NCERT Solutions for Exercise 9.4 Class 9 Maths Chapter 9 - Areas Of Parallelograms And Triangles

NCERT Solutions for Exercise 9.4 Class 9 Maths Chapter 9 - Areas Of Parallelograms And Triangles

The NCERT Solutions for Class 9 Maths exercise 9.4 is an optional (not from the examination point of view) exercise which comprises of theorems related to the previous exercises which include concepts of parallelograms and triangle on the same base and between the same parallels, having questions related to areas of figure which includes the medians for triangles.

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Few important concepts related to NCERT syllabus Class 9 Maths chapter 9 exercise 9.4 in order to solve this exercise are:

  • Parallelograms having the same base and are in between the same parallels have the same area, as because the height (distance between the two parallel lines) continues as before for the two parallelograms and they have same bases subsequently their regions are equivalent.

  • Triangle that has a same base or the equivalent base and are between same pair of parallel lines have equivalent area. This is on the grounds that we realize that the area of the triangle is a half of the result of base and height. Since the height (distance between the parallels) continues as before for the two triangles and they have same bases subsequently their area are equivalent.

Major questions in this exercise use a blend of at least two or more concepts.

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Along with NCERT book Class 9 Maths chapter 9 exercise 9.4 the following exercises are also present.

  • Areas Of Parallelograms And Triangles Exercise 9.1

  • Areas Of Parallelograms And Triangles Exercise 9.2

  • Areas Of Parallelograms And Triangles Exercise 9.3

Areas Of Parallelograms And Triangles Class 9 Chapter 9 Exercise: 9.1

Q1 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:



We have ||gm ABCD and a rectangle ABEF both lie on the same base AB such that, ar(||gm ABCD) = ar(ABEF)
for rectangle, AB = EF
and for ||gm AB = CD
CD = EF
AB + CD = AB + EF ...........(i)

SInce BEC and AFD are right angled triangle
Therefore, AD > AF and BC > BE
(BC + AD ) > (AF + BE)...........(ii)

Adding equation (i) and (ii), we get

(AB + CD)+(BC + AD) > (AB + BE) + (AF + BE)
(AB + BC + CD + DA) > (AB + BE + EF + FA)
Hence proved, perimeter of ||gm ABCD is greater than perimeter of rectangle ABEF.

Q2 In Fig. , D and E are two points on BC such that . Show that .

Answer:

In ABC, D and E are two points on BC such that BD = DE = EC

AD is the median of ABE, therefore,

area of ABD= area of AED...................(1)

AE is the median of ACD, therefore,

area of AEC= area of AED...................(2)

From (1) and (2)

area of ABD=area of AED= area of AEC
Hence proved.

Q3 In Fig. , ABCD, DCFE and ABFE are parallelograms. Show that .

Answer:

Given,

ABCD is a parallelogram. Therefore, AB = CD & AD = BC & AB || CD .......(i)

Now, ADE and BCF are on the same base AD = BC and between same parallels AB and EF.
Therefore, ar ( ADE) = ar( BCF)

Hence proved.

Q4 In Fig. , ABCD is a parallelogram and BC is produced to a point Q such that . If AQ intersects DC at P, show that . [ Hint : Join AC.]

Answer:


Given,
ABCD is a ||gm and AD = CQ. Join AC.
Since DQC and ACD lie on the same base QC and between same parallels AD and QC.
Therefore, ar( DQC) = ar( ACD).......(i)

Subtracting ar( PQC) from both sides in eq (i), we get
ar( DPQ) = ar( PAC).............(i)

Since PAC and PBC are on the same base PC and between same parallel PC and AB.
Therefore, ar( PAC) = ar( PBC)..............(iii)

From equation (ii) and eq (ii), we get

Hence proved.

Q5 (i) In Fig. , ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that

[ Hint: Join EC and AD. Show that and , etc.]

