NCERT SOLUTIONS FOR CLASS 9 MATHS CHAPTER 9 AREAS OF PARALLELOGRAMS AND TRIANGLES

The National Council of Educational Research and Training (NCERT) is an autonomous body of the Indian government that formulates the curricula for schools in India that are governed by the Central Board of Secondary Education (CBSE) and certain state boards. Therefore, students who will be taking the Class 10 tests administered by various boards should consult this NCERT Syllabus in order to prepare for those examinations, which in turn will assist those students get a passing score.

When working through the exercises in the NCERT textbook, if you run into any type of difficulty or uncertainty, you may use the swc NCERT Solutions for class 9 as a point of reference. While you are reading the theory form textbook, it is imperative that you always have notes prepared. You should make an effort to understand things from the very beginning so that you may create a solid foundation in the topic. Use the NCERT as your parent book to ensure that you have a strong foundation. After you have finished reading the theoretical section of the textbook, you should go to additional reference books.

NCERT Solutions for Class 9 Maths Exercise 9.1

Question 1. Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.

https://media.mycbseguide.com/images/static/ncert/09/maths/ch09/Ex9.1/image001.jpg

Solution:
In figure (i): NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image002.pngDPC and trap. ABCD are on the same base DC and between same parallel DC and AB.

In figure (iii): NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image002.pngRTQ and parallelogram PQRS are on the same base QR and between same parallel QR and PS.

In figure (v): Parallelogram ABCD and parallelogram APQD are on the same base AD and between the same parallels AD and BQ.

NCERT Solutions for Class 9 Maths Exercise 9.2

Question 1. In figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/Q1
Solution:
BSOWe have, AE ⊥ DC and AB = 16 cm
∵ AB = CD [Opposite sides of parallelogram]
∴ CD = 16 cm
Now, area of parallelogram ABCD = CD x AE
= (16 x 8) cm2 = 128 cm2 [∵ AE = 8 cm]
Since, CF ⊥ AD
∴ Area of parallelogram ABCD = AD x CF
⇒ AD x CF = 128 cm
⇒ AD x 10 cm = 128 cm2 [∵ CF= 10 cm]
⇒ AD = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Trianglescm = 12.8 cm 10
Thus, the required length of AD is 12.8 cm

Question 2. If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = ar (ABCD).
Solution:
Join GE and HE, where GE || BC || DA and HF || AB || DC
(∵ E, F, G and H are the mid¬points of the sides of a ||gm ABCD).
If a triangle and a parallelogram are on the same base and between the same parallels, then A E U the area of the triangle is equal to half the area of the parallelogram.
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/ A2
Now, ∆EFG and parallelogram EBCG are on the same base EG and between the same parallels EG and BC.
∴ ar(∆EFG) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles… (1)
Similarly, ar(∆EHG) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles…(2)
Adding (1) and (2), we get
ar(∆EFG) + ar(∆EHG) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles
Thus, ar(EFGH) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles

Question 3. 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).
Solution:
∵ ABCD is a parallelogram.
∴ AB || CD and BC || AD.
Now, ∆APB and parallelogram ABCD are on the same base AB and between the same parallels AB and CD.
∴ ar(∆APB) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles…….(1)
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/A3
Also, ∆BQC and parallelogram ABCD are on the same base BC and between the same parallels BGand AD.
∴ ar(∆BQC) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles…(2)
From (1) and (2), we have ar(∆APB) = ar(∆BQC).

Question 4. In figure, P is a point in the interior of a parallelogram ABCD. Show that
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/Q4
(i) ar (APB) + ar (PCD) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles
(ii) ar (APD) + ar(PBC) = ar (APB) + ar (PCD)
Solution:
We have a parallelogram ABCD, i.e., AB || CD and BC || AD. Let us draw EF || AB and HG || AD through P.
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/A4
(i) ∆APB and ||gm AEFB are on the same base AB and between the same parallels AB and EF.
∴ ar(∆APB) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles…(1)
Also, ∆PCD and parallelogram CDEF are on the same base CD and between the same parallels CD and EF.
∴ ar(APCD) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles…(2)
Adding (1) and (2), we have
ar(∆APB) + ar(∆PCD) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles
⇒ ar(∆APB) + ar(∆PCD) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles…(3)

