Class 9 - Maths - Areas Parallelograms Triangles

**Exercise 9.1**

**Question 1:**

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.

Answer:

(i) Trapezium ABCD and ΔPDC lie on the same DC and between the same parallel lines AB and

CD.

(ii) Parallelogram PQRS and trapezium SMNR lie on the same base SR but not between the

same parallel lines.

(iii) Parallelogram PQRS and ΔRTQ lie on the same base QR and between parallel lines QR and

PS.

(iv) Parallelogram ABCD and ΔPQR do not lie on the same base but between the same parallel

lines BC and AD.

(v) Quadrilateral ABQD and trapezium APCD lie on the same base AD and between the same

parallel lines AD and BQ.

(vi) Parallelogram PQRS and parallelogram ABCD do not lie on the same SR but between the

same parallel lines SR and PQ.

**Exercise 9.2**

**Question 1:**

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

Answer:

Area of parallelogram ABCD = AB * AE

= 16 * 8

= 128 cm^{2}

Also, area of parallelogram ABCD = AD * FC

=> AD * 10 = 128

=> AD = 128/10

=> AD = 12.8 cm

**Question 2:**

If E, F, G, and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar(EFGH) = ar(ABCD)/2.

Answer:

Given: A parallelogram ABCD · E, F, G, H are mid-points of sides AB, BC, CD, DA respectively.

To Prove: ar(EFGH) = ar(ABCD)/2

Construction : Join AC and HF.

Proof : In ∆ABC,

E is the mid-point of AB. F is the mid-point of BC.

=> EF is parallel to AC and EF = AC/2 ………..1

Similarly, in ∆ADC, we can show that

HG || AC and HG = AC/2 ...........2

From equation 1 and 2, we get

EF || HG and EF = HG

So, EFGH is a parallelogram. [One pour of opposite sides is equal and parallel]

In quadrilateral ABFH, we have

HA = FB and HA || FB [AD = BC => AD/2 = BC/2 => HA = FB]

So, ABFH is a parallelogram. [One pair of opposite sides is equal and parallel]

Now, triangle HEF and parallelogram HABF are on the same base HF and between the same

parallels HF and AB.

Now, Area of ∆HEF = area of HABF/2 ...........3

Similarly, area of ∆HGF = area of HFCD/2 .........4

Adding equation 3 and 4, we get

Area of ∆HEF + area of ∆HGF = (area of HABF + area of HFCD)/2

=> ar(EFGH) = ar(ABCD)

Hence, Proved.

**Question 3:**

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).

Answer:

Given: A parallelogram ABCD. P and Q are any points on DC and AD respectively.

To prove: ar(APB) = ar(BQC)

Construction: Draw PS || AD and QR || AB.

Proof: In parallelogram ABRQ, BQ is the diagonal.

So, area of ∆BQR = area of ABRQ/2 .............1

In parallelogram CDQR, CQ is a diagonal.

So, area of ∆RQC = area of CDQR/2 ............2

Adding equation 1 and 2, we get

area of ∆BQR + area of ∆RQC = (area of ABRQ + area of CDQR)/2

=> area of ∆BQC = area of ABCD/2 .........3

Again, in parallelogram DPSA, AP is a diagonal.

So, area of ∆ASP = area of DPSA/2 ..........4

In parallelogram BCPS, PB is a diagonal.

Now, area of ∆BPS = area of BCPS/2 ……….5

Adding equation 4 and 5, we get

area of ∆ASP + area of ∆BPS = (area of DPSA + area of BCPS)/2

=> area of ∆APB = (area of ABCD) ……..6

From equation 3 and 6, we have

area of ∆APB = area of ∆BQC.

Hence, Proved.

**Question 4:**

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

(i) ar(APB) + ar(PCD) = ar(ABCD)/2

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

Answer:

Given: A parallelogram ABCD. P is a point inside it.

To prove:

(i) ar(APB) + ar(PCD) = ar(ABCD)/2

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

Construction: Draw EF through P parallel to AB, and GH through P parallel to AD.

