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

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.

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Areas Of Parallelograms And Triangles Class 9 Chapter 9 Exercise: 9.1
More About NCERT Solutions for Class 9 Maths Exercise 9.4
Benefits of NCERT Solutions for Class 9 Maths Exercise 9.4
NCERT Solutions of Class 10 Subject Wise
Subject Wise NCERT Exemplar Solutions

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.

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:

1595870426562

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
$\Rightarrow$ CD = EF
$\Rightarrow$ AB + CD = AB + EF ...........(i)

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

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

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

Q2 In Fig. $\small 9.30$ , D and E are two points on BC such that $\small BD=DE=EC$ . Show that $\small ar(ABD)=ar(ADE)=ar(AEC)$ .

1640236583768 Answer:

In $\Delta$ 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. $\small 9.31$ , ABCD, DCFE and ABFE are parallelograms. Show that $\small ar(ADE)=ar(BCF)$ .

1640236638372 Answer:

Given,
1595870559171
ABCD is a parallelogram. Therefore, AB = CD & AD = BC & AB || CD .......(i)

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

Hence proved.

Q4 In Fig. $\small 9.32$ , ABCD is a parallelogram and BC is produced to a point Q such that $\small AD=CQ$ . If AQ intersects DC at P, show that $\small ar (BPC) = ar (DPQ)$ . [ Hint : Join AC.]

1640236665039 Answer:

1595870619379
Given,
ABCD is a ||gm and AD = CQ. Join AC.
Since $\Delta$ DQC and $\Delta$ ACD lie on the same base QC and between same parallels AD and QC.
Therefore, ar( $\Delta$ DQC) = ar( $\Delta$ ACD).......(i)

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

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

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

$\small ar (BPC) = ar (DPQ)$

Hence proved.

Q5 (i) In Fig. $\small 9.33$ , ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that

$\small ar(BDE)=\frac{1}{4}ar(ABC)$

[ Hint: Join EC and AD. Show that $\small BE\parallel AC$ and $\small DE\parallel AB$ , etc.]

1640236707923 Answer:

1595870673567
Let join the CE and AD and draw $EP \perp BC$ . It is given that $\Delta$ ABC and $\Delta$ BDE is an equilateral triangle.
So, AB =BC = CA = $a$ and D i sthe midpoint of BC
therefore, $BD = \frac{a}{2}= DE = BE$
(i) Area of $\Delta$ ABC = $\frac{\sqrt{3}}{4}a^2$ and
Area of $\Delta$ BDE = $\frac{\sqrt{3}}{4}(\frac{a}{2})^2= \frac{\sqrt{3}}{16}a^2$

Hence, $\small ar(BDE)=\frac{1}{4}ar(ABC)$

Q5 (ii) In Fig. $\small 9.33$ , ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
$\small ar(BDE)=\frac{1}{2}ar(BAE)$

[ Hint: Join EC and AD. Show that $\small BE\parallel AC$ and $\small DE\parallel AB$ , etc.]

1640236728299

Answer:

1595870700069
Since $\Delta$ ABC and $\Delta$ BDE are equilateral triangles.
Therefore, $\angle$ ACB = $\angle$ DBE = $60^0$
$\Rightarrow$ BE || AC

$\Delta$ BAE and $\Delta$ BEC are on the same base BE and between same parallels BE and AC.
Therefore, ar ( $\Delta$ BAE) = ar( $\Delta$ BEC)
$\Rightarrow$ ar( $\Delta$ BAE) = 2 ar( $\Delta$ BED) [since D is the meian of $\Delta$ BEC ]
$\Rightarrow$ $\small ar(BDE)=\frac{1}{2}ar(BAE)$
Hence proved.

Q5 (iii) In Fig. $\small 9.33$ , ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
$\small ar(ABC)=2ar(BEC)$

[ Hint : Join EC and AD. Show that $\small BE\parallel AC$ and $\small DE\parallel AB$ , etc.]

1640236744718

Answer:

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

$\Rightarrow$ ar( $\Delta$ ABC) = 2. ar( $\Delta$ 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
$\small ar(BFE)=ar(AFD)$
[ Hint: Join EC and AD. Show that $\small BE\parallel AC$ and $\small DE\parallel AB$ , etc.]

1640236764354

Answer:

1595870707470
Since $\Delta$ ABC and $\Delta$ BDE are equilateral triangles.
Therefore, $\angle$ ACB = $\angle$ DBE = $60^0$
$\Rightarrow$ BE || AC

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

On subtracting $\Delta$ EFD from both sides we get

$\small ar(BFE)=ar(AFD)$

Hence proved.

Q5 (v) In Fig. $\small 9.33$ , ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
$\small ar(BFE)=2ar(FED)$

[ Hint : Join EC and AD. Show that $\small BE\parallel AC$ and $\small DE\parallel AB$ , etc.]