(ii) ∆APD and ||gm ∆DGH are on the same base AD and between the same parallels AD and GH.
∴ ar(∆APD) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles…(4)
Similarly,
ar(∆PBC) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles…(5)
Adding (4) and (5), we have
ar(∆APD) + ar(∆PBC) = = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles
⇒ ar(∆APD) + ar(∆PBC) =NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles …….(6)
From (3) and (6), we have
ar(∆APD) + ar(∆PBC) = ar(∆APB) + ar(∆PCD)

Question 5. In figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = 

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/Q5
Solution:
(i) Parallelogram PQRS and parallelogram ABRS are on the same base RS and between the same parallels RS and PB.
∴ ar(PQRS) = ar(ABRS)
(ii) AAXS and ||gm ABRS are on the same base AS and between the same parallels AS and BR. *
∴ ar(AXS) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles …(1)
But ar(PQRS) = ar(ABRS) …(2) [Proved in (i) part]
From (1) and (2), we have
ar(AXS) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles

Question 6. 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.
Solution:
The farmer is having the field in the form of parallelogram PQRS and a point A is situated on RS. Join AP and AQ.
Clearly, the field is divided into three parts i.e., in ∆APS, ∆PAQ and ∆QAR.
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/ A6
Since, ∆PAQ and pt.
parallelogram PQRS are on the same base PQ and between the same parallels PQ and RS.
ar(∆PAQ) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles …(1)
⇒ ar(||gm PQRS) – ar(∆PAQ) = ar(||gm PQRS) – NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles
⇒ [ar(∆APS) + ar(∆QAR)] = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles …(2)
From (1) and (2), we have
ar(∆PAQ) = ar[(∆APS) + (∆QAR)]
Thus, the farmer can sow wheat in (∆PAQ) and pulses in [(∆APS) + (∆QAR)] or wheat in [(∆APS) + (∆QAR)] and pulses in (∆PAQ).

NCERT Solutions for Class 9 Maths Exercise 9.3

Question 1. In figure, E is any point on median AD of a ABC. Show that ar (ABE) = ar (ACE).

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image002.jpg

Solution:
In NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC, AD is a median.

ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABD) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngACD) ……….(i)

[NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image003.pngMedian divides a NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pnginto two NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngs of equal area]

Again in NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngEBC, ED is a median

ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngEBD) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngECD) ……….(ii)

Subtracting eq. (ii) from (i),

ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABD) – ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngEBD) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngACD) – ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngECD)

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABE) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngACE)

Question 2. In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = ar (ABC).

Solution: 
Given: A NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC, AD is the median and E is the mid-point of median AD.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image006.jpg

To prove: ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBED) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image005.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC)

Proof: In NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC, AD is the median.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABD) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngADC)

[NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image003.pngMedian divides a NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pnginto two NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngs of equal area]

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABD) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image008.pngar (ABC) ……….(i)

In NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABD, BE is the median.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBED) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBAE)

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBED) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image008.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABD)

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBED) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image009.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image005.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC)

Question 3. Show that the diagonals of a parallelogram divide it into four triangles of equal area.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image010.jpg

Solution: 
Let parallelogram be ABCD and its diagonals AC and BD intersect each other at O.

In NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC and NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngADC,

AB = DC [Opposite sides of a parallelogram]

BC = AD [Opposite sides of a parallelogram]

And AC = AC [Common]

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngNCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image011.pngNCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngCDA [By SSS congruency]

Since, diagonals of a parallelogram bisect each other.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngO is the mid-point of bisection.

Now in NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngADC, DO is the median.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAOD) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngCOD) ……….(i)

[Median divides a triangle into two equal areas]

Similarly, in NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC, OB is the median.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAOB) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBOC) ……….(ii)

And in NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAOB and NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAOD, AO is the median.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAOB) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAOD) ……….(iii)

From eq. (i), (ii) and (iii),

ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAOB) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAOD) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBOC) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngCOD)

Thus diagonals of parallelogram divide it into four triangles of equal area.