Proof: In parallelogram FPGA, AP is a diagonal,

So, area of ∆APG = area of ∆APF ..............1

In parallelogram BGPE, PB is a diagonal,

So, area of ∆BPG = area of ∆EPB ..............2

In parallelogram DHPF, DP is a diagonal,

area of ∆DPH = area of ∆DPF ..........3

In parallelogram HCEP, CP is a diagonal,

So, area of ∆CPH = area of ∆CPE .....4

Adding equation 1, 2, 3 and 4, we get

area of ∆APG + area of ∆BPG + area of ∆DPH + area of ∆CPH

= area of ∆APF + area of ∆EPB + area of ∆DPF + area ∆CPE

=> [area of ∆APG + area of ∆BPG] + [area of ∆DPH + area of ∆CPH]

= [area of ∆APF + area of ∆DPF] + [area of ∆EPB + area of ∆CPE]

=> area of ∆APB + area of ∆CPD = area of ∆APD + area of ∆BPC ……..5

But area of parallelogram ABCD

= area of ∆APB + area of ∆CPD + area of ∆APD + area of ∆BPC ……….6

From equation 5 and 6, we get

area of ∆APB + area of ∆PCD = area of ABCD/2

=> ar(APB) + ar(PCD) = ar(ABCD)/2

Hence, Proved.

From equation 5,

=> ar (APD) + ar (PBC) = ar (APB) + ar (CPD)

**Question 5:**

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

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

(ii) ar(AXS) = ar(PQRS)/2

Answer:

Given: PQRS and ABRS are parallelograms and X is any point on side BR.

To prove:

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

(ii) ar(AXS) = ar(PQRS)/2

(i) In ∆ASP and BRQ, we have,

∠SPA = ∠RQB [Corresponding angles] ...(1)

∠PAS = ∠QBR [Corresponding angles] ...(2)

So, ∠PSA = ∠QRB [Angle sum property of a triangle] ...(3)

Also, PS = QR [Opposite sides of the parallelogram PQRS] ...(4)

So, ∆ASP ≅ ∆BRQ [ASA axiom, using (1), (3) and (4)]

Therefore, area of ∆PSA = area of ∆QRB [Congruent figures have equal areas] ...(5)

Now, ar(PQRS) = ar(PSA) + ar(ASRQ)

= ar(QRB) + ar(ASRQ)

= ar(ABRS)

So, ar(PQRS) = ar(ABRS) Proved.

(ii) Now, ∆AXS and ||gm ABRS are on the same base AS and between same parallels AS and BR

So, area of ∆AXS = area of ABRS/2

=> area of ∆AXS = area of PQRS/2 [Since ar(PQRS) = ar(ABRS]

=> ar of (AXS) = ar of (PQRS)/2

**Question 6:**

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 fields 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?

Answer:

The field is divided in three triangles.

Since triangle APQ and parallelogram PQRS are on the same base PQ and between the same

parallels PQ and RS.

So, ar(APQ) = ar(PQRS)/2

=> 2ar(APQ) = ar(PQRS)

But ar(PQRS) = ar(APQ) + ar(PSA) + ar(ARQ)

=> 2 ar(APQ) = ar(APQ) + ar(PSA) + ar(ARQ)

=> ar(APQ) = ar(PSA) + ar(ARQ)

Hence, area of ∆APQ = area of ∆PSA + area of ∆ARQ.

To sow wheat and pulses in equal portions of the field separately, farmer sow wheat in ∆APQ

and pulses in other two triangles or pulses in ∆APQ and wheat in other two triangles.

**Exercise 9.3**

**Question 1:**

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

Answer:

AD is the median of ∆ABC. Therefore, it will divide ∆ABC into two triangles of equal areas.

So, Area (∆ABD) = Area (∆ACD) ... (1)

ED is the median of ∆EBC.

So, Area (∆EBD) = Area (∆ECD) ... (2)

On subtracting equation (2) from equation (1), we obtain

Area (∆ABD) − Area (EBD) = Area (∆ACD) − Area (∆ECD)

Area (∆ABE) = Area (∆ACE)

**Question 2:**

In a triangle ABC, E is the mid-point of median AD. Show that ar(BED) = ar(ABC)/4 .

Answer:

Given ABC is a triangle in which E is the mid-point of the median.

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

So area of Δ ABD = 1/2 * area of Δ ABC

Again BE is the median of the triangle ABD.

