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: A four side shape having equal opposite sides and whose diagonals bisect each other is called a parallelogram. Triangle is a 3 side shape having 3 sides and 3 angles. In this particular chapter, you will learn about the areas of different triangles and parallelograms. Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles will provide you the solutions to all types of triangles and parallelograms covered in the chapter. Many structures have the shape of the triangle and parallelograms. Understanding the area is mandatory in the field of calculating the area of land. If a land has an irregular shape, the land will be divided into triangles and the area will be calculated.

CBSE NCERT solutions for class 9 maths chapter 9 Areas Of Parallelograms And Triangles is also covering the solutions to the application based questions as well. There are several interesting problems in the chapter. Solving these problems will help you in improving the concepts of the chapter and also in exams like the olympiads. In total there are 4 practice exercises having 51 questions. NCERT solutions for class 9 maths chapter 9 Areas of Parallelograms and Triangles have covered all the exercises including the optional ones in a detailed manner. NCERT solutions are available for other classes and chapters as well. Four exercises of this chapter are explained below.

Exercise:9.1

Exercise:9.2

Exercise:9.3

Exercise:9.4

 

NCERT solutions for class 9 maths chapter 9 Areas Of Parallelograms And Triangles Excercise: 9.1

Q 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.

Answer:

In figure (i), (iii) and (v) we can see that. they lie on the same base and between the same parallel lines.
                               In figure (i)      figure (iii)     figure (v)
Common base            DC                   QR                AD
Two parallels         DC and AB     QR and PS    AD and BQ

 

NCERT solutions for class 9 maths chapter 9 Areas Of Parallelograms And Triangles Excercise: 9.2

Q1 In Fig. \small 9.15, ABCD is a parallelogram, \small AE\perp DC and \small CF\perp AD. If \small AB=16\hspace{1mm}cm\small AE=8\hspace{1mm}cm  and \small CF=10\hspace{1mm}cm, find AD.

            

 

Answer:


We have,
AE \perp DC and CF \perp AD
AB = 16 cm, AE = 8 cm and CF = 10 cm

Since ABCD is a parallelogram,
therefore, AB = DC = 16 cm
We know that, area of parallelogram (ABCD) = base . height
                                                                        = CD \times AE  = (16 \times 8 ) cm^2
SInce,  CF \perp AD
therefore area of parallelogram  = AD \times CF = 128 cm^2
                                                   = AD = 128/10 
                                                   = AD = 12.8 cm                           

Thus the required length of AD is 12.8 cm.

Q2 If  E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that

\small ar(EFGH)=\frac{1}{2}ar(ABCD)

Answer:

Join GE and HE,
Since F, F, G, H are the mid-points of the parallelogram ABCD. Therefore, GE || BC ||AD and HF || DC || AB.
It is known that if a triangle and the parallelogram are on the same base and between the same parallel lines. then the area of the triangle is equal to the half of the area of the parallelogram.

Now, \DeltaEFG and ||gm BEGC are on the same base and between the same parallels EG and BC.
Therefore, ar (\DeltaEFG) = 1/2 ar (||gm BEGC)...............(i)
Similarly, ar (\DeltaEHG) = 1/2 . ar(||gm AEGD)..................(ii)
By adding eq (i) and eq (ii), we get

ar (\DeltaEFG) +  ar (\DeltaEHG) = 1/2 (ar (||gm BEGC) + ar(||gm AEGD))
ar (EFGH) = 1/2 ar(ABCD)
Hence proved                                           

Q3 P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that \small ar(APB)=ar(BCQ).

Answer:


We have, 
ABCD is a parallelogram, therefore AB || CD and BC || AD.


Now,  \DeltaAPB and ||gm ABCD are on the same base AB and between two parallels AB and DC.
Therefore, ar  (\DeltaAPB) = 1/2 . ar(||gm ABCD)...........(i)

Also, \DeltaBQC and ||gm ABCD are on the same base BC and between two parallels BC and AD.
Therefore, ar(\DeltaBQC) = 1/2 . ar(||gmABCD)...........(ii)

From eq(i) and eq (ii),  we get,
ar  (\DeltaAPB) =  ar(\DeltaBQC)
Hence proved.

Q4 (i) In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that
\small ar(APB)+ar(PCD) = \frac{1}{2}ar(ABCD)

[Hint : Through P, draw a line parallel to AB.]

                

 

Answer:


We have a ||gm ABCD and AB || CD, AD || BC.  Through P, draw a line parallel to AB
 Now, \DeltaAPB and ||gm ABEFare on the same base AB and between the same parallels EF and AB.
Therefore, ar (\DeltaAPB) = 1/2 . ar(ABEF)...............(i)
Similarly, ar (\DeltaPCD ) = 1/2 . ar (EFDC) ..............(ii)
Now, by adding both equations, we get
\small ar(APB)+ar(PCD) = \frac{1}{2}ar(ABCD)

Hence proved.

