##### VMC VIQ Scholarship Test

ApplyRegister for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.

Edited By Safeer PP | Updated on Jul 29, 2022 06:22 PM IST

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.

**Scholarship Test:** **Vidyamandir Intellect Quest (VIQ)**

**Don't Miss: ****JEE Main 2027: Narayana Scholarship Test Preparation Kit for Class 9**

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.

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

** 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.

**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.

[ ** 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,

[ ** 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.

[ ** 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.

**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.

[ ** 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)

[ ** 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.

** 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.

** 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.

** 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.

** 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.

** 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.

** 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

** 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.

** 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.

** 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.

** 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.

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

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.

**Also, See**

1. What is the total fundamental concept of NCERT solutions for Class 9 Maths exercise 9.4?

This exercise deals with all the difficult questions regarding the area of triangles and parallelograms and their properties.

2. If two figures A and B are congruent to each other then what can you describe between the relation of their areas?

If figure A and figure B are congruent to each other then the relation between their areas is that they are equal to each other.

Area of (A) = Area of (B)

3. What do you mean by the term “two figures lies on the same base”?

The term “two figures on the same base” means that two figures which are said to be on the same base have one common side as their base or their commo side (base length) is equal to one other.

4. Find the area of a triangle with base as 6cm and height as 4cm?

The term “two figures on the same base” means that two figures which are said to be on the same base have one common side as their base or their commo side (base length) is equal to one other.

5. Can we consider rectangle and a square both as a parallelogram?

Yes, we can consider a rectangle and a square as a parallelogram as both of its opposite sides are parallel to each other.

6. Find the area of a parallelogram with base as 12cm and height as 10cm?

Yes, we can consider a rectangle and a square as a parallelogram as both of its opposite sides are parallel to each other.

7. Name the different types of parallelograms?

The different types of parallelograms are:

Square

Rectangle

Rhombus and

Parallelogram

8. What do you mean by figures on a similar base and between same parallels ?

Two figures are supposed to be on a same base and between same parallels if the two of them have one common (normal) side as their base and the vertices which are inverse to the normal base lie on a line parallel to the base.

Nov 08, 2024

Nov 08, 2024

Application Date:09 September,2024 - 14 November,2024

Application Date:09 September,2024 - 14 November,2024

Application Date:01 October,2024 - 14 November,2024

Admit Card Date:04 October,2024 - 29 November,2024

Admit Card Date:04 October,2024 - 29 November,2024

Get answers from students and experts

Register for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.

As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters

As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters

Accepted by more than 11,000 universities in over 150 countries worldwide

Register now for PTE & Unlock 20% OFF : Use promo code: 'C360SPL20'. Valid till 30th NOV'24! Trusted by 3,500+ universities globally

As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE

News and Notifications

Back to top