So area of Δ BDE =1/2 * area of Δ ABD

= 1/2 *1/2 * area of Δ ABC

= 1/4 * area of Δ ABC

So, area of Δ BDE = 1/4 * area of Δ ABC

**Question 3:**

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

Answer:

We know that diagonals of parallelogram bisect each other.

Therefore, O is the mid-point of AC and BD.

BO is the median in ∆ABC.

Therefore, it will divide it into two triangles of equal areas.

So, Area (∆AOB) = Area (∆BOC) ... (1)

In ∆BCD, CO is the median.

So, Area(∆BOC) = Area(∆COD) ... (2)

Similarly, Area(∆COD) = Area(∆AOD) ... (3)

From equations (1), (2), and (3), we obtain

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

Therefore, it is evident that the diagonals of a parallelogram divide it into four triangles of

equal area.

**Question 4:**

In Fig. 9.24, 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).

Answer:

Consider ∆ACD.

Line-segment CD is bisected by AB at O. Therefore, AO is the median of ∆ACD.

So, Area (∆ACO) = Area (∆ADO) ... (1)

Considering ∆BCD, BO is the median.

So, Area (∆BCO) = Area (∆BDO) ... (2)

Adding equations (1) and (2), we obtain

Area (∆ACO) + Area (∆BCO) = Area (∆ADO) + Area (∆BDO)

=> Area (∆ABC) = Area (∆ABD)

**Question 5:**

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) ar(DEF) = ar(ABC)/4

(iii) ar(BDEF) = ar(ABC)/2

Answer:

Given: D, E and F are the mid-points of the sides BC, CA and AB of ΔABC.

(i) In ΔABC,

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

So, EF || BC

In a triangle, the line segment joining the mid-points of any two sides is parallel to the third

side.

So, EF || BD ………..1

Similarly, ED || BF …….2

From equation 1 and 2, we get

BDEF is a parallelogram [a quadrilateral is a parallelogram if opposite sides are parallel]

(ii) As in (i), we can prove that

AFDE and FDCE are parallelograms.

FD is a diagonal of the parallelogram BDEF.

So, area(ΔFBD) = area(ΔDEF) ……………3

Similarly, area(ΔDEF) = area(ΔEAE) ……………4

and area(ΔDEF) = area(ΔDCE) ……………5

From equation 3, 4 and 5, we get

area(ΔFBD) = area(ΔDEF) = area(ΔFAE) = area(ΔDCE) …………6

So, ΔABC is divided into four non-overlapping triangles ΔFBD, ΔDEF, ΔFAE and ΔDCE.

Hence, area(ΔABC) = area(ΔFBD) + area(ΔDEF) + area(ΔFAE) + area(ΔDCE) + area(ΔDEF)

=> area(ΔABC) = 4 * area(ΔDEF) [From equation 6]

=> area(ΔDEF) = area(ΔABC)/4 ……..7

(iii) area(BDEF) = area(ΔFBD) + area(ΔDEF) [From equation 3]

= area(ΔDEF) + area(ΔDEF)

= 2 * area(ΔDEF)

= 2 * area(ΔABC)/4

= area(ΔABC)/2

**Question 6:**

In Fig. 9.25, 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.]

Answer:

Let us draw DN ⊥ AC and BM ⊥ AC.

(i) In ∆DON and ∆BOM,

∠DNO = ∠BMO (By construction)

∠DON = ∠BOM (Vertically opposite angles)

OD = OB (Given)

By AAS congruence rule,

∆DON ≅ ∆BOM

DN = BM ... (1)

We know that congruent triangles have equal areas.

Area (∆DON) = Area (∆BOM) ... (2)

In ∆DNC and ∆BMA,

∠DNC = ∠BMA (By construction)

CD = AB (Given)

DN = BM [Using equation (1)]

So, ∆DNC ≅ ∆BMA (RHS congruence rule)

Area (∆DNC) = Area (∆BMA) ... (3)

On adding equations (2) and (3), we obtain

Area (∆DON) + Area (∆DNC) = Area (∆BOM) + Area (∆BMA)

Therefore, Area (∆DOC) = Area (∆AOB)

(ii) We obtained,

Area (∆DOC) = Area (∆AOB)

Area (∆DOC) + Area (∆OCB) = Area (∆AOB) + Area (∆OCB) [Adding Area (∆OCB) to both sides]

Area (∆DCB) = Area (∆ACB)

(iii) We obtained,

Area (∆DCB) = Area (∆ACB)

If two triangles have the same base and equal areas, then these will lie between the same

parallels.