Q4 (ii) In Fig. \small 9.16, P is a point in the interior of a parallelogram ABCD. Show that

\small ar(APD)+ar(PBC)=ar(APB)+ar(PCD)

[Hint : Through P, draw a line parallel to AB.]

            

 

Answer:


We have a ||gm ABCD and AB || CD, AD || BC.  Through P, draw a line parallel to AB
 Now, \DeltaAPD and ||gm ADGHare on the same base AD and between the same parallels GH and AD.
Therefore, ar (\DeltaAPD) = 1/2 . ar(||gm ADGH).............(i)

Similarily, ar (\DeltaPBC) = 1/2 . ar(||gm BCGH)............(ii)

By adding the equation i and eq (ii), we get

\small ar(APD)+ar(PBC)=ar(APB)+ar(PCD)

Hence proved.

Q5 In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that

(i)   \small ar(PQRS)=ar(ABRS)
(ii)  \small ar(AXS)=\frac{1}{2}ar(PQRS)

            

 

Answer:


(i) Parallelogram PQRS and ABRS are on the same base RS and between the same parallels RS and PB.
Therefore,  \small ar(PQRS)=ar(ABRS)............(i)
Hence proved

(ii) \DeltaAXS and ||gm ABRS are on the same base AS and between same parallels AS and RB.
Therefore, ar (\DeltaAXS) = 1/2 . ar(||gm ABRS)............(ii)

Now, from equation (i) and equation (ii), we get

\small ar(AXS)=\frac{1}{2}ar(PQRS)

Hence proved.

Q6 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?

Answer:

We have a field in the form of parallelogram PQRS and a point A is on the side RS. Join AP and AQ. The field is divided into three parts i.e, \DeltaAPS, \DeltaQAR and \DeltaPAQ. 

Since \DeltaAPQ and parallelogram, PQRS is on the same base PQ and between same parallels RS and PQ.
Therefore, ar(\Delta APQ) = \frac{1}{2}ar(PQRS)............(i)
We can write above equation as,

ar (||gm PQRS) - [ar (\DeltaAPS) + ar(\DeltaQAR)] = 1/2 .ar(PQRS)
\Rightarrow ar(\Delta APS)+ar(\Delta QAR) = \frac{1}{2}ar(PQRS)
from equation (i),
\Rightarrow ar(\Delta APS)+ar(\Delta QAR) =ar(\Delta APQ)

Hence, she can sow wheat in \DeltaAPQ and pulses in [\DeltaAPS + \DeltaQAR] or wheat in [\DeltaAPS + \DeltaQAR] and pulses in  \DeltaAPQ.

 

NCERT solutions for class 9 maths chapter 9 Areas Of Parallelograms And Triangles Excercise: 9.3

Q1 In Fig. \small 9.23,  E is any point on median AD of a \small \Delta ABC. Show that. \small ar(ABE)=ar(ACE)

              

 

Answer:


We have \DeltaABC such that AD is a median. And we know that median divides the triangle into two triangles of equal areas.
Therefore, ar(\DeltaABD) = ar( \DeltaACD)............(i)

Similarly, In triangle \DeltaBEC,
ar(\DeltaBED) = ar (\DeltaDEC)................(ii)

On subtracting eq(ii) from eq(i), we get
ar(\DeltaABD) - ar(\DeltaBED) = 
\small ar(ABE)=ar(ACE)

Hence proved.

Q2 In a triangle ABC, E is the mid-point of median AD. Show that \small ar(BED)=\frac{1}{4}ar(ABC).

Answer:

We have a triangle ABC and  AD is a median. Join B and E.

Since the median divides the triangle into two triangles of equal area.
\therefore ar(\DeltaABD) = ar (\DeltaACD) = 1/2 ar(\DeltaABC)..............(i)
Now, in triangle \DeltaABD,
BE is the median [since E is the midpoint of AD]
\therefore ar (\DeltaBED) = 1/2 ar(\DeltaABD)........(ii)

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

ar (\DeltaBED) = 1/2 . (1/2 ar(ar (\DeltaABC))
ar (\DeltaBED)  = 1/4 .ar(\DeltaABC)

Hence proved.

Q3 Show that the diagonals of a parallelogram divide it into four triangles of equal area.

