∆ACD, AO is a median

∴ ar(∆OAC) = ar(∆OAD) ...(1)

Again, in ∆BCD, BO is a median

∴ ar(∆OBC) = ar(∆ODB) ...(2)

Adding (1) and (2), we have

ar(∆OAC) + ar(∆OBC) = ar(∆OAD) + ar(∆ODB)

ar(∆ABC) = ar(∆ABD)

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

(i) BDEF is a parallelogram

(ii)

(iii)

We have ∆ABC such that D, E and F are the mid-points of BC, CA and AB respectively.

(i) In ∆ABC, E and F are the mid-points of AC and AB respectively.

∴ EF ∥ BC [Mid-point theorem]

⇒ EF ∥ BD

⇒ EF = BD [D is the mid-point of BC]

Since BDEF is a quadrilateral whose one pair of opposite sides is parallel and of equal lengths.

∴ BDEF is a parallelogram.

(ii) We have proved that BDEF is a parallelogram.

Similarly, DCEF is a parallelogram and DEAF is a parallelogram.

Now, parallelogram BDEF and parallelogram DCEF are on the same base EF and between the same parallels BC & EF.

∴ ar(∥^{gm} BDEF) = ar(^{gm} DCEF)

In the figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that

(i) ar( ∆DOC) = ar( ∆AOB)

(ii) ar( ∆DCB) = ar( ∆ACB)

(iii) DA ∥ CB or ABCD is a parallelogram.

[Hint: From D and B, draw perpendiculars to AC.]

We have a quadrilateral ABCD whose diagonals AC and BD intersect at O.

We also have that OB = OD, AB = CD

Let us draw DE ⊥ AC and BF ⊥ AC

(i) In ∆DEO and ∆BFO, we have

DO = BO [Given]

∠DOE = ∠BOF [Vertically opposite angles]

∠DEO = ∠BFO [Each 90°]

∴ ∆DEO ≅ ∆BFO [By AAS congruency]

⇒ DE = BF [By C.P.C.T]

and ar(∆DEO) = ar(∆BFO) ...(1)

Now, in ∆DEC and ∆BFA, we have

∠DEC = ∠BFA [Each 90°]

DE = BF [Proved above]

DC = BA [Given]

∴ ∆DEC ≅ ∆BFA [By RHS congruency]

⇒ ar(∆DEC) = ar(∆BFA) ...(2)

and ∠1 = ∠2 ...(3) [By C.P.C.T.]

Adding (1) and (2), we have

ar(∆DEO) + ar(∆DEC) = ar(∆BFO) + ar(∆BFA)

⇒ ar(∆DOC) = ar(∆AOB)

(ii) Since ar(∆DOC) = ar(∆AOB) [Proved above] Adding ar(∆BOC), on both sides, we have

ar(∆DOC) + ar(∆BOC) = ar(∆AOB) + ar(∆BOC)

⇒ ar(∆DCB) = ar(∆ACB)

(iii) Since, ∆DCB and ∆ACB are on the same base CB and having equal areas.

∴ They lie between the same parallels CB and DA.

⇒ CB∥DA

Also ∠1 = ∠2, [By (3)]

which forms an alternate interior angles.

So, AB ∥ CD

Hence, ABCD is a parallelogram

D and E are points on sides AB and AC respectively of ∆ABC such that ar(∆DBC) = ar(∆EBC). Prove that DE ∥ BC.

SOLUTION: We have ∆ABC and points D and E are such that ar(∆DBC) = ar(∆EBC)

Since ∆DBC and ∆EBC are on the same base BC and having same area.

∴ They lie between the same parallels DE and BC.

Hence, DE ∥ BC

XY is a line parallel to side BC of a triangle ABC. If BE ∥ AC and CF ∥ AB meet XY at E and F respectively, show that ar(∆ABE) = ar(∆ACF).

SOLUTION: We have a ∆ABC such that XY ∥ BC, BE ∥ AC and CF ∥ AB.

Since, XY ∥ BC and BE ∥ CY

∴ BCYE is a parallelogram.
Now, the parallelogram BCYE and ∆ABE are on the same base BE and between the same parallels BE and AC.

∴ BCFX is a parallelogram.

Now, ∆ACF and parallelogram BCFX are on the same base CF and between the same parallels AB and FC.

Also parallelogram BCFX and parallelogram BCYE are on the same base BC and between the same parallels BC and EF.