So, DA || CB ... (4)

In quadrilateral ABCD, one pair of opposite sides is equal (AB = CD) and the other pair of

opposite sides is parallel (DA || CB).

Therefore, ABCD is a parallelogram.

**Question 7:**

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

Prove that DE || BC.

Answer:

Since ∆BCE and ∆BCD are lying on a common base BC and also have equal areas,

∆BCE and ∆BCD will lie between the same parallel lines.

So, DE || BC

**Question 8:**

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)

Answer:

It is given that XY || BC and EY || BC BE || AC and BE || CY

Therefore, EBCY is a parallelogram.

It is given that

XY || BC and XF || BC

FC || AB and FC || XB

Therefore, BCFX is a parallelogram.

Parallelograms EBCY and BCFX are on the same base BC and between the same parallels BC

and EF.

So, Area (EBCY) = Area (BCFX) ... (1)

Consider parallelogram EBCY and ∆AEB

These lie on the same base BE and are between the same parallels BE and AC.

So, Area (∆ABE) = Area (EBCY) ... (2)

Also, parallelogram BCFX and ∆ACF are on the same base CF and between the same parallels

CF and AB.

So, Area (∆ACF) = Area (BCFX) ... (3)

From equations (1), (2), and (3), we obtain

Area (∆ABE) = Area (∆ACF)

**Question 9:**

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 Fig. 9.26).

Show that ar (ABCD) = ar (PBQR).

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

Answer:

Let us join AC and PQ.

∆ACQ and ∆AQP are on the same base AQ and between the same parallels AQ and CP.

So, Area (∆ACQ) = Area (∆APQ)

=> Area (∆ACQ) − Area (∆ABQ) = Area (∆APQ) − Area (∆ABQ)

=> Area (∆ABC) = Area (∆QBP) ... (1)

Since AC and PQ are diagonals of parallelograms ABCD and PBQR respectively,

Area (∆ABC) = Area (ABCD)/2 ... (2)

Area (∆QBP) = Area (PBQR) ... (3)

From equation 1, 2 and 3, we get

Area (ABCD)/2 = Area (PBQR)/2

=> Area (ABCD) = Area (PBQR)

**Question 10:**

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

Answer:

It can be observed that ∆DAC and ∆DBC lie on the same base DC and between the same

parallels AB and CD.

Area(∆DAC) = Area(∆DBC)

Area(∆DAC) - Area(∆DOC) = Area(∆DBC) - Area(∆DOC)

Area(∆AOD) = Area(∆BOC)

**Question 11:**

In Fig. 9.27, 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)

Answer:

(i) ∆ACB and ∆ACF lie on the same base AC and are between the same parallels AC and BF.

So, Area (∆ACB) = Area (∆ACF)

(ii) It can be observed that

Area (∆ACB) = Area (∆ACF)

Area (∆ACB) + Area (ACDE) = Area (ACF) + Area (ACDE)

Area (ABCDE) = Area (AEDF)

**Question 12:**

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.

Answer:

Let quadrilateral ABCD be the original shape of the field.

The proposal may be implemented as follows.

Join diagonal BD and draw a line parallel to BD through point A. Let it meet the extended side

CD of ABCD at point E. Join BE and AD. Let them intersect each other at O. Then, portion ∆AOB

can be cut from the original field so that the new shape of the field will be ∆BCE. (See figure)

We have to prove that the area of ∆AOB (portion that was cut so as to construct Health

Centre) is equal to the area of ∆DEO (portion added to the field so as to make the area of the

new field so formed equal to the area of the original field)

It can be observed that ∆DEB and ∆DAB lie on the same base BD and are between the same

parallels BD and AE.

So, Area (∆DEB) = Area (∆DAB)

=> Area (∆DEB) − Area (∆DOB) = Area (∆DAB) − Area (∆DOB)

=> Area (∆DEO) = Area (∆AOB)

**Question 13:**

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.]

