Chapter 14 : Areas of Parallelograms and Triangles - Rd Sharma Solutions for Class 9 Maths CBSE

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Chapter 14 - Areas of Parallelograms and Triangles Excercise Ex. 14.1

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 two parallels:

Solution 1

(i) ΔAPB and trapezium ABCD are on the same base AB and between the same parallels AB and CD.

(ii) Parallelograms ABCD and APQD are on the same base AD and between the same parallels AD and BQ.

(iii) Parallelogram ABCD and ΔPQR are between the same parallels AD and BC but they are not on the same base.

(iv) ΔQRT and parallelogram PQRS are on the same base QR and between the same parallels QR and PS.

(v) Parallelogram PQRS and trapezium SMNR are on the same base SR but they are not between the same parallels.

(vi) Parallelograms PQRS, AQRD, BQRC are between the same parallels. Also, parallelograms PQRS, BPSC and APSD are between the same parallels.

Chapter 14 - Areas of Parallelograms and Triangles Excercise Ex. 14.2

Question 1
In the given figure, ABCD is parallelogram, AE  DC and CF  AD. If AB = 16 cm. AE = 8 cm and CF = 10 cm, find AD. 
 
Solution 1
In parallelogram ABCD, CD = AB = 16 cm     [Opposite sides of a parallelogram are equal]
We know that,
Area of parallelogram = Base x corresponding attitude
Area of parallelogram ABCD = CD x AE = AD x CF  
16 cm x 8 cm = AD x 10 cm
AD = cm = 12.8 cm.
Thus, the length of AD is 12.8 cm.
Question 2

In Q. No. 1, if AD = 6 cm, CF = 10 cm, and AE = 8, find AB.

Solution 2



Question 3

Solution 3



Question 4

Solution 4



Chapter 14 - Areas of Parallelograms and Triangles Excercise Ex. 14.3

Question 1

In fig., compute the area of quadrilateral ABCD.

 

 

Solution 1

Question 2

In the fig., PQRS is a square and T and U are, respectively, the mid-points of PS and QR. Find the area of ΔOTS if PQ = 8 cm.

 

Solution 2

Question 3

Compute the area of trapezium PQRS in fig.

 

 

Solution 3

Question 4

In fig., AOB = 90, AC = BC, OA = 12 cm and OC = 6.5 cm. find the area of ΔAOB

 

Solution 4

Question 5

In fig., ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.

 

 

Solution 5

Question 6

Solution 6

Question 7

In fig., ABCD is a trapezium in which AB  DC. PRove that ar (ΔAOD) = ar (ΔBOC)

 

Solution 7

Question 8

Solution 8

Question 9

In fig., ABC and ABD are two triangles on the base Ab. If the line segment CD is bisected by AB at O, show that ar (Δ ABC) = ar (Δ ABD).

 

 

 

Solution 9

Question 10

Solution 10



Question 11

Solution 11



Question 12

Solution 12



Question 13

Solution 13




Question 14

Solution 14



Question 15

In fig., D and E are two points on BC such that BD = DE = EC. Show that ar (ΔABD) = ar (ΔADE) = ar(ΔAEC).

Solution 15

Draw a line l through A parallel to BC.

Given that, BD = DE = EC.

We observe that the triangles ABD, ADE and AEC are on the equal bases and between the same parallels l and BC. Therefore, their areas are equal.

Hence, ar (ΔABD) = ar (ΔADE) = ar(ΔAEC).

Question 16

Solution 16



Question 17

Solution 17



Question 18

Solution 18



Question 19

Solution 19


(i) 

(ii) 
(iii)

Question 20

Solution 20

 

Question 21

In fig., CD  AE and CY ∥ BA.

(i) Name a triangle equal in area of ΔCBX 

(ii) Prove that ar (ΔZDE) = ar (ΔCZA)

(iii) Prove that ar (BCZY) = ar (ΔEDZ)

 

Solution 21

Question 22

In fig., PSDA is a parallelogram in which PQ = QR = RS and AP  BQ CR. Prove that ar(Δ PQE) = ar (Δ CFD).

 

Solution 22

Question 23

In fig., ABCD is a trapezium in which AB DC and DC = 40 cm and AB = 60 cm. If X and Y are, respectively, the mid - points of AD and BC, prove that:

(i) XY = 50 cm

(ii) DCYX is a trapezium

(iii) ar (trap. DCYX) = (9/11)ar (trap.(XYBA)

 


Solution 23



Question 24

Solution 24



Question 25

In fig., X and Y are the mid-points of AC and AB respectively, QP  BC and CYQ and BXP are straight lines. Prove that ar(Δ ABP) = ar (Δ ACQ)

 

Solution 25

Question 26

In fig., ABCD and AEFD are two parallelograms. Prove that

(i) PE = FQ

(ii) ar(ΔAPE) : ar(ΔPFA) = ar Δ(QFD) : ar (ΔPFD)

(iii) ar(ΔPEA) = ar (ΔQFD)

 

 

Solution 26

Question 27

In fig. ABCD is a gm. O is any point on AC. PQ ∥ AB and LM ∥ AD. Prove that ar(gm DLOP) = ar (gm BMOQ).

