Course Content
Class 9th Science
0/30
Class 9th Math
0/81
Class 9 Social Science History: India and the Contemporary World â€“ I
0/16
Class 9 Social Science Geography: Contemporary India â€“ I
0/12
Class 9 Social Science Civics (Political Science): Democratic Politics â€“ I
0/12
Class 9 Social Science Economics: Understanding Economic Development â€“ I
0/8
Class 9 English Beehive
0/22
Class 9 English Beehive Poem
0/19
Class 9 English Moments
0/20
Online Class For 9th Standard Students (CBSE) (English Medium)

### Exercise 9.4(Optional)* Page: 164

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.

Solution:

Given,

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

To prove,

Perimeter of || gm ABCD is greater than the perimeter of rectangle ABEF.

Proof,

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

, AB = DC [As ABCD is a || gm]

and, AB = EF [As ABEF is a rectangle]

, DC = EF â€¦ (i)

Adding AB on both sides, we get,

â‡’AB + DC = AB + EF â€¦ (ii)

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

, BE < BC and AF < AD

â‡’ BC > BE and AD > AF

â‡’ BC+AD > BE+AF â€¦ (iii)

Adding (ii) and (iii), we get

â‡’ AB+BC+CD+DA > AB+ BE+EF+FA

â‡’ perimeter of || gm ABCD > perimeter of rectangle ABEF.

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

Hence Proved.

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 â€˜Introductionâ€™ 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 intoÂ nÂ equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide DABC intoÂ nÂ triangles of equal areas.]

Solution:

Given,

BD = DE = EC

To prove,

ar (â–³ABD) = ar (â–³ADE) = ar (â–³AEC)

Proof,

In (â–³ABE), AD is median [since, BD = DE, given]

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

, ar(â–³ABD) = ar(â–³AED) â€”(i)

Similarly,

In (â–³ADC), AE is median [since, DE = EC, given]

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

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

Solution:

Given,

ABCD, DCFE and ABFE are parallelograms

To prove,

Proof,

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

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

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

, â–³ADE â‰… â–³BCF [Using SSS Congruence theorem]

, ar(â–³ADE) = ar(â–³BCF) [ By CPCT]

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.]

Solution:

Given:

ABCD is a parallelogram

To prove:

ar (â–³BPC) = ar (â–³DPQ)

Proof:

âˆ APD = âˆ QPC [Vertically Opposite Angles]

âˆ ADP = âˆ QCP [Alternate Angles]

, â–³ABO â‰… â–³ACD [AAS congruency]

, DP = CP [CPCT]

In â–³CDQ, QP is median. [Since, DP = CP]

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

, ar(â–³DPQ) = ar(â–³QPC) â€”(i)

In â–³PBQ, PC is median. [Since, AD = CQ and AD = BC â‡’ BC = QC]

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

, ar(â–³QPC) = ar(â–³BPC) â€”(ii)

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

ar(â–³BPC) = ar(â–³DPQ)

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) =1/4 ar (ABC)

(ii) ar (BDE) = Â½ 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)

Solution:

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

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

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

âˆ´ GH =1/2 BC and GH || BD

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

Similarly,

GD = HC = HA

HD = AG = BG

, Î”ABC is divided into 4 equal equilateral triangles Î”BGD, Î”AGH, Î”DHC and Î”GHD

We can say that,

Î”BGD = Â¼ Î”ABC

Considering, Î”BDG and Î”BDE

BD = BD (Common base)

Since both triangles are equilateral triangle, we can say that,

BG = BE

DG = DE

, Î”BDG Î”BDE [By SSS congruency]

, area (Î”BDG) = area (Î”BDE)

ar (Î”BDE) = Â¼ ar (Î”ABC)

Hence proved

(ii)

ar(Î”BDE) = ar(Î”AED) (Common base DE and DE||AB)

ar(Î”BDE)âˆ’ar(Î”FED) = ar(Î”AED)âˆ’ar (Î”FED)

ar(Î”BEF) = ar(Î”AFD) â€¦(i)