1640236770366

Answer:

1595870716157
In right angled triangle $\Delta$ ABD, we get

$\\AB^2 = AD^2+BD^2\\ AD^2 = AB^2-BD^2$
$=a^2 - \frac{a^2}{4}=\frac{3a^2}{4}$
$AD=\frac{\sqrt{3}a}{2}$

So, in $\Delta$ PED,
$PE^2 = DE^2-DP^2$
$\\=(\frac{a}{2})^2-(\frac{a}{4})^2\\ =\frac{3a^2}{16}$
So, $PE=\frac{\sqrt{3}a}{4}$

Therefore, the Area of $\Delta AFD =1/2 (FD)\frac{\sqrt{3}a}{2}$ ..........(i)

And, Area of triangle $\Delta EFD =1/2 (FD)\frac{\sqrt{3}a}{4}$ ...........(ii)

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

$\Rightarrow$ $\small ar(BFE)=2ar(FED)$

Q5 (vi) 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
$ar (FED) = \frac{1}{8} ar (AFC)$

[ Hint: Join EC and AD. Show that $BE\parallel AC$ and $DE\parallel AB$ , etc.]

1640236806124 Answer:

$ar(\Delta AFC) =$
$\\= ar(\Delta AFD) + ar(\Delta ADC)\\ = ar(\Delta BFE) + 1/2. ar(\Delta ABC)\\ = ar(\Delta BFE) + 1/2\times [4\times ar(\Delta BDE)]\\ = ar(\Delta BFE) + 2\times ar(\Delta BDE)$
$= 2\times ar (\Delta FED) + 2\times [ar(\Delta BFE) + ar(\Delta FED)]$ .....(from part (v) ar( $\Delta$ BFE) = 2. ar( $\Delta$ FED) ]
$\\=2ar(\Delta FED) + 2[3\times ar(\Delta FED)]\\ =8\times ar(\Delta FED)$

$ar (FED) = \frac{1}{8} ar (AFC)$
Hence proved.

Q6 Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that $ar (APB) \times ar (CPD) = ar (APD) \times ar (BPC).$

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

Answer:

1595870785823
Given that,
A quadrilateral ABCD such that it's diagonal AC and BD intersect at P. Draw $AM \perp BD$ and $CN \perp BD$

Now, ar( $\Delta$ APB) = $\frac{1}{2}\times BP\times AM$ and,
$ar(\Delta CDP)=\frac{1}{2}\times DP\times CN$
Therefore, ar( $\Delta$ APB) $\times$ ar( $\Delta$ CDP)=
$\\=(\frac{1}{2}\times BP\times AM).(\frac{1}{2}\times DP \times CN)\\ =\frac{1}{4}\times BP \times DP\times AM\times CN$ ....................(i)

Similarly, ar( $\Delta$ APD) $\times$ ar ( $\Delta$ BPC) =
$=\frac{1}{4}\times BP \times DP\times AM\times CN$ ................(ii)

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

$ar (APB) \times ar (CPD) = ar (APD) \times ar (BPC).$
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

$\small ar(PRQ)=\frac{1}{2}ar(ARC)$

Answer:

1595870842857
We have $\Delta$ 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 $\Delta$ APC,
Since R is the midpoint. So, RC is the median of the $\Delta$ APC
Therefore, ar( $\Delta$ ARC) = 1/2 . ar ( $\Delta$ APC)............(i)

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

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

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

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

From eq (i),
$\frac{1}{2}ar(\Delta ARC)= \frac{1}{4}ar(\Delta APC)$ ...........(vi)
Now, put the value of ar( $\Delta$ APC) from eq (ii), we get
Taking RHS;
$= \frac{1}{8}[ar(\Delta ABC)]$
$= \frac{1}{4}[ar(\Delta ABQ)]$ (from equation (iii))

$= \frac{1}{2}[ar(\Delta APQ)]$ (from equation (v))

$=ar(\Delta PRQ)$ (from equation (iv))

$\Rightarrow$ $\small ar(PRQ)=\frac{1}{2}ar(ARC)$

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
$\small ar(RQC)=\frac{3}{8}ar(ABC)$

Answer:

1595870850022
In $\Delta$ RBC, RQ is the median
Therefore ar( $\Delta$ RQC) = ar( $\Delta$ RBQ)
= ar (PRQ) + ar (BPQ)
= 1/8 (ar $\Delta$ ABC) + ar( $\Delta$ BPQ) [from eq (vi) & eq (A) in part (i)]
= 1/8 (ar $\Delta$ ABC) + 1/2 (ar $\Delta$ PBC) [ since PQ is the median of $\Delta$ BPC]
= 1/8 (ar $\Delta$ ABC) + (1/2).(1/2)(ar $\Delta$ ABC) [CP is the medain of $\Delta$ ABC]
= 3/8 (ar $\Delta$ 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

$ar (PBQ) = ar (ARC)$

Answer:

1595870859523

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

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

Hence proved.

