Chapter 9 : AREAS OF PARALLELOGRAMS AND TRIANGLES

Explanation:

A is the right Answer 

We know that the median of the triangle is a line segment joining a vertex to the mid of the opposite line. Therefore, a median divides a triangle into equal areas.

Explanation:

In figures (a), (b), and (c) two polygons are not present between the same parallels.

In Figure (d), two polygons PQRA and BQRS are present on the same base and between the same parallels. Therefore, figure (d) is the right answer.

Explanation:

The correct answer is D)

 Let ABCDis a rectangle.

 Length of rectangle ABCD= 8cm

 The breadth of rectangle ABCD= 6cm

 Let H, E, F, and G be the mid-points of sides of rectangle ABCD, then HEFG is a  rhombus.

 Then, the diagonal of rhombus HEFG are HF and EG.

 Here, EG=BC=8cm  and HF=AB=6cm

∴  Area of rhombus HEFG =  Productofdiagonals/2

​∴  Area of rhombus HEFG =  8×6/2  =4×6=24cm².

Explanation:

Area of parallelogram = Base × Corresponding altitude

= AB × DL … (1)

Since opposite sides of a parallelogram are equal,

Then we have, AB = DC

Substituting this in (1), we have,

Area of parallelogram = AB × DL= DC × DL

Hence, (C)DC × DL is the correct answer.

Explanation:

Since the Parallelogram ABCD and rectangle ABEM are of equal areas has the same base, they are within the same parallels MC & AB. 

Now, in a triangle

 AMD & BEC we have 

AM=BE       ...( opposite sides of rectangle)

AD=BC     ...(opposite sides of parallelogram) 

∠AMD=∠BEC      ...( both are right angles since they are corners of a rectangle)

Therefore triangle AMD and Triangle BEC are congruent.  

Implies MD=EC    ....(a) 

Again triangle AMD is the right one. 

(∠M=90°    being a corner of a rectangle.)

So, AD is the hypotenuse. i.e AD>AM........(b)

Now, ME=MD+DE & DC=EC+DE=MD+DE    ....(From a) 

Implies ME=DC    ........(c) 

Now, perimeter ABEM=2(AM+ME)  .........(d) 

And perimeter ABCD=2(DC+AD)=2(ME+AD)   .......(e)     ...[from c] 

Comparing (iv) & (v) AM is common to both but AD>AM.

Therefore  (e) >(d) i.e perimeter ABCD> perimeter ABEM.

Explanation:

D, E, and F are the midpoints of sides BC, AC, and AB respectively.

On joining FE, we divide △ABC into 4 triangles of equal area.

Also, a median of a triangle divides it into two triangles with equal area

ar(AFDE)=ar(triangle AFE)+ar(triangle FED)

=2ar(triangle AFE)=

=2×1/4ar(triangle ABC)=1/2ar(triangle ABC)

Hence, the correct answer is (A) 1/2 ar (ABC).

Explanation:

(B) 1:1

Area of a parallelogram = base ⨯ height

If both parallelograms stand on the same base and between the same parallels, then their heights are the same.

So, their areas will also be the same.

Explanation:

Quadrilateral  ABCD Needs not be any rectangle, rhombus, or parallelogram because if a quadrilateral ABCD is square then its diagonal AC also divides into 2 parts which are equal in area.

Explanation:

Area of triangle = 1/2 Area of parallelogram

So we get

Area of triangle/ Area of parallelogram = 1/2

Ratio will be

Area of a triangle: Area of parallelogram = 1: 2

Therefore, the ratio of the area of the triangle to the area of the parallelogram is 1: 2.

