Chapter 9 Areas of Parallelograms and Triangles Ex- 9.1 |
Chapter 9 Areas of Parallelograms and Triangles Ex- 9.2 |
Chapter 9 Areas of Parallelograms and Triangles Ex- 9.3 |

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

**Answer
1** :

Given,

|| gm ABCD and a rectangleABEF have the same base AB and equal areas.

To prove,

Perimeter of || gmABCD is greater than the perimeter of rectangle ABEF.

Proof,

We know that, theopposite sides of a|| gm and rectangle are equal.

, AB = DC [As ABCD isa || gm]

and, AB = EF [As ABEFis a rectangle]

, DC = EF … (i)

Adding AB on bothsides, we get,

⇒AB + DC = AB + EF …(ii)

We know that, theperpendicular segment is the shortest of all the segments that can be drawn toa given line from a point not lying on it.

, BE < BC and AF< AD

⇒ BC > BE and AD> AF

⇒ BC+AD > BE+AF …(iii)

Adding (ii) and (iii),we get

AB+DC+BC+AD >AB+EF+BE+AF

⇒ AB+BC+CD+DA > AB+BE+EF+FA

⇒ perimeter of || gmABCD > perimeter of rectangle ABEF.

, the perimeter of theparallelogram is greater than that of the rectangle.

Hence Proved.

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 intothree parts of equal area?

**Answer
2** :

Given,

BD = DE = EC

To prove,

ar (△ABD) = ar (△ADE) = ar (△AEC)

Proof,

In (△ABE), AD is median [since, BD = DE, given]

We know that, themedian of a triangle divides it into two parts of equal areas

, ar(△ABD) = ar(△AED) —(i)

Similarly,

In (△ADC), AE is median [since, DE = EC, given]

,ar(ADE) = ar(AEC)—(ii)

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

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

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

**Answer
3** :

Given,

ABCD, DCFE and ABFEare parallelograms

To prove,

ar (△ADE) = ar (△BCF)

Proof,

In △ADE and △BCF,

AD = BC [Since, theyare the opposite sides of the parallelogram ABCD]

DE = CF [Since, theyare the opposite sides of the parallelogram DCFE]

AE = BF [Since, theyare the opposite sides of the parallelogram ABFE]

, △ADE ≅ △BCF [Using SSS Congruence theorem]

, ar(△ADE) = ar(△BCF) [ By CPCT]

In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Qsuch that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ).

[Hint : Join AC.]

**Answer
4** :

Given:

ABCD is aparallelogram

AD = CQ

To prove:

ar (△BPC) = ar (△DPQ)

Proof:

In △ADP and △QCP,

∠APD = ∠QPC [Vertically Opposite Angles]

∠ADP = ∠QCP [Alternate Angles]

AD = CQ [given]

, △ABO ≅ △ACD [AAS congruency]

, DP = CP [CPCT]

In △CDQ, QP is median. [Since, DP = CP]

Since, median of atriangle divides it into two parts of equal areas.

, ar(△DPQ) = ar(△QPC) —(i)

In △PBQ, PC is median. [Since, AD = CQ and AD = BC⇒ BC = QC]

Since, median of atriangle divides it into two parts of equal areas.

, ar(△QPC) = ar(△BPC) —(ii)

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

ar(△BPC) = ar(△DPQ)

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

(i) ar (BDE) =1/4 ar (ABC)

(ii) ar (BDE) = ½ ar (BAE)

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

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

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

(vi) ar (FED) = 1/8 ar (AFC)

**Answer
5** :

(i) Assume that G andH are the mid-points of the sides AB and AC respectively.

Join the mid-pointswith line-segment GH. Here, GH is parallel to third side.

, BC will be half ofthe length of BC by mid-point theorem.