Question 4. In figure, ABC and ABD are two triangles on the same base AB. If line-segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD).

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image012.jpg

Solution: 
Draw CMNCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image013.pngAB and DNNCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image013.pngAB.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image014.png

In NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngCMO and NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDNO,

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image015.pngCMO = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image015.pngDNO = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image016.png[By construction]

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image015.pngCOM = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image015.pngDON [Vertically opposite]

OC = OD [Given]

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngNCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngCMO NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image011.pngNCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDNO [By ASA congruency]

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image003.pngAM = DN [By CPCT] ……(i)

Now ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image017.png……….(ii)

ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngADB) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image018.png……….(iii)

Using eq. (i) and (iii),

ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngADB) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image017.png……….(iv)

From eq. (ii) and (iv),

ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngADB)

Question 5. D, E and F are respectively the mid-points of the sides BC, CA and AB of a ABC. Show that:

(i) BDEF is a parallelogram.

(ii) ar (DEF) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image005.pngar (ABC)

(iii) ar (BDEF) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image008.pngar (ABC)

Solution: 
(i) F is the mid-point of AB and E is the mid-point of AC.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image019.jpg

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngFENCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pngBC and FE = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image008.pngBD

[NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image003.png Line joining the mid-points of two sides of a triangle is parallel to the third and half of it]

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngFENCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pngBD [BD is the part of BC]

And FE = BD

Also, D is the mid-point of BC.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngBD = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image008.pngBC

And FENCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pngBC and FE = BD

Again E is the mid-point of AC and D is the mid-point of BC.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngDENCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pngAB and DE = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image008.pngAB

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngDENCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pngAB [BF is the part of AB]

And DE = BF

Again F is the mid-point of AB.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngBF = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image008.pngAB

But DE = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image008.pngAB

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngDE = BF

Now we have FENCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pngBD and DENCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pngBF

And FE = BD and DE = BF

Therefore, BDEF is a parallelogram.

(ii) BDEF is a parallelogram.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBDF) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDEF) ……….(i)

[diagonals of parallelogram divides it in two triangles of equal area]

DCEF is also parallelogram.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDEF) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDEC) ……….(ii)

Also, AEDF is also parallelogram.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAFE) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDEF) ……….(iii)

From eq. (i), (ii) and (iii),

ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDEF) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBDF) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDEC) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAFE) ……….(iv)

Now, ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDEF) + ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBDF) + ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDEC) + ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAFE) ……….(v)

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDEF) + ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDEF) + ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDEF) + ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDEF)

[Using (iv) & (v)]

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC) = 4 NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image021.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDEF)

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDEF) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image005.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC)

(iii) ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pnggm BDEF) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBDF) + ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDEF) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDEF) + ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDEF) [Using (iv)]

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pnggm BDEF) = 2 ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDEF)

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pnggm BDEF) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image022.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC)

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pnggm BDEF) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image008.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC)

Question 6. In figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that :

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image023.png

(i) ar (DOC) = ar (AOB)

(ii) ar (DCB) = ar (ACB)

(iii) DANCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pngCB or ABCD is a parallelogram.

Solution: 
(i) Draw BM NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image013.pngAC and DN NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image013.pngAC.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image024.jpg

In NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDON and NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBOM,

OD = OB [Given]

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image015.pngDNO = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image015.pngBMO = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image016.png[By construction]

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image015.pngDON = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image015.pngBOM [Vertically opposite]

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngNCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDON NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image025.pngBOM [By RHS congruency]

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngDN = BM [By CPCT]

Also ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDON) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBOM) ……….(i)

Again, In NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDCN and NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABM,

CD = AB [Given]

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image015.pngDNC = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image015.pngBMA = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image016.png[By construction]

DN = BM [Prove above]

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngNCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDCN NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image025.pngBAM [By RHS congruency]

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDCN) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBAM) ……….(ii)

Adding eq. (i) and (ii),

ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDON) + ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDCN) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBOM) + ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBAM)

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDOC) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAOB)

(ii) Since ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDOC) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAOB)

Adding ar NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBOC both sides,

ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDOC) + ar NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBOC = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAOB) + ar NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBOC

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDCB) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngACB)

(iii) Since ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDCB) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngACB)

Therefore, these two triangles in addition to be on the same base CB lie between two same parallels CB and DA.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngDANCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pngCB

Now AB = CD and DANCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pngCB

Therefore, ABCD is a parallelogram.