Answer:


Let join the CE and AD and draw . It is given that ABC and BDE is an equilateral triangle.
So, AB =BC = CA = and D i sthe midpoint of BC
therefore,
(i) Area of ABC = and
Area of BDE =

Hence,


Q5 (ii) In Fig. , ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that

[ Hint: Join EC and AD. Show that and , etc.]


Answer:


Since ABC and BDE are equilateral triangles.
Therefore, ACB = DBE =
BE || AC

BAE and BEC are on the same base BE and between same parallels BE and AC.
Therefore, ar ( BAE) = ar( BEC)
ar( BAE) = 2 ar( BED) [since D is the meian of BEC ]

Hence proved.

Q5 (iii) In Fig. , ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that

[ Hint : Join EC and AD. Show that and , etc.]


Answer:

We already proved that,
ar( ABC) = 4.ar( BDE) (in part 1)
and, ar( BEC) = 2. ar( BDE) (in part ii )

ar( ABC) = 2. ar( BEC)

Hence proved.

Q5 (iv) 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

[ Hint: Join EC and AD. Show that and , etc.]


Answer:


Since ABC and BDE are equilateral triangles.
Therefore, ACB = DBE =
BE || AC

BDE and AED are on the same base ED and between same parallels AB and DE.
Therefore, ar( BED) = ar( AED)

On subtracting EFD from both sides we get

Hence proved.

Q5 (v) In Fig. , ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that

[ Hint : Join EC and AD. Show that and , etc.]


Answer:


In right angled triangle ABD, we get



So, in PED,


So,

Therefore, the Area of ..........(i)

And, Area of triangle ...........(ii)

From eq (i) and eq (ii), we get
ar( AFD) = 2. ar( EFD)
Since ar( AFD) = ar( BEF)


Q5 (vi) In Fig. , ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that

[ Hint: Join EC and AD. Show that and , etc.]

Answer:



.....(from part (v) ar( BFE) = 2. ar( FED) ]


Hence proved.

Q6 Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that

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

Answer:


Given that,
A quadrilateral ABCD such that it's diagonal AC and BD intersect at P. Draw and

Now, ar( APB) = and,

Therefore, ar( APB) ar( CDP)=
....................(i)

Similarly, ar( APD) ar ( BPC) =
................(ii)

From eq (i) and eq (ii), we get


Hence proved.

Q7 (i) 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


Answer:


We have ABC such that P, Q and R are the midpoints of the side AB, BC and AP respectively. Join PQ, QR, AQ, PC and RC as shown in the figure.
Now, in APC,
Since R is the midpoint. So, RC is the median of the APC
Therefore, ar( ARC) = 1/2 . ar ( APC)............(i)

Also, in ABC, P is the midpoint. Thus CP is the median.
Therefore, ar( APC) = 1/2. ar ( ABC)............(ii)

Also, AQ is the median of ABC
Therefore, 1/2. ar ( ABC) = ar (ABQ)............(iii)

In APQ, RQ is the median.
Therefore, ar ( PRQ) = 1/2 .ar ( APQ).............(iv)

In ABQ, PQ is the median
Therefore, ar( APQ) = 1/2. ar( ABQ).........(v)


From eq (i),
...........(vi)
Now, put the value of ar( APC) from eq (ii), we get
Taking RHS;

(from equation (iii))

(from equation (v))

(from equation (iv))

Hence proved.

Q7 (ii) 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

Answer:


In RBC, RQ is the median
Therefore ar( RQC) = ar( RBQ)
= ar (PRQ) + ar (BPQ)
= 1/8 (ar ABC) + ar( BPQ) [from eq (vi) & eq (A) in part (i)]
= 1/8 (ar ABC) + 1/2 (ar PBC) [ since PQ is the median of BPC]
= 1/8 (ar ABC) + (1/2).(1/2)(ar ABC) [CP is the medain of ABC]
= 3/8 (ar ABC)

Hence proved.

Q7 (iii) 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

Answer:

QP is the median of ABQ
Therefore, ar( PBQ) = 1/2. (ar ABQ)

= (1/2). (1/2) (ar ABC) [since AQ is the median of ABC
= 1/4 (ar ABC)
= ar ( ARC) [from eq (A) of part (i)]

Hence proved.