∴ GH =1/2 BC and GH ||BD

∴ GH = BD = DC and GH|| BD (Since, D is the mid-point of BC)

Similarly,

GD = HC = HA

HD = AG = BG

, ΔABC is divided into4 equal equilateral triangles ΔBGD, ΔAGH, ΔDHC and ΔGHD

We can say that,

ΔBGD = ¼ ΔABC

Considering, ΔBDG andΔBDE

BD = BD (Common base)

Since both trianglesare equilateral triangle, we can say that,

BG = BE

DG = DE

, ΔBDG ΔBDE [By SSScongruency]

, area (ΔBDG) = area(ΔBDE)

ar (ΔBDE) = ¼ ar(ΔABC)

Hence proved

(ii)

ar(ΔBDE) = ar(ΔAED)(Common base DE and DE||AB)

ar(ΔBDE)−ar(ΔFED) =ar(ΔAED)−ar (ΔFED)

ar(ΔBEF) = ar(ΔAFD)…(i)

Now,

ar(ΔABD) =ar(ΔABF)+ar(ΔAFD)

ar(ΔABD) =ar(ΔABF)+ar(ΔBEF) [From equation (i)]

ar(ΔABD) = ar(ΔABE)…(ii)

AD is the median ofΔABC.

ar(ΔABD) = ½ ar (ΔABC)

= (4/2) ar (ΔBDE)

= 2 ar (ΔBDE)…(iii)

From (ii) and (iii),we obtain

2 ar (ΔBDE) = ar(ΔABE)

ar (BDE) = ½ ar (BAE)

Hence proved

(iii) ar(ΔABE) =ar(ΔBEC) [Common base BE and BE || AC]

ar(ΔABF) + ar(ΔBEF) =ar(ΔBEC)

From eq^{n} (i),we get,

ar(ΔABF) + ar(ΔAFD) =ar(ΔBEC)

ar(ΔABD) = ar(ΔBEC)

½ ar(ΔABC) = ar(ΔBEC)

ar(ΔABC) = 2 ar(ΔBEC)

Hence proved

(iv) ΔBDE and ΔAED lieon the same base (DE) and are in-between the parallel lines DE and AB.

∴ar (ΔBDE) = ar (ΔAED)

Subtracting ar(ΔFED)from L.H.S and R.H.S,

We get,

∴ar (ΔBDE)−ar (ΔFED) =ar (ΔAED)−ar (ΔFED)

∴ar (ΔBFE) = ar(ΔAFD)

Hence proved

(v) Assume that h isthe height of vertex E, corresponding to the side BD in ΔBDE.

Also assume that H isthe height of vertex A, corresponding to the side BC in ΔABC.

While solving Question(i),

We saw that,

ar (ΔBDE) = ¼ ar (ΔABC)

While solving Question(iv),

We saw that,

ar (ΔBFE) = ar (ΔAFD).

∴ar (ΔBFE) = ar (ΔAFD)

= 2 ar (ΔFED)

Hence, ar (ΔBFE) = 2ar (ΔFED)

Hence proved

(vi) ar (ΔAFC) = ar(ΔAFD) + ar(ΔADC)

= 2 ar (ΔFED) + (1/2)ar(ΔABC) [using (v)

= 2 ar (ΔFED) + ½[4ar(ΔBDE)] [Using result of Question (i)]

= 2 ar (ΔFED) +2ar(ΔBDE)

Since, ΔBDE and ΔAEDare on the same base and between same parallels

= 2 ar (ΔFED) +2 ar(ΔAED)

= 2 ar (ΔFED) +2 [ar(ΔAFD) +ar (ΔFED)]

= 2 ar (ΔFED) +2 ar(ΔAFD) +2 ar (ΔFED) [From question (viii)]

= 4 ar (ΔFED) +4 ar(ΔFED)

⇒ar (ΔAFC) = 8 ar(ΔFED)

⇒ar (ΔFED) = (1/8) ar(ΔAFC)

Hence proved

6. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.Show that

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

**Answer
6** :

Given:

The diagonal AC and BDof the quadrilateral ABCD, intersect each other at point E.