Explanation:

AB parallel to  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 triangle ABD,

E is the midpoint of AD and EM parallel to AB

M is the midpoint of BD

Using the midpoint theorem

EM = 1/2 AB … (a)

In triangle CBD

MF = 1/2 CD … (b)

From (a) and (b)

EM + MF = 1/2 AB + 1/2 CD

EF = 1/2 (AB + CD)

EF = 1/2 (a + b)

Here, Area of trapezium ABFE = 1/2 [sum of parallel sides] × [distance between parallel sides]

Substituting the values

= 1/2 [a + 1/2 (a + h)] × h

= 1/4 [3a + b] h

Similarly

Area of trapezium EFCD = 1/2 [b + 1/2 (a + h)] × h = 1/4 [3b + a] h

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

By substituting the values

= 1/4 [3a + b] h/ 1/4 [3b + a] h

= [3a + b]/ [3b + a]

= [3a + b]: [3b + a]

Therefore, the ratio of ar (ABFE) and ar (EFCD) is (3a + b) : (a + 3b)

Explanation:

False, because ar (ABD) = ar (ACD) and ar (PBD) = ar (PCD), therefore, ar (ABP) = ar (ACP).

Explanation: 

PQRS and EFRS are on the same base SR and between the same parallels EF and SR

The areas will be equal

ar (PQRS) = ar (EFRS) …. (a)

ar (triangle MFR) = 1/2 ar (EFRS) …. (b)

From both the equations

ar (MFR) = 1/2 ar (PQRS)

Therefore, the statement is true.

Explanation:

In triangle ABC, X is the midpoint of AB

Considering the given question,

Hence ar(triangle  AXC) = ar(triangle BXC) = ½ ar(triangle ABC)

Therefore,

ar(triangle AXC) = ar(triangle BXC) = 12 cm²

Area of ||gm ABCD = 2 x ar(triangle ABC)

= 2 x 24 = 48 cm²

But, according to the question,

area of ||gm ABCD = ar(AXCD) + ar(triangle BXC)

= 24 + 12 = 36 cm²

Hence, we find a contradiction here.

So, If ar (AXCD) = 24 cm², then ar (ABC) ≠ 24 cm².

Eplanation:

The  statement can be true if PA is equal to PS.

Given,

PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm.

A is any point on PQ

PA<PQ

ar (triangle PQR) = ½ ×PQ×QR

= ½ ×12×5

= 30cm²

PS=5 cm

Suppose PA<PQ,

ar(triangle PAS) < ar(triangle PQR)

ar(triangle PAS) < 30 cm²

Suppose PA=PQ,

ar (triangle PAS) = ½ ×PQ×PS

= ½ ×12×5

= 30cm².

Explanation:

Given, the area of parallelogram PQRS is 180 cm² and QS is it's diagonal which divides it into two triangles of equal areas. 

Therefore Area of the triangle is PQS = 90 cm² Now, A is any point on QS.

Therefore Area of triangle ASR < Area of triangle SRQ.

Explanation:

triangle ABC and triangle  BDE are two equilateral triangles

Area (triangle  ABC) = √3/4 × (side)²

In an equilateral triangle ABC, AB = BC = AC

= √3/4 × (BC)² …. (1)

D is the midpoint of BC

BD = DC = 1/2 BC … (2)

Area of triangle  BDE = √3/4 × (side)²

In an equilateral triangle BDE, BD = DE = BE

= √3/4 × (BD)²

= √3/4 × (1/2 BC)²

= √3/4 × 1/4 BC²

= 1/4 [√3/4 BC²]

Area of triangle  BDE = 1/4 Area of triangle  ABC

Therefore, the statement is true.

Explanation:

Given: ABCD and EFGD are ||gm and G is the midpoint of CD.

To prove: ar (triangle DPC) =  1/2   ar (EFGD)

 Proof: ar (triangle DPC) =  1/2  ar (||gm ABCD) 

(Because triangle PDC and parallelogram ABCD are on the same base DC and between the same parallels AB and CD)

Because G is the mid-point of DC.

Therefore DG =  1/2 DC 

Therefore ar (EFGD) =  1/2  ar (ABCD) …(ii)

 Implies  ar (triangle DPC) = ar (EFGD) 

The given statement is false.

Explanation: 

T and U are midpoints of PS and QR respectively (Given)

Therefore, TU parallel to PQ => TO parallel to PQ

In triangle PQS,

T is the mid-point of PS and TO parallel to PQ

So, TO = 1/2 PQ = 4 cm

(PQ = 8 cm given)

Also, TS = 1/2 PS = 4 cm

[PQ = PS, As PQRS is a square)

Now,

Area of triangle OTS = 1/2(TO×TS)

= 1/2(4×4) cm²

= 8cm²

Area of triangle OTS is 8 cm².