Now,

ar(Î”ABD) = ar(Î”ABF)+ar(Î”AFD)

ar(Î”ABD) = ar(Î”ABF)+ar(Î”BEF) [From equation (i)]

ar(Î”ABD) = ar(Î”ABE) â€¦(ii)

AD is the median of Î”ABC.

ar(Î”ABD) = Â½ ar (Î”ABC)

= (4/2) ar (Î”BDE)

= 2 ar (Î”BDE)â€¦(iii)

From (ii) and (iii), we obtain

2 ar (Î”BDE) = ar (Î”ABE)

ar (BDE) = Â½ ar (BAE)

Hence proved

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

ar(Î”ABF) + ar(Î”BEF) = ar(Î”BEC)

From eqnÂ (i), we get,

ar(Î”ABF) + ar(Î”AFD) = ar(Î”BEC)

ar(Î”ABD) = ar(Î”BEC)

Â½ ar(Î”ABC) = ar(Î”BEC)

ar(Î”ABC) = 2 ar(Î”BEC)

Hence proved

(iv) Î”BDE and Î”AED lie on the same base (DE) and are in-between the parallel lines DE and AB.

âˆ´ar (Î”BDE) = ar (Î”AED)

Subtracting ar(Î”FED) from L.H.S and R.H.S,

We get,

âˆ´ar (Î”BDE)âˆ’ar (Î”FED) = ar (Î”AED)âˆ’ar (Î”FED)

âˆ´ar (Î”BFE) = ar(Î”AFD)

Hence proved

(v) Assume that h is the height of vertex E, corresponding to the side BD in Î”BDE.

Also assume that H is the height of vertex A, corresponding to the side BC in Î”ABC.

While solving Question (i),

We saw that,

ar (Î”BDE) = Â¼ ar (Î”ABC)

While solving Question (iv),

We saw that,

ar (Î”BFE) = ar (Î”AFD).

âˆ´ar (Î”BFE) = ar (Î”AFD)

= 2 ar (Î”FED)

Hence, ar (Î”BFE) = 2 ar (Î”FED)

Hence proved

(vi) ar (Î”AFC) = ar (Î”AFD) + ar(Î”ADC)

= 2 ar (Î”FED) + (1/2) ar(Î”ABC) [using (v)

= 2 ar (Î”FED) + Â½ [4ar(Î”BDE)] [Using result of Question (i)]

= 2 ar (Î”FED) +2 ar(Î”BDE)

Since, Î”BDE and Î”AED are on the same base and between same parallels

= 2 ar (Î”FED) +2 ar (Î”AED)

= 2 ar (Î”FED) +2 [ar (Î”AFD) +ar (Î”FED)]

= 2 ar (Î”FED) +2 ar (Î”AFD) +2 ar (Î”FED) [From question (viii)]

= 4 ar (Î”FED) +4 ar (Î”FED)

â‡’ar (Î”AFC) = 8 ar (Î”FED)

â‡’ar (Î”FED) = (1/8) ar (Î”AFC)

Hence proved

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

ar (APB)Ã—ar (CPD) = ar (APD)Ã—ar (BPC).

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

Solution:

Given:

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

Construction:

From A, draw AM perpendicular to BD

From C, draw CN perpendicular to BD

To Prove,

ar(Î”AED) ar(Î”BEC) = ar (Î”ABE) Ã—ar (Î”CDE)

Proof,

ar(Î”ABE) = Â½ Ã—BEÃ—AMâ€¦â€¦â€¦â€¦.. (i)

ar(Î”AED) = Â½ Ã—DEÃ—AMâ€¦â€¦â€¦â€¦.. (ii)

Dividing eq. ii by i , we get,

ar(AED)/ar(ABE) = DE/BEâ€¦â€¦.. (iii)

Similarly,

ar(CDE)/ar(BEC) = DE/BE â€¦â€¦. (iv)

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

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

, ar(Î”AED)Ã—ar(Î”BEC) = ar(Î”ABE)Ã—ar (Î”CDE)

Hence proved.