Question 7. D and E are points on sides AB and AC respectively of ABC such that ar (DBC) = ar (EBC). Prove that DEBC.

Solution:
Given: ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDBC) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngEBC)

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image026.png

Since two triangles of equal area have common base BC.

Therefore, DENCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pngBC [NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image003.pngTwo triangles having same base (or equal bases) and equal areas lie between the same parallel]

Question 8. XY is a line parallel to side BC of triangle ABC. If BEAC and CFAB meet XY at E and F respectively, show that ar (ABE) = ar (ACF).

Solution:
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABE and parallelogram BCYE lie on the same base BE and between the same parallels BE and AC.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image027.png

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABE) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image008.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pnggm BCYE) ……….(i)

Also NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngACF and NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pnggm BCFX lie on the same base CF and between same parallel BX and CF.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngACF) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image008.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pnggm BCFX) ……….(ii)

But NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pnggm BCYE and NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pnggm BCFX lie on the same base BC and between the same parallels BC and EF.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pnggm BCYE) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pnggm BCFX) ……….(iii)

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

ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABE) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngACF)

Question 9. The side AB of 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. Show that ar (ABCD) = ar (PBQR).

Solution: 
Given: ABCD is a parallelogram, CPNCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pngAQ and PBQR is a parallelogram.

To prove: ar (ABCD) = ar (PBQR)

Construction: Join AC and QP.

Proof: Since AQNCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pngCP

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image028.jpg

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAQC) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAQP)

[Triangles on the same base and between the same parallels are equal in area]

Subtracting ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABQ) from both sides, we get

ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAQC – ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABQ) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAQP) – ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABQ)

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngQBP) ……….(i)

Now ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image008.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pnggm ABCD)

[Diagonal divides a parallelogram in two parts of equal area]

And ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngPQB) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image008.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pnggm PBQR)

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

ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pnggm ABCD) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pnggm PBQR)

Question 10. Diagonals AC and BD of a trapezium ABCD with ABDC intersect each other at O. Prove that ar(AOD) = ar (BOC).

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image029.jpg

Solution:
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABD and NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC lie on the same base AB and between the same parallels AB and DC.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image030.png

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABD) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC)

Subtracting ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAOB) from both sides,

ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABD) – ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAOB)

= ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC) – ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAOB)

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAOD) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBOC)

Question 11. In figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that :

(i) ar (ACB = ar (ACF)

(ii) ar (AEDF) = ar (ABCDE)

Solution: 
(i) Given that BFNCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pngAC

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngACB and NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngACF lie on the same base AC and between the same parallels AC and BF.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngACB) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngACF) ……….(i)

(ii) Now ar (ABCDE) = ar (trap. AEDC) + ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC) ……….(ii)

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngar (ABCDE) = ar (trap. AEDC) + ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngACF) = ar (quad. AEDF) [Using (i)]

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngar (AEDF) = ar (ABCDE)

Question 12. A villager Itwaari has a plot of land of the shape of quadrilateral. The Gram Panchyat of two villages decided to take over some portion of his plot from one of the corners to construct a health centre. Itwaari agrees to the above personal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.

Solution:
Let Itwari has land in shape of quadrilateral PQRS.

Draw a line through 5 parallel to PR, which meets QR produced at M.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image031.jpg

Let diagonals PM and RS of new formed quadrilateral intersect each other at point N.

We have PRNCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pngSM [By construction]

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngPRS) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngPMR)

[Triangles on the same base and same parallel are equal in area]

Subtracting ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngPNR) from both sides,

ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngPRS) – ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngPNR) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngPMR) – ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngPNR)

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngPSN) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngMNR)

It implies that Itwari will give corner triangular shaped plot PSN to the Grampanchayat for health centre and will take equal amount of land (denoted by NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngMNR) adjoining his plot so as to form a triangular plot PQM.