∴ ar(∥gm BCFX) = ar(∥gm BCYE) ...(3)

From (1), (2) and (3) we get

ar(∆ABE) = ar(∆ACF)

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see figure). Show

that ar(ABCD) = ar(PBQR).
[Hint: Join AC and PQ. Now compare ar(ACQ) and ar(APQ)]

Let us join AC and PQ.
ABCD is a parallelogram [Given] and AC is its diagonal, we know that diagonal of a parallelogram divides it into two triangles of equal areas.

Diagonals AC and BD of a trapezium ABCD with AB ∥ DC intersect each other at O. Prove that ar(AOD) = ar(BOC).

SOLUTION: We have a trapezium ABCD having AB ∥ CD and its diagonals AC and BD are joined.

Since, triangles on the same base and between the same parallels have equal areas.

∆ABD and ∆ABC are on the same base AB and between the same parallels AB and DC

∴ ar(∆ABD) = ar(∆ABC)

Subtracting ar(∆AOB) from both sides, we get

ar(∆ABD) – ar(∆AOB) = ar(∆ABC) –ar(∆AOB)

⇒ ar(∆AOD) = ar(∆BOC)

In the figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that

(i) ar(ACB) = ar(ACF)

(ii) ar(AEDF) = ar(ABCDE)

We have a pentagon ABCDE in which BF ∥ AC and DC is produced to F.

(i) Since, the triangles between the same parallels and on the same base are equal in area.

∆ACB and ∆ACF are on the same base AC and between the same parallels AC and BF.

∴ ar(∆ACB) = ar(∆ACF)

(ii) Since ar(∆ACB) = ar(∆ACF) [Proved above]

Adding ar(quad. AEDC) to both sides, we get

⇒ ar(∆ACB) + ar(quad. AEDC)

= ar(∆ACF) + ar(quad. AEDC)

∴ ar(ABCDE) = ar(AEDF)

A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a

Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain

how this proposal will be implemented.

We have a plot in the form of a quadrilateral ABCD.

Let us draw DF ∥ AC and join A and F.

Now, ∆DAF and ∆DCF are on the same base DF and between the same parallels AC and DF.

∴ ar(∆DAF) = ar(∆DCF)

Subtracting ar(∆DEF) from both sides, we get

ar(∆DAF) – ar(∆DEF)

= ar(∆DCF) – ar(∆DEF)

⇒ ar(∆ADE) = ar(∆CEF)

The portion ∆ADE can be taken over by the Gram Panchyat by adding the land (∆CEF) to his (Itwaari) land so as to form a triangular plot, i.e. ∆ABF.

Let us prove that ar(∆ABF) = ar(quad. ABCD), we have

ar(∆CEF) = ar(∆ADE) [Proved above]

Adding ar(quad. ABCE) to both sides, we get

ar(∆CEF) + ar(quad. ABCE)

= ar(∆ADE) + ar (quad. ABCE)

⇒ ar(∆ABF) = ar (quad. ABCD)

ABCD is a trapezium with AB ∥ DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar(∆ADX) = ar(∆ACY). [Hint : Join CX]

SOLUTION: We have a trapezium ABCD such that AB ∥ DC. XY ∥ AC meets AB at X and BC at Y.

Let us join CX.

∵ ∆ADX and ∆ACX are on the same base AX and between the same parallels AB and DC.

∴ ar(∆ADX) = ar(∆ACX) ...(1)

∵ ∆ACX and ∆ACY are on the same base AC and between the same parallels AC and XY.

∴ ar(∆ACX) = ar(∆ACY) ...(2)

From (1) and (2), we have

ar(∆ADX) = ar(∆ACY)

In the figure, AP ∥ BQ ∥ CR. Prove that ar(∆AQC) = ar(∆PBR).

We have AP ∥ BQ ∥ CR

∵ ∆BCQ and ∆BQR are on the same base BQ and between the same parallels BQ and CR.

∴ ar(∆BCQ) = ar(∆BQR) ...(1)

∵ ∆ABQ and ∆PBQ are on the same base BQ and between the same parallels AP and BQ.

∴ ar(∆ABQ) = ar(∆PBQ) ...(2)

Adding (1) and (2), we have

ar(∆BCQ) + ar(∆ABQ) = ar(∆BQR) + ar(∆PBQ)

⇒ ar(∆AQC) = ar(∆PBR)

Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(∆AOD) = ar(∆BOC). Prove that ABCD is a trapezium.

SOLUTION: We have a quadrilateral ABCD and its diagonals AC and BD intersect at O such that

ar(∆AOD) = ar(∆BOC) [Given]

Adding ar(∆AOB) to both sides, we have

ar(∆AOD) + ar(∆AOB) = ar(∆BOC) + ar(∆AOB)

⇒ ar(∆ABD) = ar(∆ABC)

Also they are on the same base AB.