Q8 (i) In Fig. , 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 meets BC at Y. Show that:

Answer:


We have, a ABC such that BCED, ACFG and ABMN are squares on its side BC, CA and AB respec. Line segment meets BC at Y

(i) [each 90]
Adding on both sides, we get


In ABD and MBC, we have
AB = MB
BD = BC

Therefore, By SAS congruency
ABD MBC

Hence proved.

Q8 (ii) In Fig. , 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 meets BC at Y. Show that:

Answer:


SInce ||gm BYXD and ABD are on the same base BD and between same parallels BD and AX
Therefore, ar( ABD) = 1/2. ar(||gm BYXD)..........(i)

But, ABD MBC (proved in 1st part)
Since congruent triangles have equal areas.
Therefore, By using equation (i) we get

Hence proved.

Q8 (iii) In Fig. , 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 meets BC at Y. Show that:

Answer:


Since, ar (||gm BYXD) = 2 .ar ( MBC) ..........(i) [already proved in 2nd part]
and, ar (sq. ABMN) = 2. ar ( MBC)............(ii)
[Since ABMN and AMBC are on the same base MB and between same parallels MB and NC]
From eq(i) and eq (ii), we get

Q8 (iv) In Fig. , 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 meets BC at Y. Show that:

Answer:


[both 90]
By adding ABC on both sides we get
ABC + FCA = ABC + BCE
FCB = ACE

In FCB and ACE
FC = AC [sides of square]
BC = AC [sides of square]
FCB = ACE
FCB ACE

Hence proved.

Q8 (v) In Fig. , 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 meets BC at Y. Show that:

Answer:


Since ||gm CYXE and ACE lie on the same base CE and between the same parallels CE and AX.
Therefore, 2. ar( ACE) = ar (||gm CYXE)
B
ut, FCB ACE (in iv part)
Since the congruent triangle has equal areas. So,

Hence proved.

Q8 (vi) In Fig. , 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 meets BC at Y. Show that:

Answer:


Since ar( ||gm CYXE) = 2. ar( ACE) {in part (v)}...................(i)
Also, FCB and quadrilateral ACFG lie on the same base FC and between the same parallels FC and BG.
Therefore, ar (quad. ACFG) = 2.ar( FCB ) ................(ii)

From eq (i) and eq (ii), we get


Hence proved.

Q8 (vii) In Fig. , 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 meets BC at Y. Show that:

Answer:


We have,
ar(quad. BCDE) = ar(quad, CYXE) + ar(quad. BYXD)

= ar(quad, CYXE) + ar (quad. ABMN) [already proved in part (iii)]
Thus,

Hence proved.

More About NCERT Solutions for Class 9 Maths Exercise 9.4

NCERT solutions Class 9 Maths exercise 9.4 includes some important concepts from the previous exercises that may prove to be crucial in order to solve some miscellanies problems from NCERT solutions Class 9 Maths exercise 9.4 we get to know another idea which manages the opposite of the above theorem. We get to realize that in case two triangles or parallelograms have a similar base (or on the other hand in case they have an equivalent base) and in case they have the equivalent area then they lie on the same parallel line.

Apart from all the above-mentioned concepts, the vice versa of these concepts (theorem) are very critical in solving some questions.

Also Read| Areas Of Parallelograms And Triangles Class 9 Notes

Benefits of NCERT Solutions for Class 9 Maths Exercise 9.4

  • Exercise 9.4 Class 9 Maths, is based on AREAS OF PARALLELOGRAMS AND TRIANGLES and most of its crucial properties.

  • From Class 9 Maths chapter 9 exercise 9.4 we get to revise the entire concepts of this chapter in a single exercise.

  • Understanding the concepts from Class 9 Maths chapter 9 exercise 9.4 will make the concepts and questions from higher standards (like Class10) easier for us.

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