Q8 (i) In Fig. $\small 9.34$ , 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 $\small AX\perp DE$ meets BC at Y. Show that: $\small \Delta MBC\cong \Delta ABD$

1640236844952

Answer:

1595870953001
We have, a $\Delta$ ABC such that BCED, ACFG and ABMN are squares on its side BC, CA and AB respec. Line segment $\small AX\perp DE$ meets BC at Y

(i) $\angle CBD = \angle MBA$ [each 90]
Adding $\angle ABC$ on both sides, we get
$\angle ABC + \angle CBD = \angle MBA + \angle ABC$
$\Rightarrow \angle ABD = \angle MBC$
In $\Delta$ ABD and $\Delta$ MBC, we have
AB = MB
BD = BC
$\angle ABD = \angle MBC$
Therefore, By SAS congruency
$\Delta$ ABD $\cong$ $\Delta$ MBC

Hence proved.

Q8 (ii) In Fig. $\small 9.34$ , 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 $\small AX\perp DE$ meets BC at Y. Show that: $\small ar(BYXD)=2ar(MBC)$

Answer:

1595870964455
SInce ||gm BYXD and $\Delta$ ABD are on the same base BD and between same parallels BD and AX
Therefore, ar( $\Delta$ ABD) = 1/2. ar(||gm BYXD)..........(i)

But, $\Delta$ ABD $\cong$ $\Delta$ MBC (proved in 1st part)
Since congruent triangles have equal areas.
Therefore, By using equation (i) we get
$\small ar(BYXD)=2ar(MBC)$
Hence proved.

Q8 (iii) In Fig. $\small 9.34$ , 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 $\small AX\perp DE$ meets BC at Y. Show that: $ar(BYXD)=ar(ABMN)$

Answer:

1595870972199
Since, ar (||gm BYXD) = 2 .ar ( $\Delta$ MBC) ..........(i) [already proved in 2nd part]
and, ar (sq. ABMN) = 2. ar ( $\Delta$ 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

$\small ar(BYXD)=ar(ABMN)$

Q8 (iv) In Fig. $\small 9.34$ , 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 $\small AX\perp DE$ meets BC at Y. Show that: $\small \Delta FCB\cong \Delta ACE$

Answer:

1595870980947
$\angle FCA = \angle BCE$ [both 90]
By adding $\angle$ ABC on both sides we get
$\angle$ ABC + $\angle$ FCA = $\angle$ ABC + $\angle$ BCE
$\angle$ FCB = $\angle$ ACE

In $\Delta$ FCB and $\Delta$ ACE
FC = AC [sides of square]
BC = AC [sides of square]
$\angle$ FCB = $\angle$ ACE
$\Rightarrow$ $\Delta$ FCB $\cong$ $\Delta$ ACE

Hence proved.

Q8 (v) In Fig. $\small 9.34$ , 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 $\small AX\perp DE$ meets BC at Y. Show that: $\small ar(CYXE)=2ar(FCB)$

Answer:

1595870988524
Since ||gm CYXE and $\Delta$ ACE lie on the same base CE and between the same parallels CE and AX.
Therefore, 2. ar( $\Delta$ ACE) = ar (||gm CYXE)
B ut, $\Delta$ FCB $\cong$ $\Delta$ ACE (in iv part)
Since the congruent triangle has equal areas. So,
$\small ar(CYXE)=2ar(FCB)$
Hence proved.

Q8 (vi) In Fig. $\small 9.34$ , 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 $\small AX\perp DE$ meets BC at Y. Show that: $\small ar(CYXE)=ar(ACFG)$

Answer:

1595871000670
Since ar( ||gm CYXE) = 2. ar( $\Delta$ ACE) {in part (v)}...................(i)
Also, $\Delta$ FCB and quadrilateral ACFG lie on the same base FC and between the same parallels FC and BG.
Therefore, ar (quad. ACFG) = 2.ar( $\Delta$ FCB ) ................(ii)

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

$\small ar(CYXE)=ar(ACFG)$
Hence proved.

Q8 (vii) In Fig. $\small 9.34$ , 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 $\small AX\perp DE$ meets BC at Y. Show that: $\small ar(BCED)=ar(ABMN)+ar(ACFG)$

Answer:

1595871008289
We have,
ar(quad. BCDE) = ar(quad, CYXE) + ar(quad. BYXD)

= ar(quad, CYXE) + ar (quad. ABMN) [already proved in part (iii)]
Thus, $\small ar(BCED)=ar(ABMN)+ar(ACFG)$

Hence proved.

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.

Also, See

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

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.

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

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

More About NCERT Solutions for Class 9 Maths Exercise 9.4

Benefits of NCERT Solutions for Class 9 Maths Exercise 9.4

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

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Upcoming School Exams

Himachal Pradesh Board of Secondary Education 10th Examination

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National Institute of Open Schooling 12th Examination

National Institute of Open Schooling 10th examination

National Means Cum-Merit Scholarship

Popular Questions

Colleges After 12th

Popular Course After 12th