 

 

Solution 27

Question 28

Solution 28



Question 29

In fig., ABC and BDE are two equilateral triangles such that D is the mid-point of BC. AE intersects BC in F. Prove that

(i) ar (BDE) = ar (ABC)

(ii) ar(BDE) = ar(BAE)

(iii) ar (BFE) = ar(AFD)

(iv) ar(ABC) = 2 ar(BEC)

(v) ar (FED) = ar(AFC)

(vi) ar(BFE) = 2 ar (EFD)

 

Solution 29

Given, ABC and BDE are two equilateral triangles.

Let AB = BC = CA = x. Then, BD = = DE = BE

(i) We have,

ar(ABC) = x2

ar (BDE) =

ar(BDE) = ar (ABC)

(ii) It is given that triangles ABC and BED are equilateral triangles.

ACB = DBE = 60o

BE||AC(Since, alternate angles are equal)

Triangles 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)

[ ED is a median of EBC ar(BEC) = 2ar(BDE)]

ar (BDE) = ar(BAE)

(iii) Since ABC and BDE are equilateral triangles.

ABC = 60o and BDE = 60o

ABC = BDE

AB||DE(Since, alternate angles are equal)

Triangles BED and AED are on the same base ED and between the same parallels AB and DE.

ar (BED) = ar(AED)

ar (BED) ar(EFD) = ar(AED) ar(EFD)

ar(BEF) = ar(AFD)

(iv) Since ED is a median of BEC

ar (BEC) = 2 ar (BDE)

ar (BEC) = ar (ABC)[From (i), ar (BDE) = ar (ABC)]

ar(BEC) = ar (ABC)

ar (ABC) = 2 ar (BEC)

(v) Let h be the height of vertex E, corresponding to the side BD in triangle BDE.

Let H be the height of vertex A, corresponding to the side BC in triangle ABC.

From part (i),

ar(BDE) = ar (ABC)

From part (iii),

ar (BFE) = ar (AFD)

(vi) ar (AFC) = ar (AFD) + ar (ADC)

= ar (BFE) + ar (ABC)

(Using part (iii); and AD is the median of ABC)

= ar (BFE) + 4 ar (BDE)(Using part (i))

= ar (BFE) + 2 ar (BDE) (2)

Now, from part (v),

ar (BFE) = 2ar (FED) (3)

ar (BDE) = ar (BFE) + ar (FED)

= 2 ar (FED) + ar (FED)

= 3 ar (FED) (4)

From (2), (3) and (4), we get,

ar (AFC) = 2ar (FED) + 2 3 ar (FED) = 8 ar (FED)

Hence, ar (FED) = ar(AFC)


Now, from
Question 30

If fig., ABC is a right triangle right angled at A, BCED, ACFG and ABMN are square on the sides BC, CA and AB respectively. Line segment AX DE meets BC at Y. Show that

(i) MBC ABD

(ii) ar (BYXD) = 2ar(MBC)

(iii) ar(BYXD) = ar(ABMN)

(iv) FCB ACE

(v) ar(CYXE) = 2ar (FCB)

(vi) ar(CYXE) = ar (ACFG)

(vii) ar(BCED) = ar (ABMN) + ar (ACFG)

Solution 30

(i) In MBC and ABD, we have

MB = AB

BC = BD

And MBC = ABD

[MBC and ABD are obtained by adding ABC to a right angle]

So, by SAS congruence criterion, we have

MBC ABD

ar (MBC) = ar(ABD) (1)

(ii) Clearly, triangle ABD and rectangle BYXD are on the same base BD and between the same parallels AX and BD.

ar(ABD) = ar (rect. BYXD)

ar (rect. BYXD) = 2 ar(ABD)

ar (rect. BYXD) = 2 ar (MBC)...(2)

[ ar (ABD) = ar (MBC), from (1)]

(iii) Since triangle MBC and square MBAN are on the same base MB and between the same parallels MB and NC.

2 ar (MBC) = ar (MBAN) (3)

From (2) and (3), we have

ar (sq. MBAN) = ar(rec. BYXD)

(iv) In triangles FCB and ACE, we have

FC = AC

CB = CE

And, FCB = ACE

[FCB and ACE are obtained by adding ACB to a right angle]

So, by SAS congruence criterion, we have

FCB ACE

(v) We have,

FCB ACE

ar (FCB) = ar (ACE)

Clearly, ACE and rectangle CYXE are on the same base CE ad between the same parallels CE and AX.

2 ar (ACE) = ar (CYXE)

2 ar (FCB) = ar (CYXE) (4)

(vi) Clearly, FCB and rectangle FCAG are on the same base FC and between the same parallels FC and BG.