Answer:

It can be observed that ∆ADX and ∆ACX lie on the same base AX and are between the same

parallels AB and DC.

So, Area (∆ADX) = Area (∆ACX) ………..1

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

So, Area (∆ACY) = Area (∆ACX) ………..2

From equation 1 and 2, we get

So, Area (∆ADX) = Area (∆ACY)

**Question 14:**

In Fig.9.28, AP || BQ || CR. Prove that ar(AQC) = ar(PBR).

Answer:

Since ∆ABQ and ∆PBQ lie on the same BQ and are between the same parallels AP and BQ.

So, Area(∆ABQ) = Area(∆PBQ) ………1

Again ∆BCQ and ∆BRQ lie on the same BQ and are between the same parallels BQ and CR.

So, Area(∆BCQ) = Area(∆BRQ) ………2

On adding equation 1 and 2, we get

Area(∆ABQ) + Area(∆BCQ) = Area(∆PBQ) + Area(∆BRQ)

=> Area(∆AQC) = Area(∆PBR)

**Question 15:**

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.

Answer:

It is given that

Area (∆AOD) = Area (∆BOC)

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

Area (∆ADB) = Area (∆ACB)

We know that triangles on the same base having areas equal to each other lie between the

same parallels.

Therefore, these triangles, ∆ADB and ∆ACB, are lying between the same parallels.

i.e., AB || CD

Therefore, ABCD is a trapezium.

**Question 16:**

In Fig.9.29, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the

quadrilaterals ABCD and DCPR are trapeziums.

Answer:

It is given that

Area (∆DRC) = Area (∆DPC)

As ∆DRC and ∆DPC lie on the same base DC and have equal areas, therefore, they

must lie between the same parallel lines.

So, DC || RP

Therefore, DCPR is a trapezium.

It is also given that

Area (∆BDP) = Area (∆ARC)

=> Area (BDP) − Area (∆DPC) = Area (∆ARC) − Area (∆DRC)

=> Area (∆BDC) = Area (∆ADC)

Since ∆BDC and ∆ADC are on the same base CD and have equal areas, they must lie between

the same parallel lines.

So, AB || CD

Therefore, ABCD is a trapezium.

**Exercise 9.4 (Optional)**

**Question 1:**

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.

Answer:

As the parallelogram and the rectangle have the same base and equal area, therefore, these

will also lie between the same parallels.

Consider the parallelogram ABCD and rectangle ABEF as follows.

Here, it can be observed that parallelogram ABCD and rectangle ABEF are between the same parallels AB and CF.

We know that opposite sides of a parallelogram or a rectangle are of equal lengths.

Therefore,

AB = EF [For rectangle]

AB = CD [For parallelogram]

So, CD = EF

=> AB + CD = AB + EF ... (1)

Of all the line segments that can be drawn to a given line from a point not lying on it, the

perpendicular line segment is the shortest.

So, AF < AD

Similarly, BE < BC

So, AF + BE < AD + BC ... (2)

From equations (1) and (2), we obtain

AB + EF + AF + BE < AD + BC + AB + CD

Hence, perimeter of rectangle ABEF < perimeter of parallelogram ABCD

**Question 2:**

In Fig. 9.30, D and E are two points on BC

such that BD = DE = EC. Show that

ar (ABD) = ar (ADE) = ar (AEC)

Can you now answer the question that you have left in the ‘Introduction’ of this chapter, whether

the field of Budhia has been actually divided into three parts of equal area?

[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas.

In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC,

you can divide ∆ABC into n triangles of equal areas.]

Answer:

Let us draw a line segment AM ⊥ BC.

We know that,

Area of a triangle = 1/2 * Base * Altitude

Area(ΔADE) = 1/2 * DE * AM

Area(ΔABD) = 1/2 * BD * AM

Area(ΔAEC) = 1/2 * EC * AM

It is given that DE = BD = EC

So, 1/2 * DE * AM = 1/2 * BD * AM = 1/2 * EC * AM

So, Area (∆ADE) = Area (∆ABD) = Area (∆AEC)

It can be observed that Budhia has divided her field into 3 equal parts.