Answer:


Let ABCD is a parallelogram. So, AB || CD and AD || BC and we know that Diagonals bisects each other. Therefore, AO = OC and BO = OD

Since OD = BO 
Therefore, ar (\DeltaBOC) = ar (\DeltaDOC)...........(a)   ( since OC is the median of triangle CBD)

Similarly, ar(\DeltaAOD) = ar(\DeltaDOC) ............(b)     ( since OD is the median of triangle ACD)

and, ar (\DeltaAOB) = ar(\DeltaBOC)..............(c)           ( since OB  is the median of triangle ABC)

From eq (a), (b) and eq (c), we get

  ar (\DeltaBOC) = ar (\DeltaDOC)= ar(\DeltaAOD) =  (\DeltaAOB)

Thus, the diagonals of ||gm divide it into four equal triangles of equal area.

Q4 In Fig. \small 9.24,  ABC and ABD are two triangles on the same base AB. If line- segment CD is bisected by AB at O, show that  \small ar(ABC)=ar(ABD).

                

 

Answer:


We have, \DeltaABC and \DeltaABD on the same base AB. CD is bisected by AB at point O.
\therefore OC = OD
Now, in \DeltaACD, AO is median
\therefore ar (\DeltaAOC) = ar (\DeltaAOD)..........(i)

Similarly, in \DeltaBCD, BO is the median
\therefore ar (\DeltaBOC) = ar (\DeltaBOD)............(ii)

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

ar (\DeltaAOC) + ar (\DeltaBOC) =  ar (\DeltaAOD) + ar (\DeltaBOD)

\small ar(ABC)=ar(ABD)

Hence proved.

Q5 (i) D, E and F are respectively the mid-points of the sides BC, CA and AB of a  \small \Delta ABC. Show that BDEF is a parallelogram.

Answer:

We have a triangle \DeltaABC such that D, E and F are the midpoints of the sides BC, CA and AB respectively.

Now, in \DeltaABC, 
F and E are the midpoints of the side AB and AC.
Therefore according to mid-point theorem, the line segment connecting the midpoints of two sides of a triangle is parallel to the third side and half of the third side.
\therefore EF || BC or EF || BD

also, EF = 1/2 (BC)
\Rightarrow EF = BD [ D is the midpoint of BC]
Similarly, ED || BF and ED = FB
Hence BDEF is a parallelogram.

Q5 (ii) D, E and F are respectively the mid-points of the sides BC, CA and AB of a \small \Delta ABC. Show  that 

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

 

Answer:


We already proved that BDEF is a ||gm.
Similarly, DCEF and DEAF are also parallelograms.
Now, ||gm BDEF and ||gm DCEF is on the same base EF and between same parallels BC and EF 
\therefore  Ar (BDEF) = Ar (DCEF)
\Rightarrow Ar(\DeltaBDF) = Ar (\DeltaDEF) .............(i)

It is known that diagonals of ||gm divides it into two triangles of equal area.
\therefore Ar(DCE) = Ar (DEF).......(ii)

and, Ar(\DeltaAEF) = Ar (\DeltaDEF)...........(iii)

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

 Ar(\DeltaBDF) =  Ar(DCE) = Ar(\DeltaAEF) =  Ar (\DeltaDEF)

Thus, Ar (\DeltaABC) =  Ar(\DeltaBDF) +  Ar(DCE) + Ar(\DeltaAEF) +  Ar (\DeltaDEF)
           Ar (\DeltaABC) = 4 . Ar(\DeltaDEF)
         \Rightarrow ar(\Delta DEF) = \frac{1}{4}ar(\Delta ABC)

Hence proved.

Q5 (iii) D, E and F are respectively the mid-points of the sides BC, CA and AB of a \small \Delta ABC. Show that

\small ar(BDEF)=\frac{1}{2}ar(ABC)

 

Answer:


Since we already proved that,
ar(\DeltaDEF) = ar (\Delta BDF).........(i)

So, ar(||gm BDEF) = ar(\DeltaBDF) + ar (\DeltaDEF)
                               = 2 . ar(\DeltaDEF)    [from equation (i)]
                               \\= 2[\frac{1}{4}ar(\Delta ABC)]\\ =\frac{1}{2} ar(\Delta ABC)

Hence proved.

Q6 (i) In Fig. \small 9.25, diagonals AC and BD of quadrilateral ABCD intersect at O such that. If \small AB=CD, then show that:

\small ar(DOC)=ar(AOB)

[Hint: From D and B, draw perpendiculars to AC.]

            

 

Answer:


We have ABCD is quadrilateral whose diagonals AC and BD intersect at O. And OB = OD, AB = CD
Draw DE \perp AC and FB \perpAC

In \DeltaDEO and \Delta BFO
\angleDOE = \angleBOF [vertically opposite angle]
\angleOED = \angleBFO [each 90^0]
OB = OD [given]

Therefore, by AAS congruency
\DeltaDEO \cong  \Delta BFO
\Rightarrow DE = FB [by CPCT]

and ar( \DeltaDEO) =  ar(\DeltaBFO) ............(i)

Now, In  \DeltaDEC and  \DeltaABF
\angleDEC = \angleBFA [ each 90^0]
DE = FB
DC = BA [given]
So, by RHS congruency 
\DeltaDEC \cong  \Delta BFA
\angle1 = \angle2 [by CPCT]
and, ar( \DeltaDEC) =  ar(\DeltaBFA).....(ii)

By adding equation(i) and (ii), we get
\small ar(DOC)=ar(AOB)
Hence proved.