Question 13. ABCD is a trapezium with ABDC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).

Solution:
Join CX, NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngADX and ACX lie on the same base XA and between the same parallels XA and DC.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image032.jpg

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngADX) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngACX) ……….(i)

Also NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngACX and NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngACY lie on the same base

AC and between same parallels CY and XA.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngACX) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngACY) ……….(ii)

From (i) and (ii),

ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngADX) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngACY)

Question 14. In figure, APBQCR. Prove that ar (AQC) = ar (PBR).

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image033.png

Solution:
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABQ and BPQ lie on the same base BQ and between same parallels AP and BQ.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABQ) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBPQ) ……….(i)

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBQC and NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBQR lie on the same base BQ and between same parallels BQ and CR.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBQC) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBQR) ……….(ii)

Adding eq (i) and (ii), ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABQ) + ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBQC) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBPQ) + ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBQR)

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAQC) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngPBR)

Question 15. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar(BOC). Prove that ABCD is a trapezium.

Solution:
Given that ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAOD) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBOC)

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image034.png

Adding NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAOB both sides,

ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAOD) + ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAOB) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBOC) + ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngAOB)

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABD) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC)

Since if two triangles equal in area, lie on the same base then, they lie between same parallels. We have NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABD and NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngABC lie on common base AB and are equal in area.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngThey lie in same parallels AB and DC.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngABNCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pngDC

Now in quadrilateral ABCD, we have ABNCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pngDC

Therefore, ABCD is trapezium. [NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image003.pngIn trapezium one pair of opposite sides is parallel]

Question 16. In figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image035.png

Solution:
Given that NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDRC and NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDPC lie on the same base DC and ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDPC) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDRC) …..(i)

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image007.pngDCNCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pngRP

[If two triangles equal in area, lie on the same base then, they lie between same parallels]

Therefore, DCPR is trapezium. [ In trapezium one pair of opposite sides is parallel]

Also ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBDP) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngARC) ……….(ii)

Subtracting eq. (i) from (ii),

ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBDP) – ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDPC) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngARC) – ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngDRC)

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image004.pngar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngBDC) = ar (NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image001.pngADC)

Therefore, ABNCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/image020.pngDC [If two triangles equal in area, lie on the same base then, they lie between same parallels]

Therefore, ABCD is trapezium.

NCERT Solutions for Class 9 Maths Exercise 9.4

Question 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:
We have a parallelogram ABCD and rectangle ABEF such that
ar(||gm ABCD) = ar( rect. ABEF)
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/A1
AB = CD [Opposite sides of parallelogram]
and AB = EF [Opposite sides of a rectangle]
⇒ CD = EF
⇒ AB + CD = AB + EF … (1)
BE < BC and AF < AD [In a right triangle, hypotenuse is the longest side] ⇒ (BC + AD) > (BE + AF) …(2)
From (1) and (2), we have
(AB + CD) + (BC+AD) > (AB + EF) + BE + AF)
⇒ (AB + BC + CD + DA) > (AB + BE + EF + FA)
⇒ Perimeter of parallelogram ABCD > Perimeter of rectangle ABEF.

Question 2. In figure, D and E are two points on BC such that BD = DE = EC. Show that ar(ABD) = ar(ADE) = ar(AEC).
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/Q2
Solution:
Let us draw AF, perpendicular to BC
such that AF is the height of ∆ABD, ∆ADE and ∆AEC.
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/A2

Question 3. In figure, ABCD, DCFE and ABFE are parallelograms. Show that ar(ADE) = ax(BCF).
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/Q3
Solution:
Since, ABCD is a parallelogram [Given]
∴ Its opposite sides are parallel and equal.
i.e., AD = BC …(1)
Now, ∆ADE and ∆BCF are on equal bases AD = BC [from (1)] and between the same parallels AB and EF.
So, ar(∆ADE) = ar(∆BCF).