**Question 3:**

In Fig. 9.31, ABCD, DCFE and ABFE are parallelograms. Show that: ar(ADE) = ar(BCF)

Answer:

It is given that ABCD is a parallelogram. We know that opposite sides of a parallelogram are

equal.

So, AD = BC ....... (1)

Similarly, for parallelograms DCEF and ABFE, it can be proved that

DE = CF ................ (2)

and, EA = FB ....... (3)

In ∆ADE and ∆BCF,

AD = BC [Using equation (1)]

DE = CF [Using equation (2)]

EA = FB [Using equation (3)]

So, ∆ADE ≅ BCF [SSS congruence rule]

Hence, Area (∆ADE) = Area (∆BCF)

**Question 4:**

In Fig. 9.32, 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.]

Answer:

It is given that ABCD is a parallelogram.

AD || BC and AB || DC [Opposite sides of a parallelogram are parallel to each other]

Join point A to point C.

Consider ∆APC and ∆BPC,

∆APC and ∆BPC are lying on the same base PC and between the same parallels PC and AB.

Therefore,

Area (∆APC) = Area (∆BPC) ..........(1)

In quadrilateral ACDQ, it is given that

AD = CQ

Since ABCD is a parallelogram,

AD || BC (Opposite sides of a parallelogram are parallel)

CQ is a line segment which is obtained when line segment BC is produced.

So, AD || CQ

We have,

AC = DQ and AC || DQ

Hence, ACQD is a parallelogram.

Consider ∆DCQ and ∆ACQ

These are on the same base CQ and between the same parallels CQ and AD.

Therefore,

Area (∆DCQ) = Area (∆ACQ)

=> Area (∆DCQ) − Area (∆PQC) = Area (∆ACQ) − Area (∆PQC)

=> Area (∆DPQ) = Area (∆APC) ... (2)

From equations (1) and (2), we obtain

Area (∆BPC) = Area (∆DPQ)

**Question 5:**

In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:

(i) ar(BDE) =ar (ABC)/4

(ii) ar(BDE) = ar(BAE)/2

(iii) ar(ABC) = 2 ar(BEC)

(iv) ar(BFE) = ar(AFD)

(v) ar(BFE) = 2 ar(FED)

(vi) ar (FED) = ar(AFC)/8

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

Answer:

(i) Let G and H be the mid-points of side AB and AC respectively.

Line segment GH is joining the mid-points. Therefore, it will be parallel to third side BC and

also its length will be half of the length of BC (mid-point theorem).

GH = BC/2 and GH || BD

GH = BD = DC and GH || BD [D is the mid-point of BC]

Consider quadrilateral GHDB,

GH ||BD and GH = BD

Two line segments joining two parallel line segments of equal length will also be equal and

parallel to each other.

Therefore, BG = DH and BG || DH

Hence, quadrilateral GHDB is a parallelogram.

We know that in a parallelogram, the diagonal bisects it into two triangles of equal area.

Hence, Area (∆BDG) = Area (∆HGD)

Similarly, it can be proved that quadrilaterals DCHG, GDHA, and BEDG are parallelograms and

their respective diagonals are dividing them into two triangles of equal area.

ar(∆GDH) = ar(∆CHD) [For parallelogram DCHG]

ar(∆GDH) = ar(∆HAG) [For parallelogram GDHA]

ar(∆BDE) = ar(∆DBG) [For parallelogram BEDG]

ar(∆ABC) = ar(∆BDG) + ar(∆GDH) + ar(∆DCH) + ar(∆AGH)

ar(∆ABC) = 4 * ar(∆BDE)

Hence, ar(∆BDE) = ar(∆ABC)/4

(ii)Area (∆BDE) = Area (∆AED) [Common base DE and DE||AB]

Area (∆BDE) − Area (∆FED) = Area (∆AED) − Area (∆FED)

Area (∆BEF) = Area (∆AFD) ………….(1)

Area (∆ABD) = Area (∆ABF) + Area (∆AFD)

Area (∆ABD) = Area (∆ABF) + Area (∆BEF) [From equation (1)]

Area (∆ABD) = Area (∆ABE) …………(2)

AD is the median in ∆ABC.