Explanation:

As we know  that area of triangles on the same base  and between the same parallels lines are equal

Therefore , ar(APC)=ar(BPC)... (a)

Now In quadrilateral ACQD, we have

AD=CQ and,  AD∥CQ [Given]

Therefore, this quadrilateral ADQC with one pair of opposite sides is equal and parallel is a parallelogram.

Therefore ADQC is a parallelogram.

this impliesAP=PQ and CP=DP

[Since diagonals of a|| gm bisect each other]

In Δs APC and DPQ, we have

AP=PQ [Proved above]

∠APC=∠DPQ [Vertically opp. ∠s] and,

PC=PD [Proved above]

Therefore by the SAS criterion of congruence,

Triangle APC≅ triangle DPQ

this implies (APC)=ar(DPQ) ... (2)

[Since congruent Δs have equal area]

Therefore ar(BPC)=ar(DPQ)  [From (1)]

Hence , ar(BPC)=ar(DPQ).

Explanation:

In parallelogram PABQ,

PQ Parallel AB [Because in parallelogram PSDA, PS || AD]

PA parallel QB [given]

So, PABQ is a parallelogram.

Therefore PQ = AB (a)

Similarly, QBCR is also a parallelogram.

Therefore QR = BC (d)

RCDS is a parallelogram.

RS = CD (c

Now, PQ = QR = RS (d)

From (a), (b), (c) and (d),we get

PQ = CD

In triangle PQE and triangle DCF, ∠ QPE =∠ FDC

[Since PS parallel AD and PD is transversal, then alternate interior angles are equal]

Implies ∠PQE=∠FCD=90∘

Therefore (ΔPQE)≅(ΔCFD) 

 [By AAS congruence criteria]

Therefore ar(ΔPQE)=ar(ΔCFD)

[Since congruent figures have equal area].

Explanation:

LX=XY=YN(given)

Therefore LM∥XZ

this implies  c Draw ZL⊥ rLY 

this implies  In ΔLZYXT∥YZ

this implies ar ΔLZY= 1/2 (LY)XZL=(2XY)× 1/2 ×(ZL)=(ZL)×(XY)

this implies In  quadrilateral(MZYX)=(base×height)

this implies ar(MZYX)=(XY)×(ZL)

This implies  Hence,ar(ΔLZY)=ar MZYX Hence proved.

Explanation:

According to the question,

Area of parallelogram, ABCD = 90 cm²

(i)

We know that,

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

Here,

The parallelograms ABCD and ABEF are on same base AB and between the same parallels AB and CF.

Therefore, ar (ABEF) = ar (ABCD) = 90 cm²

(ii)

We know that,

If a triangle and a parallelogram are on the same base and between the same parallels,

then area of triangle is equal to half of the area of the parallelogram.

Here,

ΔABD and parallelogram ABCD are on the same base AB and between the same parallels AB and CD.

Therefore, ar (ΔABD) = ½ ar (ABCD)

= ½ x 90 = 45 cm²

(iii)

We know that,

If a triangle and a parallelogram are on the same base and between the same parallels,

then area of triangle is equal to half of the area of the parallelogram.

Here,

Triangle BEF & parallelogram ABEF are on the same base EF and between the same parallels AB and EF.

Therefore, ar (ΔBEF) = ½ ar (ABEF)

= ½ x 90 = 45 cm².

Explanation:

In ABC, D is the midpoint of AB and P is any point on BC. If CQ || PD me...

In triangle ABC, D is the midpoint of AB and P is any point on BC. If CQ || PD meets AB in Q then prove that ar (BPQ) =1/2ar (ABC)

Construction :- Join DC

To prove :- Since D is the midpoint of AB. so, in ∆ ABC, CD is the median.

ar(∆ BCD = 1/2 ar (∆ ABC) ..... (1)

Since, ∆PDQ and ∆PDC are on the same base PD and between the same parallels lines PD QC.

therefore, ar(∆ PDQ) = ar(∆ PDC) ................ (2)

From (1) and (2)

ar(∆ BCD) = 1/2 ar(∆ ABC)

= ar(∆ BPD) + ar(∆ PDC) = 1/2ar(∆ ABC)

= ar(∆ BPD) + ar(∆ PDQ) = 1/2ar(∆ ABC) { area of triangle PDC = PDQ}

= ar(∆BPQ) = 1/2 ar(∆ ABC).