Since, the triangles on the same base and having equal area.

∴ They lie between the same parallels.

∴ AB ∥ DC

Now, ABCD is a quadrilateral having a pair of opposite sides parallel.

So, ABCD is a trapezium.

In the figure, ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

We have ar(∆DRC) = ar(∆DPC) [Given]
And they are on the same base DC.

∴ ∆DRC and ∆DPC must lie between the same parallels.

So, DC ∥ RP i.e., a pair of opposite sides of quadrilateral DCPR is parallel.

∴ Quadrilateral DCPR is a trapezium.

Again, we have

ar(∆BDP) = ar(∆ARC) [Given] ...(1)

Also ar(∆DPC) = ar(∆DRC) [Given] ...(2)

Subtracting (2) from (1), we get

ar (∆BDP) – ar(∆DPC) = ar(∆ARC) – ar(∆DRC)

⇒ ar(∆BDC) = ar(∆ADC)]

And they are on the same base DC.

∴ ∆BDC and ∆ADC must lie between the same parallels.

So, AB ∥ DC i.e. a pair of opposite sides of quadrilateral ABCD is parallel.

∴ Quadrilateral ABCD is a trapezium.

Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

SOLUTION: We have a parallelogram ABCD and rectangle ABEF such that
ar(∥gm ABCD) = ar(rect. ABEF)

AB = CD [Opposite sides of parallelogram]

and AB = EF [Opposite sides of a rectangle]

⇒ CD = EF

⇒ AB + CD = AB + EF ...(1)

∵ BE < BC and AF < AD

[In a right triangle, hypotenuse is the longest side]

⇒ (BC + AD) > (BE + AF) ...(2)

From (1) and (2), we have

(AB + CD) + (BC + AD) > (AB + EF) + (BE + AF)

⇒ (AB + BC + CD + DA) > (AB + BE + EF + FA)

⇒ Perimeter of parallelogram ABCD > Perimeter of rectangle ABEF

In the figure, D and E are two points on BC such that BD = DE = EC. Show that ar(ABD) = ar (ADE) = ar(AEC)

Let us draw AF, perpendicular to BC such that AF is the height of ∆ABD, ∆ADE and ∆AEC.

In the figure, ABCD, DCFE and ABFE are parallelograms. Show that ar(∆ADE) = ar(∆BCF).

We have parallelograms ABCD, DCFE and ABFE.

ABCD is a parallelogram [Given]

∴ Its opposite sides are parallel and equal.

Now, ∆ADE and ∆BCF are on equal bases AD = BC [from (1)] and between the same parallels AB and EF

So, ar(∆ADE) = ar(∆BCF).

In the figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar(∥ BPC) = ar(∥ DPQ)

[Hint: Join AC]

We have a parallelogram ABCD and AD = CQ. Let us join AC.

We know that triangles on the same base and between the same parallels are equal in area. Since ∆QAC and ∆QDC are on the same base QC and between the same parallels AD and BQ,

ar(∆QAC) = ar(∆QDC)

Subtracting ar(∆QPC)from both sides, we have

ar(∆QAC) – ar(∆QPC) = ar(∆QDC) – ar(∆QPC)

ar(∆PAC) = ar(∆QDP) ...(1)

Since, ∆PAC and ∆PBC are on the same base PC and between the same parallels AB and CD

∴ ar(∆PAC) = ar(∆PBC) ...(2)

From (1) and (2), we get

ar(∆PBC) = ar(∆QDP)

In figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that

[Hint : Join EC and AD. Show that BE ∥ AC and DE ∥ AB, etc.]

Let us join EC and AD. Draw EP ⊥ BC.

Let AB = BC = CA = a, then

(ii) Since ∆ABC and ∆BED are equilateral triangles

⇒ ∠ACB = ∠DBE = 60°

⇒ BE∥AC

∆BAE and ∆BEC are on the same base BE and between the same parallels BE and AC

∴ ar(∆BAE) = ar(∆BEC)

⇒ ar(∆BAE) = 2 ar(∆BDE)

[DE is median of∆EBC ∴ ar (∆BEC) = 2 ar(∆BDE)]

(iii) ar(∆ABC) = 4 ar(∆BDE) [Proved in (i) part]

ar(∆BEC) = 2 ar(∆BDE)

[DE is median of ∆BEC]

⇒ ar(∆ABC) = 2 ar(∆BEC)

(iv) Since ∆ABC and ∆BDE are equilateral triangles

⇒ ∠ABC = ∠BDE = 60°

⇒ AB ∥ DE

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

∴ ar(∆BED) = ar(∆AED)

Subtracting ar(∆EFD) from both sides, we get

⇒ ar(∆BED) = ar(∆EFD) ar(∆AED) = ar(∆EFD)

⇒ ar(∆BEF) = ar(∆AFD)

(v) In right angled ∆ABD, we get

(vi) ar(∆AFC) = ar(∆AFD) + ar(∆ADC)

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that [ar(APB) × ar(CPD)] = ar(APD) × ar(BPC).

