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.
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
Q1 In Fig. , ABCD is a parallelogram, and . If , and , find AD.
We have,
AE DC and CF 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 AE = (16 8 )
SInce, CF AD
therefore area of parallelogram = AD CF = 128
= 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
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, EFG and ||gm BEGC are on the same base and between the same parallels EG and BC.
Therefore, ar (EFG) = ar (||gm BEGC)...............(i)
Similarly, ar (EHG) = 1/2 . ar(||gm AEGD)..................(ii)
By adding eq (i) and eq (ii), we get
ar (EFG) + ar (EHG) = 1/2 (ar (||gm BEGC) + ar(||gm AEGD))
ar (EFGH) = 1/2 ar(ABCD)
Hence proved
We have,
ABCD is a parallelogram, therefore AB || CD and BC || AD.
Now, APB and ||gm ABCD are on the same base AB and between two parallels AB and DC.
Therefore, ar (APB) = 1/2 . ar(||gm ABCD)...........(i)
Also, BQC and ||gm ABCD are on the same base BC and between two parallels BC and AD.
Therefore, ar(BQC) = 1/2 . ar(||gmABCD)...........(ii)
From eq(i) and eq (ii), we get,
ar (APB) = ar(BQC)
Hence proved.
Q4 (i) In Fig. , P is a point in the interior of a parallelogram ABCD. Show that
[Hint : Through P, draw a line parallel to AB.]
We have a ||gm ABCD and AB || CD, AD || BC. Through P, draw a line parallel to AB
Now, APB and ||gm ABEFare on the same base AB and between the same parallels EF and AB.
Therefore, ar (APB) = 1/2 . ar(ABEF)...............(i)
Similarly, ar (PCD ) = 1/2 . ar (EFDC) ..............(ii)
Now, by adding both equations, we get
Hence proved.
Q4 (ii) In Fig. , P is a point in the interior of a parallelogram ABCD. Show that
[Hint : Through P, draw a line parallel to AB.]
We have a ||gm ABCD and AB || CD, AD || BC. Through P, draw a line parallel to AB
Now, APD and ||gm ADGHare on the same base AD and between the same parallels GH and AD.
Therefore, ar (APD) = 1/2 . ar(||gm ADGH).............(i)
Similarily, ar (PBC) = 1/2 . ar(||gm BCGH)............(ii)
By adding the equation i and eq (ii), we get
Hence proved.
Q5 In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) Parallelogram PQRS and ABRS are on the same base RS and between the same parallels RS and PB.
Therefore, ............(i)
Hence proved
(ii) AXS and ||gm ABRS are on the same base AS and between same parallels AS and RB.
Therefore, ar (AXS) = 1/2 . ar(||gm ABRS)............(ii)
Now, from equation (i) and equation (ii), we get
Hence proved.
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, APS, QAR and PAQ.
Since APQ and parallelogram, PQRS is on the same base PQ and between same parallels RS and PQ.
Therefore, ............(i)
We can write above equation as,
ar (||gm PQRS) - [ar (APS) + ar(QAR)] = 1/2 .ar(PQRS)
from equation (i),
Hence, she can sow wheat in APQ and pulses in [APS + QAR] or wheat in [APS + QAR] and pulses in APQ.
Q1 In Fig. , E is any point on median AD of a . Show that.
We have ABC such that AD is a median. And we know that median divides the triangle into two triangles of equal areas.
Therefore, ar(ABD) = ar( ACD)............(i)
Similarly, In triangle BEC,
ar(BED) = ar (DEC)................(ii)
On subtracting eq(ii) from eq(i), we get
ar(ABD) - ar(BED) =
Hence proved.
Q2 In a triangle ABC, E is the mid-point of median AD. Show that .
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.
ar(ABD) = ar (ACD) = 1/2 ar(ABC)..............(i)
Now, in triangle ABD,
BE is the median [since E is the midpoint of AD]
ar (BED) = 1/2 ar(ABD)........(ii)
From eq (i) and eq (ii), we get
ar (BED) = 1/2 . (1/2 ar(ar (ABC))
ar (BED) = 1/4 .ar(ABC)
Hence proved.
Q3 Show that the diagonals of a parallelogram divide it into four triangles of equal area.
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 (BOC) = ar (DOC)...........(a) ( since OC is the median of triangle CBD)
Similarly, ar(AOD) = ar(DOC) ............(b) ( since OD is the median of triangle ACD)
and, ar (AOB) = ar(BOC)..............(c) ( since OB is the median of triangle ABC)
From eq (a), (b) and eq (c), we get
ar (BOC) = ar (DOC)= ar(AOD) = (AOB)
Thus, the diagonals of ||gm divide it into four equal triangles of equal area.
We have, ABC and ABD on the same base AB. CD is bisected by AB at point O.
OC = OD
Now, in ACD, AO is median
ar (AOC) = ar (AOD)..........(i)
Similarly, in BCD, BO is the median
ar (BOC) = ar (BOD)............(ii)
Adding equation (i) and eq (ii), we get
ar (AOC) + ar (BOC) = ar (AOD) + ar (BOD)
Hence proved.