Explanation:

ABCD is a square

AB = BC = CD = DA 

E and F are mid-points of BC and CD respectively.

In ∆ABE and ∆ADF 

AB = AD BE = DF 

∠ABE = ∠ADF = 90° 

∆ABE = ∆ADF (by SAS congruency) 

⇒ AE = EF (by c.p.c.t) 

⇒ ∆AEF is an isosceles triangle,

 R is mid-point of EF.

 ⇒ ∆ARE = ∆ARF.

  (∵ AE = AF, EF = RF, ∠ARE = ∠ARF = 90°) 

⇒ ar (∆AER) = ar (∆AFR) Hence proved.

Explanation:

To prove: ar (∆PSO) = ar (∆PQO) 

Join SQ to intersect PR at B.

 We know that the diagonals of a parallelogram bisect each other, so B is the mid-point of SQ.

 Thus PB is a median of ∆QPS and median of a triangle divides it into two triangles of equal area ar (∆BPQ) = ar (∆BPS) …(i)

 Also OB is a median of ∆OSQ ar (∆OBQ) = ar (∆OBS) …(ii) 

On adding equation (i) and (ii), 

we get ar (∆BPQ) + ar (∆OBQ)

 = ar (∆BPS) + ar (∆OSB) 

⇒ ar (∆PQO) = ar (∆PSO)

 Hence proved.

Explanation:

Given,

ABCD is a parallelogram and CE = BC i.e., C is the midpoint of BE.

Also 

ar(ΔDFB)=3cm2

Now,ΔADF and ΔDFB are on the same base DF and between parallels CD and AB.

Then,ar(ΔADF)=ar(ΔDFB)=3cm2 ….(i)

In ΔABE, by the converse of mid-point theorem,

EF = AF [since, C is midpoint of BE] …(ii)

In ΔADF and ΔECF,∠AFD=∠CFE

 [vertically opposite angles]

AF = EF [from Eq. (ii)]

∠DAF=∠CEF

[BE || AD and AE is transversal, then alternate interior angles are equal]

ΔADF≅ΔECF [by ASA congruence rule]

Then,ar(ΔADF)=ar(ΔCFE) ...(iii)

[since, congruent figures have equal area]

ar(ΔCFE)=ar(ΔADF)=3cm2 ...(iv)

Now,ar(ΔBDC)=ar(ΔDFB)+ar(ΔBFC)

=3+3=6cm2

[from Eqs. (i) and (iv)]

We know that, diagonal of a parallelogram divides it into two congruent triangles of equal areas.

∴ Area of parallelogram ABCD

=2×Area of ΔBDC

=2×6

=12 cm2

Hence, the area of parallelogram ABCD is 12 cm2.

Explanation:

ASAB∥DC, so AB∥DQ

In ΔCLQ and ΔBLP, we have

∴∠QCL=∠LBP [[Alt. ∠S]

CL=LB[∵L is the mid-point of BC]

∠CLQ=∠BLP [ Vertically opposite ∠S]

∴ΔQCL≅∠BLP [ By ASA congruence rule]

⇒ar(ΔCLQ)=ar(ΔBLP) .......(1) [Congruence ΔS are equal in area]

Adding ar(APLCD) to both sides of (1), we get

ar(ΔCLQ)+ar(APLCD)=ar(ΔBLP)+ar(APLCD)

⇒ar(APQD)=ar(ABCD)

Hence, ar(ABCD)=ar(APQD).

Explanation:

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

BR is the median of the triangle BDA.

So, ar(BRA) = 1/2 ar(BDA) --------------- (1)

RS is the median of the triangle BRA.