Question 4. In figure, 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) = ax(DPQ).[Hint Join AC.]
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/ Q4
Solution:
We have a parallelogram ABCD and AD = CQ. Let us join AC.
We know that triangles on the same base and between the same parallels are equal in area.
Since, ∆QAC and ∆QDC are on the same base QC and between the same parallels AD and BQ.
∴ ar(∆QAC) = ar(∆QDC)
Subtracting ar(∆QPC) from both sides, we have
ar(∆QAQ – ar(∆QPC) = ar(∆QDC) – ar(∆QPC)
⇒ ar(∆PAQ = ar(∆QDP) …(1)
Since, ∆PAC and ∆PBC are on the same base PC and between the same parallels AB and CD.
∴ ar(∆PAC) = ar(∆PBC) …(2)
From (1) and (2), we get
ar(∆PBC) = ar(∆QDP)

Question 5. In figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, Show that
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/ Q5
[Hint Join EC and AD. Show that BE || AC and DE || AB, etc.]

Solution:
Let us join EC and AD. Draw EP ⊥ BC.
Let AB = BC = CA = a, then
BD = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles= DE = BE
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/ A5

(ii) Since, ∆ABC and ∆BED are equilateral triangles.
⇒ ∠ACB = ∠DBE = 60°
⇒ BE || AC
∆BAE and ∆BEC are on the same base BE and between the same parallels BE and AC.
ar(∆BAE) = ar(∆BEC)
⇒ ar(∆BAE) = 2 ar(∆BDE) [ DE is median of ∆EBC. ∴ ar(∆BEC) = || ar(∆BDE)]
⇒ ar(ABDE) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Trianglesar(∆BAE)

(iii) ar(∆ABC) = 4 ar(∆BDE)[Proved in (i) part]
ar(∆BEC) = 2 ar(∆BDE)
[ ∵ DE is median of ∆BEC]
⇒ ar(∆ABC) = 2 ar(∆BEC)

(iv) Since, ∆ABC and ∆BDE are equilateral triangles.
⇒ ∠ABC = ∠BDE = 60°
⇒ AB || DE
∆BED and ∆AED are on the same base ED and between the same parallels AB and DE.
∴ ar(∆BED) = ar(∆AED)
Subtracting ar(AEFD) from both sides, we get
⇒ ar(∆BED) – ar(∆EFD) = ar(∆AED) – ar(∆EFD)
⇒ ar(∆BEE) = ar(∆AFD)

(v) In right angled ∆ABD, we get
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/A5A

From (1) and (2), we get
ar(∆AFD) = 2 ar(∆EFD)
ar(∆AFD) = ar(∆BEF) [From (iv) part]
⇒ ar(∆BFE) = 2 ar(∆EFD)

(vi) ar(∆AFC) = ar(∆AFD) + ar(∆ADC)
= ar(∆BFE) + NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Trianglesar(∆ABC) [From (iv) part]
= ar(∆BFE) + NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Trianglesx 4 x ar(∆BDE) [From (i) part]
= ar(∆BFE) + 2ar(∆BDE)
= 2ar(∆FED) + 2[ar(∆BFE) + ar(∆FED)]
= 2ar(∆FED) + 2[2ar(∆FED) + ar(∆FED)] [From (v) part]
= 2ar(∆FED) + 2[3ar(∆FED)]
= 2ar(∆FED) + 6ar(∆FED)
= 8ar(∆FED)
∴ ar(∆FED) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/ar(∆AFC)

Question 6. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that
ar(APB) x ar(CPD) = ar(APD) x ar(BPC).

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

Solution:
We have a quadrilateral ABCD such that its diagonals AC and BD intersect at P.
Let us draw AM ⊥ BD and CN ⊥ BD.
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/A6

Question 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
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/Q7
Solution:
We have a ∆ABC such that P is the mid-point of AB and Q is the mid-point of BC.
Also, R is the mid-point of AP. Let us join AQ, RQ, PC and PC.