[Hint: From A and C, draw perpendiculars to BD.]

We have a quadrilateral ABCD such that its diagonals AC and BD intersect at P.

Let us draw AM ⊥ BD and CN ⊥ BD.

From (1) and (2), we get

ar(∆APB) × ar(∆CPD) = ar(∆APD) × ar(∆BPC)

P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that

(i)

(ii)

(iii) ar(PBQ) = ar(ARC)

We have a ∆ABC such that P is the mid-point of AB and Q is the mid-point of BC.

Also, R is the mid-point of AP. Let us join AQ and PC.

(i) In ∆APQ, R is the mid-point of AP [Given]

In ∆ABQ, P is the mid-point of AB

∴ QP is a median of ABQ

From (1) and (2), we get

(ii) In ∆RBC, RQ is a median.

(iii) QP is a median of ∆ABQ.

Thus, ar(∆PBQ) = ar(∆ARC)

In the figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:

(i) ∆MBC ≅ ∆ABD

(ii) ar(BYXD) ≅ 2ar(∆MBC)

(iii) ar(BYXD) = ar(ABMN)

(iv) ∆FCB ≅ ∆ACE

(v) ar(CYXE) = 2ar(FCB)

(vi) ar(CYXE) = ar(ACFG)

(vii) ar(BCED) = ar(ABMN) + ar(ACFG)

We have a right ∆ABC such that BCED, ACFG and ABMN are squares on its sides BC, CA and AB respectively. Line segment AX ⊥ DE is also drawn such that it meets BC at Y.

(i) In ∆ABD and ∆MBC, we

∠CBD = ∠MBA [Each 90°]

∴ ∠CBD + ∠ABC = ∠MBA + ∠ABC

[By adding ∠ABC on both sides]

or ∠MBC = ∠ABD

∴ ∆MBC ≅ ∆ABD [By SAS congruency]

(ii) Since parallelogram BYXD and ∆ABD are on the same base BD and between the same parallels BD and AX.

But ar(∆ABD) = ar(∆MBC)

[Congruent triangles have equal areas]

⇒ ar(∥gm BYXD) = 2ar(∆MBC)

(iii) Since ar(BYXD) = 2ar(∆MBC) ...(1)

[From (ii) part]

and ar(square ABMN) = 2ar(∆MBC) ...(2)

[ABMN and ∆MBC are on the same base MB and between the same parallels MB and NC]

From (1) and (2), we have

ar(BYXD) = ar(ABMN)

(iv) In ∆FCB and ∆ACE,

FC = AC [Sides of a square]

CB = CE [Sides of a square]

∠FCA = ∠BCE [Each 90°]

or ∠FCA + ∠ACB = ∠BCE + ∠ACB

[By adding ∠ACB on both sides.]

⇒ ∠FCB = ∠ACE

⇒ ∆FCB ≅ ∆ACE [By SAS congruency]

(v) Since ∥gm CYXE and ∆ACE are on the same base CE and between the same parallels CE and AX.

∴ ar(∥gm CYXE) = 2ar(∆ACE)

But ∆ACE ≅ ∆FCB

Since, congruent triangles are equal in areas.

∴ ar (∥gm CYXE) = 2ar(∆FCB)

(vi) Since, ar(∥gm CYXE) = 2ar(∆FCB) ...(3)

[From (v) part]

Also (quad. ACFG) and ∆FCB are on the same base FC and between the same parallels FC and BG.

⇒ ar(quad. ACFG) = 2ar(∆FCB) ...(4)

From (3) and (4), we get

ar(quad. CYXE) = ar(quad. ACFG) ...(5)

(vii) We have ar(quad. BCED)

= ar(quad. CYXE) + ar(quad. BYXD)

= ar(quad. CYXE) + ar(quad. ABMN)

[From (iii) part]

Now, ar(quad. BCED)

= ar(quad. ABMN) + ar(quad. CYXE)

Thus,ar (quad. BCED)

= ar(quad. ABMN) + ar(quad. ACFG)

[From (5)]