So, ar(ASR) = 1/2 ar(BRA) --------------- (2)

From (1) and (2),

ar(ASR) = 1/2 (1/2 ar(BDA)

ar(ASR) = 1/4 ar(BDA) --------------------- (3)

Similarly, ar(CFP) = 1/4 ar(BCD) -------- (4)

Adding (3) and (4),

ar(ASR) + ar(CFP) = 1/4 ar(BDA) + 1/4 ar(BCD)

ar(ASR) + ar(CFP) = 1/4 ar(BCDA) --------- (5)

Similarly, ar(DRF) + ar(BSP) = 1/4 ar(BCDA) ------------ (6)

Adding (5) and (6),

ar(ASR) + ar(CFP) + ar(DRF) + ar(BSP) = 2(1/4 ar(BCDA))

ar(ASR) + ar(CFP) + ar(DRF) + ar(BSP) = 1/2 ar(BCDA) ------------- (7)

From the figure,

ar(ASR) + ar(CFP) + ar(DRF) + ar(BSP) + ar(PFRS) = ar(BCDA)

From (7),

1/2 ar(BCDA) + ar(PFRS) = ar(BCDA)

ar(PFRS) = ar(BCDA) - 1/2 ar(BCDA)

Therefore, ar(PFRS) = 1/2 ar(BCDA).

Explanation:

Through P and Q, draw PR and QS parallel to AB. Now PQRS is a parallelogram and its base PQ = 1/3 BC. 

ar (APD) = 1/2 ar (ABCD) [Same base BC and BC || AD] (1)  

ar (AQD) = 1/2 ar (ABCD) (2)  

From (1) and (2), 

we have ar (APD) = ar (AQD) (3) 

 Subtracting ar (AOD) from both sides,

 we have ar (APD) – ar (AOD) = ar (AQD) – ar (AOD) (4) 

 ar (APO) = ar (OQD),  Adding ar (OPQ) on both sides in (4), 

we have ar (APO) + ar (OPQ) = ar (OQD) + ar (OPQ)  

ar (APQ) = ar (DPQ) 

 Since, ar (APQ) = 1/2 ar (PQRS),

  therefore ar (DPQ) = 1/2 ar (PQRS) 

 Now, ar (PQRS) = 1/3 ar (ABCD) 

 Therefore, ar (APQ) = ar (DPQ)  

= 1/2 ar (PQRS)

= 1/2 × 1/3 ar (ABCD)  

= 1/6 ar (ABCD).

Explanation:

We know that the area of triangles on the same base and between the same parallel lines are equal.

Triangles ADF and AQD are on the same base AD and between the same parallel lines AD and n.

So, ar(ADF) = ar(DFB) ------- (1)

Adding ar (ABCD) on both sides in (1),

ar (APD) + ar (ABCD) = ar (AQD) + ar (ABCD)

From the figure,

ar(APD) + ar(ABCD) = ar(ABCDP)

So, ar(ABCDP) = ar(AQD) + ar(ABCD)

From the figure,

ar(AQD) + ar(ABCD) = ar(ABCQ)

Therefore, ar(ABCDP) = ar(ABCQ).

Explanation:

E is the mid point of CA

=> CE = CA/2

BD = CA/2

=> BD = CE

BD ║ CA => BD ║CE

BD = CE &  BD ║CE

=> BCED is a parallelogram

Diagonal of parallelogram divided it into two equal area traingle

=> Area of ΔDBE = Area of ΔCBE = (1/2) Area of  BCED

Area of ΔDBC = Area of ΔEBC = (1/2) Area of  BCED

=>  Area of ΔCBE =  Area of ΔDBC

in ΔABC  , BE is the median

Hence

Area of ΔCBE = (1/2) Area of ΔABC

=> Area of ΔDBC  = (1/2) Area of ΔABC

=> Area of ΔABC = 2  Area of ΔDBC

QED Hence proved.

Explanation:

We have : ABCD is a parallelogram. A point E is taken on the side BC,AE and DC are produced to meet at F.