ar(∆ABD) = ar(∆ABC)/2

= 4 * ar(∆BDE)/2 [As proved earlier]

= 2 * ar(∆BDE)

=> ar(∆ABD) = 2 * ar(∆BDE) ………..(3)

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

2 * ar(∆BDE) = ar(∆ABE)

=> ar(∆BDE) = ar(∆ABE)/2

(iii) ar(∆ABE) = ar(∆BEC) [Common base BE and BE||AC]

ar(∆ABF) + ar(∆BEF) = ar(∆BEC)

Using equation (1), we obtain

ar(∆ABF) + ar(∆AFD) = ar(∆BEC)

=> ar(∆ABD) = ar(∆BEC)

=> ar(∆ABC)/2 = ar(∆BEC)

=> ar(∆ABC) = 2 * ar(∆BEC)

(iv) It is seen that ∆BDE and ar ∆AED lie on the same base (DE) and between the parallels DE

and AB.

ar (∆BDE) = ar (∆AED)

ar (∆BDE) − ar (∆FED) = ar (∆AED) − ar (∆FED)

ar (∆BFE) = ar (∆AFD)

(v) Let h be the height of vertex E, corresponding to the side BD in ∆BDE.

Let H be the height of vertex A, corresponding to the side BC in ∆ABC.

In (i), it was shown that ar(∆BDE) = ar(∆ABC)/4

=> 1/2 * BD * h = (1/2 * BC * H)/4

=> BD * h = (2BD * H)/4

=> h = H/2

In (iv), it was shown that ar (∆BFE) = ar (∆AFD).

ar (∆BFE) = ar (∆AFD)

= 1/2 * FD * H

= 1/2 * FD * 2h

= 2(1/2 * FD * h)

= 2 * ar (∆FED)

Hence, ar (∆BFE) = 2 * ar (∆FED)

(vi) Area (AFC) = area (AFD) + area (ADC)

[in (iv), ar(∆BFE) = ar(∆AFD), AD is the median of ΔABC]

= ar (∆BFE) + ar (∆ABC)/2

= ar (∆BFE) + 4 * ar (∆BDE)/2

= ar (∆BFE) + 2 * ar (∆BDE) …………..(5)

Now, by (v), ar (∆BFE) = 2 * ar (∆FED) …………..(6)

ar (∆BDE) = ar (∆BFE) + ar (∆FED) = 2 * ar (∆FED) + ar (∆FED) = 3 * ar (∆FED) ………7

From equation (5), (6) and (7), we get

ar (∆AFC) = 2 * ar (∆FED) + 2 * 3 * ar (∆FED)

= 2 * ar (∆FED) + 6 * ar (∆FED)

= 8 * ar (∆FED)

Hence, ar (∆FED) = ar (∆AFC)/8

**Question 6:**

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.]

Answer:

Construction: From A and C, draw perpendiculars AM and CN to BD.

ar (∆APB) * ar (∆CPD) = 1/2 * BP * AM * 1/2 * PD * CN …………..1

ar (∆APD) * ar (∆BPC) = 1/2 * PD * AM * 1/2 * BP * CN …………..2

From equation 1 and 2, we get

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

**Question 7:**

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) ar (PRQ) = ar (ARC)/2

(ii) ar (RQC) = 3ar (ABC)/8

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

Answer:

P and Q are respectively the mid-points of sides AB and BC of ΔABC and R is the mid-point of AP.

Join AQ and PC.

(i) We have,

ar(ΔPQR) = ar(ΔAPQ)/2 [Since QR is a median and ΔAPQ and it divides the triangle into two

other triangles of equal area]

= 1/2 * 1/2 * ar(ΔABQ) [Since QP is a median of ΔABQ]

= 1/4 * ar(ΔABQ)

= 1/4 * 1/2 * ar(ΔABQ) [Since AQ is a median of ΔABQ]

= 1/8 * ar(ΔABQ) ……………..1

Again, ar(ΔARC) = 1/2 * ar(ΔAPC) [Since CR is a median of ΔAPC]

= 1/4 * 1/2 * ar(ΔABQ) [Since CP is a median of ΔABC]

= 1/8 * ar(ΔABQ) ………2

From equation 1 and 2, we get

ar(ΔPQR) = 1/8 * ar(ΔABC)