Q6 (ii) In Fig. \small 9.25, diagonals AC and BD of quadrilateral ABCD intersect at O such that \small OB=OD. If \small AB=CD, then show that: \small ar(DCB)=ar(ACB)

[Hint: From D and B, draw perpendiculars to AC.]

            

 

Answer:


We already proved that,
ar(\Delta DOC)=ar(\Delta AOB)
Now, add ar(\Delta BOC) on both sides we get

\\ar(\Delta DOC)+ar(\Delta BOC)=ar(\Delta AOB)+ar(\Delta BOC)\\ ar(\Delta DCB) = ar (\Delta ACB)
Hence proved.

Q6 (iii) In Fig. \small 9.25, diagonals AC and BD of quadrilateral ABCD intersect at O such that \small OB=OD. If \small AB=CD, then show that:

\small DA\parallel CB or ABCD is a parallelogram.

[Hint : From D and B, draw perpendiculars to AC.]

            

 

Answer:


Since \DeltaDCB and \DeltaACB both lie on the same base BC and having equal areas.
Therefore, They lie between the same parallels BC and AD
\Rightarrow CB || AD 
also \angle1 = \angle2 [ already proved]
So, AB ||  CD 
Hence ABCD is a || gm

Q7 D and E are points on sides AB and AC respectively of  \small \Delta ABC such that  \small ar(DBC)=ar(EBC). Prove that \small DE\parallel BC.

Answer:

We have  \Delta ABC and points D and E are on the sides AB and AC such that ar(\DeltaDBC ) = ar (\DeltaEBC)


Since \DeltaDBC and \DeltaEBC are on the same base BC and having the same area.
\therefore They must lie between the same parallels DE and BC
Hence DE || BC.

Q8 XY is a line parallel to side BC of a triangle ABC. If  \small BE\parallel AC and   \small CF\parallel AB meet XY at E and F respectively, show that \small ar(ABE)=ar(ACF)

 

Answer:

We have a \DeltaABC  such that BE || AC and CF || AB
Since XY || BC and BE || CY 
Therefore, BCYE is a ||gm

Now, The ||gm BCEY and \DeltaABE are on the same base BE and between the same parallels AC and BE.
\therefore ar(\DeltaAEB) = 1/2 .ar(||gm BEYC)..........(i)
Similarly, ar(\DeltaACF) = 1/2 . ar(||gm BCFX)..................(ii)

Also, ||gm BEYC and ||gmBCFX are on the same base BC and between the same parallels BC and EF.
\therefore ar (BEYC) = ar (BCFX).........(iii)

From eq (i), (ii) and (iii),  we get

ar(\Delta ABE) = ar(\DeltaACF)
Hence proved.

Q9 The side AB of a 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 (see Fig. \small 9.26). Show that  \small ar(ABCD)=ar(PBQR). [Hint: Join AC and PQ. Now compare  \small ar(ACQ) and  \small ar(APQ).]

                    

 

Answer:

Join the AC and PQ.

It is given that ABCD is a ||gm and AC is a diagonal of ||gm
Therefore, ar(\DeltaABC) = ar(\DeltaADC) = 1/2 ar(||gm ABCD).............(i)

Also, ar(\DeltaPQR) = ar(\DeltaBPQ) = 1/2 ar(||gm PBQR).............(ii)

Since \DeltaAQC and \DeltaAPQ are on the same base AQ and between same parallels AQ and CP.
\therefore  ar(\DeltaAQC) = ar (\DeltaAPQ)

Now, subtracting \DeltaABQ from both sides we get,

ar(\DeltaAQC) - ar (\DeltaABQ) = ar (\DeltaAPQ) - ar (\DeltaABQ)
ar(\DeltaABC) = ar (\DeltaBPQ)............(iii)

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

\small ar(ABCD)=ar(PBQR)

Hence proved.

Q10 Diagonals AC and BD of a trapezium ABCD with  \small AB\parallel DC  intersect each other at O. Prove that  \small ar(AOD)=ar(BOC).

Answer:

We have a trapezium ABCD such that AB || CD and it's diagonals AC and BD intersect each other at O 
\DeltaABD and \Delta ABC are on the same base AB and between same parallels AB and CD 
\therefore ar(\DeltaABD) = ar (\DeltaABC)

Now, subtracting \DeltaAOB from both sides we get

ar (\Delta AOD) = ar (\Delta BOC)

Hence proved.