7.Â 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) = Â½ ar (ARC)

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

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

Solution:

(i)

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

PC is the median of ABC.

Ar (Î”BPC) = ar (Î”APC) â€¦â€¦â€¦.(i)

RC is the median of APC.

Ar (Î”ARC) = Â½ ar (Î”APC) â€¦â€¦â€¦.(ii)

PQ is the median of BPC.

Ar (Î”PQC) = Â½ ar (Î”BPC) â€¦â€¦â€¦.(iii)

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

ar (Î”PQC) = Â½ ar (Î”APC) â€¦â€¦â€¦.(iv)

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

ar (Î”PQC) = ar (Î”ARC) â€¦â€¦â€¦.(v)

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

PQ||AC

and, PA = Â½ AC

Since, triangles between same parallel are equal in area, we get,

ar (Î”APQ) = ar (Î”PQC) â€¦â€¦â€¦.(vi)

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

ar (Î”APQ) = ar (Î”ARC) â€¦â€¦â€¦.(vii)

R is the mid-point of AP.

, RQ is the median of APQ.

Ar (Î”PRQ) = Â½ ar (Î”APQ) â€¦â€¦â€¦.(viii)

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

ar (Î”PRQ) = Â½ ar (Î”ARC)

Hence Proved.

(ii) PQ is the median of Î”BPC

ar (Î”PQC) = Â½ ar (Î”BPC)

= (Â½) Ã—(1/2 )ar (Î”ABC)

= Â¼ ar (Î”ABC) â€¦â€¦â€¦.(ix)

Also,

ar (Î”PRC) = Â½ ar (Î”APC) [From (iv)]

ar (Î”PRC) = (1/2)Ã—(1/2)ar ( ABC)

= Â¼ ar(Î”ABC) â€¦â€¦â€¦.(x)

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

ar (Î”PQC) + ar (Î”PRC) = (1/4)Ã—(1/4)ar (Î”ABC)

ar (quad. PQCR) = Â¼ ar (Î”ABC) â€¦â€¦â€¦.(xi)

Subtracting ar (Î”PRQ) from L.H.S and R.H.S,

ar (quad. PQCR)â€“ar (Î”PRQ) = Â½ ar (Î”ABC)â€“ar (Î”PRQ)

ar (Î”RQC) = Â½ ar (Î”ABC) â€“ Â½ ar (Î”ARC) [From result (i)]

ar (Î”ARC) = Â½ ar (Î”ABC) â€“(1/2)Ã—(1/2)ar (Î”APC)

ar (Î”RQC) = Â½ ar (Î”ABC) â€“(1/4)ar (Î”APC)

ar (Î”RQC) = Â½ ar (Î”ABC) â€“(1/4)Ã—(1/2)ar (Î”ABC) [ As, PC is median of Î”ABC]

ar (Î”RQC) = Â½ ar (Î”ABC)â€“(1/8)ar (Î”ABC)

ar (Î”RQC) = [(1/2)-(1/8)]ar (Î”ABC)

ar (Î”RQC) = (3/8)ar (Î”ABC)

(iii) ar (Î”PRQ) = Â½ ar (Î”ARC) [From result (i)]

2ar (Î”PRQ) = ar (Î”ARC) â€¦â€¦â€¦â€¦â€¦..(xii)

ar (Î”PRQ) = Â½ ar (Î”APQ) [RQ is the median of APQ] â€¦â€¦â€¦.(xiii)

But, we know that,

ar (Î”APQ) = ar (Î”PQC) [From the reason mentioned in eq. (vi)] â€¦â€¦â€¦.(xiv)

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

ar (Î”PRQ) = Â½ ar (Î”PQC) â€¦â€¦â€¦.(xv)

At the same time,

ar (Î”BPQ) = ar (Î”PQC) [PQ is the median of Î”BPC] â€¦â€¦â€¦.(xvi)

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

ar (Î”PRQ) = Â½ ar (Î”BPQ) â€¦â€¦â€¦.(xvii)

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

2Ã—(1/2)ar(Î”BPQ)= ar (Î”ARC)

âŸ¹ ar (Î”BPQ) = ar (Î”ARC)

Hence Proved.