Construction:

From A, draw AMperpendicular to BD

From C, draw CNperpendicular to BD

To Prove,

ar(ΔAED) ar(ΔBEC) = ar(ΔABE) ×ar (ΔCDE)

Proof,

ar(ΔABE) = ½×BE×AM………….. (i)

ar(ΔAED) = ½×DE×AM………….. (ii)

Dividing eq. ii by i ,we get,

ar(AED)/ar(ABE) =DE/BE…….. (iii)

Similarly,

ar(CDE)/ar(BEC) =DE/BE ……. (iv)

From eq. (iii) and(iv) , we get

ar(AED)/ar(ABE) =ar(CDE)/ar(BEC)

, ar(ΔAED)×ar(ΔBEC) =ar(ΔABE)×ar (ΔCDE)

Hence proved.

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

(i) ar (PRQ) = ½ ar (ARC)

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

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

**Answer
7** :

(i)

We know that, mediandivides the triangle into two triangles of equal area,

PC is the median ofABC.

Ar (ΔBPC) = ar (ΔAPC)……….(i)

RC is the median ofAPC.

Ar (ΔARC) = ½ ar(ΔAPC) ……….(ii)

PQ is the median ofBPC.

Ar (ΔPQC) = ½ ar(ΔBPC) ……….(iii)

From eq. (i) and(iii), we get,

ar (ΔPQC) = ½ ar(ΔAPC) ……….(iv)

From eq. (ii) and(iv), we get,

ar (ΔPQC) = ar (ΔARC)……….(v)

P and Q are themid-points of AB and BC respectively [given]

PQ||AC

and, PA = ½ AC

Since, trianglesbetween same parallel are equal in area, we get,

ar (ΔAPQ) = ar (ΔPQC)……….(vi)

From eq. (v) and (vi),we obtain,

ar (ΔAPQ) = ar (ΔARC)……….(vii)

R is the mid-point ofAP.

, RQ is the median ofAPQ.

Ar (ΔPRQ) = ½ ar(ΔAPQ) ……….(viii)

From (vii) and (viii),we get,

ar (ΔPRQ) = ½ ar (ΔARC)

Hence Proved.

(ii) PQ is the medianof ΔBPC

ar (ΔPQC) = ½ ar(ΔBPC)

= (½) ×(1/2 )ar (ΔABC)

= ¼ ar (ΔABC) ……….(ix)

Also,

ar (ΔPRC) = ½ ar(ΔAPC) [From (iv)]

ar (ΔPRC) =(1/2)×(1/2)ar ( ABC)

= ¼ ar(ΔABC) ……….(x)

Add eq. (ix) and (x),we get,

ar (ΔPQC) + ar (ΔPRC)= (1/4)×(1/4)ar (ΔABC)

ar (quad. PQCR) = ¼ ar(ΔABC) ……….(xi)

Subtracting ar (ΔPRQ)from L.H.S and R.H.S,

ar (quad. PQCR)–ar(ΔPRQ) = ½ ar (ΔABC)–ar (ΔPRQ)

ar (ΔRQC) = ½ ar(ΔABC) – ½ ar (ΔARC) [From result (i)]

ar (ΔARC) = ½ ar(ΔABC) –(1/2)×(1/2)ar (ΔAPC)

ar (ΔRQC) = ½ ar(ΔABC) –(1/4)ar (ΔAPC)

ar (ΔRQC) = ½ ar(ΔABC) –(1/4)×(1/2)ar (ΔABC) [ As, PC is median of ΔABC]

ar (ΔRQC) = ½ ar(ΔABC)–(1/8)ar (ΔABC)

ar (ΔRQC) =[(1/2)-(1/8)]ar (ΔABC)

ar (ΔRQC) = (3/8)ar(ΔABC)

(iii) ar (ΔPRQ) = ½ ar(ΔARC) [From result (i)]