(i) In ∆APQ, R is the mid-point of AP. [Given] B
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/A7
∴RQ is a median of ∆APQ.
⇒ ar(∆PRQ) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Trianglesar(∆APQ) …(1)
In ∆ABQ, P is the mid-point of AB.
∴ QP is a median of ∆ABQ.
∴ ar(∆APQ) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Trianglesar(∆ABQ) …(2)

NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/ A7b
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/A7c

Question 8. In figure, 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
NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Triangles/Q8
(i) ∆MBC = ∆ABD
(ii) ar(BYXD) = 2 ar(MBC)
(iii) ar(BYXD) = ax(ABMN)
(iv) ∆FCB ≅ ∆ACE
(v) ar(CYXE) = 2 ar(FCB)
(vi) ar(CYXE) = ax(ACFG)
(vii) ar(BCED) = ar(ABMN) + ar(ACFG)

Solution:
We have a right ∆ABC such that BCED, ACFG and ABMN are squares on its sides BC, CA and AB respectively. Line segment AX 1 DE is also drawn such that it meets BC at Y.

(i) ∠CBD = ∠MBA [Each90°]
∴ ∠CBD + ∠ABC = ∠MBA + ∠ABC
(By adding ∠ABC on both sides)
or ∠ABD = ∠MBC
In ∆ABD and ∆MBC, we have
AB = MB [Sides of a square]
BD = BC
∠ABD = ∠MBC [Proved above]
∴ ∆ABD = ∆MBC [By SAS congruency]

(ii) Since parallelogram BYXD and ∆ABD are on the same base BD and between the same parallels BD and AX.
∴ ar(∆ABD) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Trianglesar(||gm BYXD)
But ∆ABD ≅ ∆MBC [From (i) part]
Since, congruent triangles have equal
areas.
∴ ar(∆MBC) = NCERT Solutions for Class 9 Maths Chapter-9 Areas of Parallelograms and Trianglesar(||gm BYXD)
⇒ ar(||gm BYXD) = 2ar(∆MBC)

(iii) Since, ar(||gm BYXD) = 2ar(∆MBC) …(1) [From (ii) part]
and or(square ABMN) = 2or(∆MBC) …(2)
[ABMN and AMBC are on the same base MB and between the same parallels MB and NC]
From (1) and (2), we have
ar(BYXD) = ar(ABMN) .

(iv) ∠FCA = ∠BCE (Each 90°)
or ∠FCA+ ∠ACB = ∠BCE+ ∠ACB
[By adding ∠ACB on both sides]
⇒ ∠FCB = ∠ACE
In ∆FCB and ∆ACE, we have
FC = AC [Sides of a square]
CB = CE [Sides of a square]
∠FCB = ∠ACE [Proved above]
⇒ ∆FCB ≅ ∆ACE [By SAS congruency]

(v) Since, ||gm CYXE and ∆ACE are on the same base CE and between the same parallels CE and AX.
∴ ar(||gm CYXE) = 2ar(∆ACE)
But ∆ACE ≅ ∆FCB [From (iv) part]
Since, congruent triangles are equal in areas.
∴ ar (||<gm CYXE) = 2ar(∆FCB)

(vi) Since, ar(||gm CYXE) = 2ar(∆FCB) …(3)
[From (v) part]
Also (quad. ACFG) and ∆FCB are on the same base FC and between the same parallels FC and BG.
⇒ ar(quad. ACFG) = 2ar(∆FCB) …(4)
From (3) and (4), we get
ar(quad. CYXE) = ar(quad. ACFG) …(5)

(vii) We have ar(quad. BCED)
= ar(quad. CYXE) + ar(quad. BYXD)
= ar(quad. CYXE) + ar(quad. ABMN)
[From (iii) part]
Thus, ar (quad. BCED)
= ar(quad. ABMN) + ar(quad. ACFG)
[From (vi) part]

Conclusions for NCERT SOLUTIONS FOR CLASS 9 MATHS CHAPTER 9 AREAS OF PARALLELOGRAMS AND TRIANGLES

SWC academic staff has developed NCERT answers for this chapter of the ninth grade mathematics curriculum. We have solutions prepared for all of the exercises of this chapter. The answers, broken down into steps, to all of the questions included in the NCERT textbook’s chapter are provided here. Read this chapter on theory. Be certain that you have read the theory section of this chapter of the NCERT textbook and that you have learnt the formulas for the chapter that you are studying. 

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