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) Î”MBCÂ â‰…Â Î”ABD

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

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

(iv) Î”FCBÂ â‰…Â Î”ACE

(v) ar(CYXE) = 2ar(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.

Solution:

(i) We know that each angle of a square is 90Â°. Hence, âˆ ABM = âˆ DBC = 90Âº

âˆ´âˆ ABM+âˆ ABC = âˆ DBC+âˆ ABC

âˆ´âˆ MBC = âˆ ABD

In âˆ†MBC and âˆ†ABD,

âˆ MBC = âˆ ABD (Proved above)

MB = AB (Sides of square ABMN)

BC = BD (Sides of square BCED)

âˆ´ âˆ†MBC â‰… âˆ†ABD (SAS congruency)

(ii) We have

âˆ†MBC â‰… âˆ†ABD

âˆ´ar (âˆ†MBC) = ar (âˆ†ABD) â€¦ (i)

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

âˆ´ BD || AX (Two lines perpendicular to same line are parallel to each other)

âˆ†ABD and parallelogram BYXD are on the same base BD and between the same parallels BD and AX.

Area (âˆ†YXD) = 2 Area (âˆ†MBC) [From equation (i)] â€¦ (ii)

(iii) âˆ†MBC and parallelogram ABMN are lying on the same base MB and between same parallels MB and NC.

2 ar (âˆ†MBC) = ar (ABMN)

ar (âˆ†YXD) = ar (ABMN) [From equation (ii)] â€¦ (iii)

(iv) We know that each angle of a square is 90Â°.

âˆ´âˆ FCA = âˆ BCE = 90Âº

âˆ´âˆ FCA+âˆ ACB = âˆ BCE+âˆ ACB

âˆ´âˆ FCB = âˆ ACE

In âˆ†FCB and âˆ†ACE,

âˆ FCB = âˆ ACE

FC = AC (Sides of square ACFG)

CB = CE (Sides of square BCED)

âˆ†FCB â‰… âˆ†ACE (SAS congruency)

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

Hence,

CE || AX (Two lines perpendicular to the same line are parallel to each other)

Consider BACE and parallelogram CYXE

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

âˆ´ar (âˆ†YXE) = 2ar (âˆ†ACE) â€¦ (iv)

âˆ´ âˆ†FCB â‰… âˆ†ACE

ar (âˆ†FCB) â‰… ar (âˆ†ACE) â€¦ (v)

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

ar (CYXE) = 2 ar (âˆ†FCB) â€¦ (vi)

(vi) Consider BFCB and parallelogram ACFG

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

âˆ´ar (ACFG) = 2 ar (âˆ†FCB)

âˆ´ar (ACFG) = ar (CYXE) [From equation (vi)] â€¦ (vii)

(vii) From the figure, we can observe that

ar (âˆ†CED) = ar (âˆ†YXD)+ar (CYXE)

âˆ´ar (âˆ†CED) = ar (ABMN)+ar (ACFG) [From equations (iii) and (vii)].

Â

### Extra Questions for Class 9 Maths Chapter 9

1. 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).
2. PQRS and ABRS are parallelograms and X is any point on side BR. Show that:
• ar(PQRS) = ar(ABRS)
• ar (AXS) = Â½ ar(PQRS)
3. 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?
4. In Figure, E is any point on median AD of a Î”ABC. Show that ar (ABE) =
1. ar(ACE).
2. Show that the diagonals of a parallelogram divide it into four triangles of equal area.
3. D and E are points on sides AB and AC respectively of Î”ABC such that ar(DBC) = ar(EBC). Prove that DE || BC.
4. 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 in the figure). Show that:
• ar(ABCD) = ar(PBQR).
• [Hint: Join AC and PQ. Now compare ar(ACQ) and ar(APQ).]Â
1. Diagonals AC and BD of aÂ trapeziumÂ ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).