Exercise 9.4, which can be found in page 164 of your NCERT textbook, is an optional one and is mainly given to assess your understanding of the subject and the concepts that are discussed so far in the textbook. NCERT Solutions for Class 9 Maths Chapter 9 Area of Parallelograms and Triangles Exercise 9.4 gives a clear and step by step detailed solution.

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### Access Answers to NCERT Class 9 Maths Chapter 9-Area of Parallelograms and Triangles Exercise 9.4

**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

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

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 into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide DABC into n 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]

,ar(ADE) = ar(AEC) â€”(ii)

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

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

**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,

ar (â–³ADE) = ar (â–³BCF)

Proof,

In â–³ADE and â–³BCF,

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

AD = CQ

To prove:

ar (â–³BPC) = ar (â–³DPQ)

Proof:

In â–³ADP and â–³QCP,

âˆ APD = âˆ QPC [Vertically Opposite Angles]

âˆ ADP = âˆ QCP [Alternate Angles]

AD = CQ [given]

, 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 eq^{n} (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.** **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”MBC â‰… Î”ABD**

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

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

**(iv) Î”FCB 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)

We had proved that

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)].

This will also help you to understand this topic. Summary of the concepts discussed in the NCERT Solution for Class 9 Maths Chapter 9 Area of Parallelograms and Triangles are as follow: 1. Area of a figure is a number (in some units) associated with the part of the plane enclosed by that figure. 2. Two congruent figures have equal areas but the converse need not be true. 3. If a planar region formed by a figure T is made up of two non-overlapping planar regions formed by figures P and Q, then are (T) = ar (P) + ar (Q), where (X) denotes the area of figure X. 4. Two figures are said to be on the same base and between the same parallels, if they have a common base (side) and the vertices, (or the vertex) opposite to the common base of each figure lie on a line parallel to the base. 5. Parallelograms on the same base (or equal bases) and between the same parallels are equal in area. 6.

Area of a parallelogram is the product of its base and the corresponding altitude. 7. Parallelograms on the same base (or equal bases) and having equal areas lie between the same parallels. 8. If a parallelogram and a triangle are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram. 9. Triangles on the same base (or equal bases) and between the same parallels are equal in area. 10. Area of a triangle is half the product of its base and the corresponding altitude. 11. Triangles on the same base (or equal bases) and having equal areas lie between the same parallels. 12. A median of a triangle divides it into two triangles of equal areas.

Even if this is given as an optional exercise, this exercise 9.4 from chapter 9 of NCERT Solutions for Class 9 Maths is important as it contains various questions covering the important concepts of Chapter 9. The exercise contains 4 Short Type Answer, 1 Long Type Answer and 3 Very Long Type Answers. Solving these exercises from NCERTSolutions for Class 9 will help you:

- To learn the topic very well
- Get practice and gain confidence
- Understand the concepts more clearly
- Self-assess what they know about the concept