We have a triangle ABC such that D, E and F are the midpoints of the sides BC, CA and AB respectively.
Now, in ABC,
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.
EF || BC or EF || BD
also, EF = 1/2 (BC)
[ 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 . Show that
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
Ar (BDEF) = Ar (DCEF)
Ar(BDF) = Ar (DEF) .............(i)
It is known that diagonals of ||gm divides it into two triangles of equal area.
Ar(DCE) = Ar (DEF).......(ii)
and, Ar(AEF) = Ar (DEF)...........(iii)
From equation(i), (ii) and (iii), we get
Ar(BDF) = Ar(DCE) = Ar(AEF) = Ar (DEF)
Thus, Ar (ABC) = Ar(BDF) + Ar(DCE) + Ar(AEF) + Ar (DEF)
Ar (ABC) = 4 . Ar(DEF)
Hence proved.
Q5 (iii) D, E and F are respectively the mid-points of the sides BC, CA and AB of a . Show that
Since we already proved that,
ar(DEF) = ar ( BDF).........(i)
So, ar(||gm BDEF) = ar(BDF) + ar (DEF)
= 2 . ar(DEF) [from equation (i)]
Hence proved.
Q6 (i) In Fig. , diagonals AC and BD of quadrilateral ABCD intersect at O such that. If , then show that:
[Hint: From D and B, draw perpendiculars to AC.]
We have ABCD is quadrilateral whose diagonals AC and BD intersect at O. And OB = OD, AB = CD
Draw DE AC and FB AC
In DEO and BFO
DOE = BOF [vertically opposite angle]
OED = BFO [each ]
OB = OD [given]
Therefore, by AAS congruency
DEO BFO
DE = FB [by CPCT]
and ar( DEO) = ar(BFO) ............(i)
Now, In DEC and ABF
DEC = BFA [ each ]
DE = FB
DC = BA [given]
So, by RHS congruency
DEC BFA
1 = 2 [by CPCT]
and, ar( DEC) = ar(BFA).....(ii)
By adding equation(i) and (ii), we get
Hence proved.
Q6 (ii) In Fig. , diagonals AC and BD of quadrilateral ABCD intersect at O such that . If , then show that:
[Hint: From D and B, draw perpendiculars to AC.]
We already proved that,
Now, add on both sides we get
Hence proved.
Q6 (iii) In Fig. , diagonals AC and BD of quadrilateral ABCD intersect at O such that . If , then show that:
[Hint : From D and B, draw perpendiculars to AC.]
Since DCB and ACB both lie on the same base BC and having equal areas.
Therefore, They lie between the same parallels BC and AD
CB || AD
also 1 = 2 [ already proved]
So, AB || CD
Hence ABCD is a || gm
Q7 D and E are points on sides AB and AC respectively of such that . Prove that .
We have ABC and points D and E are on the sides AB and AC such that ar(DBC ) = ar (EBC)
Since DBC and EBC are on the same base BC and having the same area.
They must lie between the same parallels DE and BC
Hence DE || BC.
We have a ABC such that BE || AC and CF || AB
Since XY || BC and BE || CY
Therefore, BCYE is a ||gm
Now, The ||gm BCEY and ABE are on the same base BE and between the same parallels AC and BE.
ar(AEB) = 1/2 .ar(||gm BEYC)..........(i)
Similarly, ar(ACF) = 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.
ar (BEYC) = ar (BCFX).........(iii)
From eq (i), (ii) and (iii), we get
ar( ABE) = ar(ACF)
Hence proved.
Join the AC and PQ.
It is given that ABCD is a ||gm and AC is a diagonal of ||gm
Therefore, ar(ABC) = ar(ADC) = 1/2 ar(||gm ABCD).............(i)
Also, ar(PQR) = ar(BPQ) = 1/2 ar(||gm PBQR).............(ii)
Since AQC and APQ are on the same base AQ and between same parallels AQ and CP.
ar(AQC) = ar (APQ)
Now, subtracting ABQ from both sides we get,
ar(AQC) - ar (ABQ) = ar (APQ) - ar (ABQ)
ar(ABC) = ar (BPQ)............(iii)
From eq(i), (ii) and (iii) we get
Hence proved.
Q10 Diagonals AC and BD of a trapezium ABCD with intersect each other at O. Prove that .
We have a trapezium ABCD such that AB || CD and it's diagonals AC and BD intersect each other at O
ABD and ABC are on the same base AB and between same parallels AB and CD
ar(ABD) = ar (ABC)
Now, subtracting AOB from both sides we get
ar ( AOD) = ar ( BOC)
Hence proved.
Q11 In Fig. , ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
We have a pentagon ABCDE in which BF || AC and CD is produced to F.
(i) Since ACB and ACF are on the same base AC and between same parallels AC and FB.
ar(ACB) = ar (ACF)..................(i)
(ii) Adding the ar (AEDC) on both sides in equation (i), we get
ar(ACB) + ar(AEDC) = ar (ACF) + ar(AEDC)
Hence proved.