= 1/2 * 1/2 * ar(ΔABC)

= 1/2 * ar(ΔARC)

(ii) We have,

ar(ΔRQC) = ar(ΔRQA) + ar(ΔAQC) - ar(ΔARC) …………..3

Now, ar(ΔRQA) = 1/2 * ar(ΔPQA) [Since RQ is a median of ΔPQA]

= 1/2 * 1/2 * ar(ΔAQB) [Since PQ is a median of ΔAQB]

= 1/4 * ar(ΔAQB)

= 1/4 * 1/2 * ar(ΔABC) [Since AQ is a median of ΔABC]

= 1/8 * ar(ΔABC) ………..4

ar(ΔAQC) = 1/2 * ar(ΔABC) ……5 [Since AQ is a median of ΔABC]

ar(ΔARC) = 1/2 * ar(ΔAPC) [Since CR is a median of ΔAPC]

= 1/2 * 1/2 * ar(ΔABC) [Since CP is a median of ΔABC]

= 1/4 * ar(ΔABC) ………6

From equation 3, 4, 5 and 6, we get

ar(ΔRQC) = 1/8 * ar(ΔABC) + 1/2 * ar(ΔABC) ** - **1/4 * ar(ΔABC)

= 3/8 * ar(ΔABC)

(iii) We have,

ar(ΔPBQ) = 1/2 * ar(ΔABQ) [Since PQ is a median of ΔABQ]

= 1/2 * 1/2 * ar(ΔABC) [Since AQ is a median of ΔABC]

= 1/4 * ar(ΔABC)

= ar(ΔARC) [From equation 6]

**Question 8:**

In Fig. 9.34, 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) = 2 ar(MBC)

(iii) ar(BYXD) = ar(ABMN) (iv) ∆ FCB ≅ ∆ ACE

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

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

Note: Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.

Answer:

(i) In ΔMBC and ΔABD, we have

BC = BD [Sides of the square BCED]

MB = AB [Sides of the square ABMN]

∠MBC = ∠ABD [since each 90^{0} + ∠ABC]

Therefore, BY SAS criterion of congruence, we have

∆ MBC ≅ ∆ ABD

(ii) ΔABD and square BYXD have the same base BD and are between the same parallels BD and

AX.

Therefore, ΔABD = 1/2 * ar(BYXD)

But ∆ MBC ≅ ∆ ABD [Proved in part (i)]

So, ar(∆ MBC) ≅ ar(∆ ABD)

Therefore, ar(∆ MBC) ≅ ar(∆ ABD) = 1/2 * ar(BYXD)

=> ar(BYXD) = 2 * ar(∆ MBC)

(iii) Square ABMN and ∆ MBC have the same base MB and are between the same parallels MB

and NAC.

ar(∆ MBC) = 1/2 * ar(ABMN)

=> ar(ABMN) = 2 * ar(∆ MBC) = ar(BYXD) [Using part (ii)]

(iv) In ΔACE and ΔBCF, we have

CE = BC [Sides of the square BCED]

AC = CF [Sides of the square ACFG]

∠ACE = ∠BCF [since each 90^{0} + ∠BCA]

Therefore, BY SAS criterion of congruence, we have

∆ ACE ≅ ∆ BCF

(v) ΔACE and square CYXE have the same base CE and are between same parallels CF and AYX.

Therefore, ar(∆ ACE) = 1/2 * ar(CYXE)

=> ar(∆ FCB) = 1/2 * ar(CYXE) [Since ∆ ACE ≅ ∆ BCF in part (iv)]

=> ar(CYXE) = 2 * ar(∆ FCB)

(vi) Square ACFG and ∆ BCF have the same base CF and are between the same parallels CF

and BAG.

ar(∆ BCF) = 1/2 * ar(ACFG)

=> 1/2 *ar(CYXE) = 1/2 * ar(ACFG) [Using part (v)]

=> ar(CYXE) = ar(ACFG)

(vii) From part (iii) and (iv), we have

ar(BYXD) = ar(ABMN)

and ar(CYXE) = ar(ACFG)

On adding, we get

ar(BYXD) + ar(CYXE) = ar(ABMN) + ar(ACFG)

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

.