2ar (ΔPRQ) = ar (ΔARC)……………..(xii)

ar (ΔPRQ) = ½ ar(ΔAPQ) [RQ is the median of APQ] ……….(xiii)

But, we know that,

ar (ΔAPQ) = ar (ΔPQC)[From the reason mentioned in eq. (vi)] ……….(xiv)

From eq. (xiii) and(xiv), we get,

ar (ΔPRQ) = ½ ar(ΔPQC) ……….(xv)

At the same time,

ar (ΔBPQ) = ar (ΔPQC)[PQ is the median of ΔBPC] ……….(xvi)

From eq. (xv) and(xvi), we get,

ar (ΔPRQ) = ½ ar(ΔBPQ) ……….(xvii)

From eq. (xii) and(xvii), we get,

2×(1/2)ar(ΔBPQ)= ar(ΔARC)

⟹ ar (ΔBPQ) = ar (ΔARC)

Hence Proved.

In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFGand ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ^DE meets BC at Y. Show that:

(i) ΔMBC ≅ ΔABD

(ii) ar(BYXD) = 2ar(MBC)

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

(iv) ΔFCB ≅ ΔACE

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

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

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

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

**Answer
8** :

(i) We know that eachangle of a square is 90°. Hence, ∠ABM = ∠DBC = 90º

∴∠ABM+∠ABC = ∠DBC+∠ABC

∴∠MBC = ∠ABD

In ∆MBC and ∆ABD,

∠MBC = ∠ABD (Proved above)

MB = AB (Sides ofsquare ABMN)

BC = BD (Sides ofsquare BCED)

∴ ∆MBC ≅ ∆ABD (SAS congruency)

(ii) We have

∆MBC ≅ ∆ABD

∴ar (∆MBC) = ar (∆ABD)… (i)

It is given that AX ⊥ DE and BD ⊥ DE (Adjacent sides of square BDEC)

∴ BD || AX (Two linesperpendicular to same line are parallel to each other)

∆ABD and parallelogramBYXD are on the same base BD and between the same parallels BD and AX.

Area (∆YXD) = 2 Area(∆MBC) [From equation (i)] … (ii)

(iii) ∆MBC andparallelogram ABMN are lying on the same base MB and between same parallels MBand NC.

2 ar (∆MBC) = ar(ABMN)

ar (∆YXD) = ar (ABMN)[From equation (ii)] … (iii)

(iv) We know that eachangle of a square is 90°.

∴∠FCA = ∠BCE = 90º

∴∠FCA+∠ACB = ∠BCE+∠ACB

∴∠FCB = ∠ACE

In ∆FCB and ∆ACE,

∠FCB = ∠ACE

FC = AC (Sides ofsquare ACFG)

CB = CE (Sides ofsquare BCED)

∆FCB ≅ ∆ACE (SAS congruency)

(v) AX ⊥ DE and CE ⊥ DE (Adjacent sides of square BDEC) [given]

Hence,

Consider BACE andparallelogram CYXE

BACE and parallelogramCYXE are on the same base CE and between the same parallels CE and AX.

∴ar (∆YXE) = 2ar (∆ACE)… (iv)

We had proved that

∴ ∆FCB ≅ ∆ACE

ar (∆FCB) ≅ ar (∆ACE) … (v)

From equations (iv)and (v), we get

ar (CYXE) = 2 ar(∆FCB) … (vi)

(vi) Consider BFCB andparallelogram ACFG

BFCB and parallelogramACFG are lying on the same base CF and between the same parallels CF and BG.

∴ar (ACFG) = 2 ar(∆FCB)

∴ar (ACFG) = ar (CYXE)[From equation (vi)] … (vii)

(vii) From the figure,we can observe that

ar (∆CED) = ar(∆YXD)+ar (CYXE)

∴ar (∆CED) = ar(ABMN)+ar (ACFG) [From equations (iii) and (vii)].

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