Proving : Since ABCD is a parallelogram and diagonal AC divides it into two triangles of equal area, we have

Refer image,

ar(ΔADC)=ar(ΔABC)..........(1)

As DC∣∣AB, So CF∣∣AB

Since triangles on the same base and between the same parallels are equal in area, so we have

ar(ΔACF)=ar(ΔBCF)..........(2)

Adding (1) and (2), we get

ar(ΔADC)+ar(ΔACF)= ar(ΔABC)+ar(ΔBCF)

⇒ar(ΔADF)=ar(ABFC)

Hence proved.

Explanation:

∵AC is a diagonal of the || gm ABCD

Refer image,

∴ar(triangle ACD)= 1/2 ar(gmABCD).......(1)

Now, in triangle AOP and ΔCOQ

AO=CO [∵ Diagonals of || gm bisects each other]

∠AOP=∠COQ [Vert, opp ∠s]

∠OAP=∠OCQ [Alt. ∠s;AB∣∣CD]

triangle AOP≅  triangle COQ [By ASA cong. Rule]

Hence, ar(triangle AOP)=  ar(triangle COQ) [Cong. Area axiom]......(2)

Adding ar(quad.AOQD) to both sides of (2), we get

ar(quad.AOQD)+ar(ΔAOP)=ar(quad.AOQD)+ar(triangle COQ)

⇒ar(quad.APQD)=ar(triangle ACD)

But, ar(triangle ACD)=  ar∣∣(gmABCD) [From (1)]

Hence, ar(quad.APQD)= 1/2 ar(∣∣gmABCD)

So, PQ divides the parallelogram into two parts of equal area.

Explanation:

According to the question,

We have,

BE & CF are medians

E is the midpoint of AC

F is the midpoint of AB

∴ triangle BCE = triangle BEA … ( i )

triangle BCF = triangle CAF

Construct:

Join EF,

By midpoint theorem,

We get FE || BC

We know that,

triangle  on the same base and between same parallels are equal in area

∴ triangle FBC = triangle BCE

triangle FBC - triangle GBC = triangle BCE - triangle GBC

⇒ triangle FBG = triangle CGE (triangle GBC is common)

⇒ triangle CGE = triangle FBG …( ii )

Subtracting equation (ii) from (i)

We get,

triangle BCE - triangle CGE = triangle BEA - triangle FBG

∴ triangle BGC =quadrilateral AFGE.

Explanation:

CD∥AE and CY∥BA. We have to prove that 

ar(triangle CBX)=ar(triangle AXY).

Since triangle on the same base and between the same parallels are equal in area, so we have

ar(triangle ABC)=ar(triangle ABY)

⇒ar(triangle CBX)+ar(triangle ABX)=ar(triangle ABX)+ar(triangle AXY)

Hence, ar(triangle CBX)=ar(triangle AXY) [Cancelling ar(triangle ABX) from both sides].

Explanation:

AB is produced to P

Join DY and extend it to meet P

Considering triangles DCY and PBY,

Since Y is the midpoint of BC

CY = BY

We know that the alternate interior angles are equal

∠DCY = ∠PBY

We know that the vertically opposite angles are equal.

∠2 = ∠3 By ASA criteria, the triangles DCY and PBY are congruent.

By CPCTC,DC = BP

Given, DC = 30 cm

So, BP = 30 cm

Now, AP = AB + BP

AP = 50 + 30

AP = 80 cm

Considering triangle ADP,

By midpoint theorem,

XY = 1/2 AP

XY = 1/2 (80)

XY = 40 cm

Let the distance between AB, XY and DC be h cm.

Area of trapezium = 1/2 × (sum of parallel sides) × (distance between parallel sides)

Area of trapezium DCYX = 1/2 × (30 + 40) × h

= 1/2 × 70 × h

= 35h square cm

Similarly, area of trapezium XYBA = 1/2 × (40 + 50) × h

= 1/2 × 90 × h

= 45h square cm

Now, ar(DCYX)/ar(XYBA) = 35h/45h

ar(DCYX)/ar(XYBA) = 7/9

Therefore, ar(DCYX) = 7/9 ar(XYBA).

Explanation:

MB and LC intersect each other at O. 

Since, triangle LBC and triangle MBC both lie on the same base BC and between the same parallel lines BC and LM 

ar (triangle LBC) = ar (triangle MBC) 

⇒ ar (triangle LOB) + ar (triangle BOC) = ar(triangle MOC) + ar(triangle BOC) 

⇒ ar (triangle LOB) = ar (triangle MOC)

 [Eliminating ar (triangle BOC) from both sides] Hence proved.

Explanation:

We know that, triangles on the same base and between the same parallels are equal in area. 

Here, triangle ADQ and triangle ADE lie on the same base AD and between the same parallels AD and EQ.

 So, ar (triangle ADQ) = ar (triangle ADE) ...(i) 

Similarly, triangle ACP and triangle ACB lie on the same base AC and between the same parallels AC and BP. 

So, ar (triangle ACP) = ar (triangle ACB) …(ii) 

On adding Eqs. (i) and (ii), 

we have ar (triangle ADQ) + ar (triangle ACP) = ar (triangle ADE) + ar (triangle ACB)

 On adding ar (triangle ACD) both sides, 

we have ar (triangle ADQ) + ar (triangle ACP) + ar (triangle ACD) = ar (triangle ADE) + ar (triangle ACB) + ar (triangle ACD) 

=> ar (triangle APQ) = ar (ABCDE)  

Hence proved.

Explanation:

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

In triangle ABC, AD is the median.  

∴ ar(triangle ABD) = ar(triangle ACD) ...(i)  

In triangle GBC, GD is the median. 

 ∴ ar(triangle GBD) = ar(triangle GCD) ...(ii)  

From (i) and (ii), 

we get  ar(triangle ABD) - ar(triangle GBD) = ar(triangle ACD) -ar(triangle GCD)  

∴ a(triangle AGB) = ar(triangle AGC).  

Similarly,  ar(triangle AGB) =ar(triangle AGC) =ar(triangle BGC) ....(iii)  

But, ar(ABC) = ar(triangle AGB) + ar(triangle AGC) + ar(triangle BGC)  = 3 ar(triangle AGB) [Using (iii)] 

 ∴ ar(triangle AGB) = 1/3 ar(triangle ABC).  

Hence, ar(triangle AGB) = ar(triangle AGC) = ar (triangle BGC) = 1/3 ar(triangle ABC).

Explanation:

Since X and Y are the mid points AC and AB respectively.

∴ XY∥BC

Clearly, triangles BYC and BXC are on the same base BC and between the same parallel XY and BC

∴ ar(triangle BYC)=ar(triangle BXC)

implies ar(triangle BYC)−ar(triangle BOC)=ar(triangle BXC)−ar(triangle BOC)

Implies ar(triangle BOY)=ar(triangle COX)

implies ar(triangle BOY)+ar(triangle XOY)=ar(triangle COX)+ar(triangle XOY)

implies ar(triangle BXY)+ar(triangle CXY)...(i)

implies ar(triangle BYX)+ar(triangle AXY)=ar(triangle AXY)+ar(triangle AYX)

implies ar(triangle BAX)=ar(triangle CAY)...(ii)

In triangle BAP, XY= 1/2 AP

In triangle CAQ, XY=1/2 AQ

∴ AP=AQ

Clearly, triangles XAP and YAQ are between the same parallels and their bases AP and AQ are equal.

∴ ar(triangle XAP)=ar(triangle YAQ).....(iii)

Adding (ii) and (iii), we obtain

ar(triangle XAP)+ar(triangle BAX)=ar(triangle YAQ)ar(triangle CAY)

implies ar(triangle ABP)=ar(triangle ACQ) 

Hence Proved.

Explanation:

In ΔAPE and ΔDQF

∠APE and ∠DQF   [Corresponding angles]

⇒AE=DF [Opposite sites of parallelogram]

⇒∠AEP=∠DFQ   [Corresponding angles]

∴ΔAPE≅ΔDQF  [By ASA congruence test]

∴PE=FQ   [C.P.C.T]

∴ar(ΔAPE)=ar(ΔDQF) [Area of congruent triangles are equal]

i.e. ∴ar(ΔPEA)=ar(ΔQFD).