2ar (FCB) = ar(FCAG) (5)

From (4) and (5), we get

ar(CYXE) = ar (ACFG)

(vii) Applying Pythagoras theorem in ACB, we have

BC2 = AB2 + AC2

Chapter 14 - Areas of Parallelograms and Triangles Excercise 14.60

Question 1

Two parallelograms are on the same base and between the same parallels. The ratio of their areas is

(a) 1 : 2

(b) 2 : 1

(c) 1 : 1

(d) 3 : 1

Solution 1

Area of parallelogram = Base × height

Base = Length of base

Height = distance between Base and Side parallel to it

In figure, there are two Parallelograms.

Base of both is same, and because both lie under same parallels that's why height is also same.

Thus, the Ratio of Areas of both parallelogram = 1 : 1

Hence, correct option is (c).

Question 2

A triangle and a parallelogram are on the same base and between the same parallels.

The ratio of the areas of triangle and parallelogram is

(a) 1 : 1

(b) 1 : 2

(c) 2 : 1

(d) 1 : 3

Solution 2

begin mathsize 12px style Area space of space straight a space triangle space DFC space equals space 1 half cross times base space cross times height space equals 1 half cross times DC cross times FG space equals 1 half cross times DC space cross times space straight h
Area space of space Parallelogram space ABCD space equals space base space cross times height space equals space DC space cross times AE space equals space DC space cross times space straight h
Required space Ratio space equals space fraction numerator begin display style 1 half cross times up diagonal strike space DC cross times straight h end strike end style over denominator up diagonal strike DC cross times straight h end strike end fraction equals 1 space colon space 2
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 3

Let ABC be a triangle of area 24 sq. units and PQR be the triangle formed by the mid-points of sides of ΔABC.

Then the area of ΔPQR is

(a) 12 sq. units

(b) 6 sq. units

(c) 4 sq. units

(d) 3 sq. units

Solution 3

When a triangle is formed by joining the mid-points of sides of a triangle, the triangle formed is Congruent to triangles formed around that.

i.e. ΔPQR is congruent to ΔRPA, ΔQBP & ΔCQR.

Hence, Area of all four triangles formed inside ΔABC is same.

So (4 × Area of any one Δ) = Area of ΔABC

              4 × (Area of ΔPQR) = 24 sq. units

                    Area of ΔPQR = 6 sq. units

Hence, correct option is (b).

Question 4

The median of a triangle divides it into two

(a) congruent triangles

(b) isosceles triangles

(c) right triangles

(d) triangles of equal areas

Solution 4

A median divides the base in two equal parts but height of a triangle remains the same.

Now, since bases and heights are equal, areas of both Δs are equal.

Hence, correct option is (d).

Question 5

In a ΔABC, D, E, F are the mid-points of sides BC, CA and AB respectively. If ar(ΔABC) = 16 cm2 , then ar(trapezium FBCE) =

(a) 4 cm2

(b) 8 cm2

(c) 12 cm2

(d) 10 cm2

Solution 5

begin mathsize 12px style Area space of space triangle ABC space equals space Area space of space triangle AEF space plus space Area space of space trapazium space FBCE
We space know space that space any space triangle space formed space by space joining space the space mid minus points space of space sides space of space triangle comma
has space area space equals space 1 fourth cross times left parenthesis Parent space triangle right parenthesis
rightwards double arrow Area space of space triangle AEF space equals space 1 fourth cross times Ar left parenthesis triangle ABC right parenthesis space equals space 1 fourth space cross times space 16 space cm squared space equals space 4 space cm squared
rightwards double arrow Area space of space Trapezium space equals space left parenthesis 16 space minus space 4 right parenthesis space cm squared space equals space 12 space cm squared space

Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 6

ABCD is a parallelogram. P is any point on CD. If ar (ΔDPA) = 15 cm2 and ar(ΔAPC) = 20 cm2, then ar(ΔAPB) =

(a) 15 cm2

(b) 20 cm2

(c) 35 cm2

(d) 30 cm2

Solution 6

begin mathsize 12px style Area space of space trapazium space ABCP space equals space Area space of space triangle APB space plus space ar space of space triangle BPC space space... left parenthesis 1 right parenthesis
triangle APC space and space triangle BPC space have space same space base space PC space and space are space between space same space parallels.
rightwards double arrow Area space of space triangle APC space equals space Area space of space triangle BPC space equals space 20 space cm to the power of 2 space space end exponent space space space... left parenthesis 2 right parenthesis
From space figure comma space Ar open parentheses triangle ADP close parentheses space plus space Ar left parenthesis triangle APC right parenthesis space equals space 1 half Ar space left parenthesis parallel to to the power of gm space ABCD right parenthesis
rightwards double arrow Ar left parenthesis parallel to to the power of gm space ABCD right parenthesis space equals space 2 left parenthesis 20 space plus space 15 right parenthesis space equals space 70 space cm squared
Area space of space trapazium space ABCP space equals space Ar space left parenthesis parallel to to the power of gm space ABCD right parenthesis space space minus space Ar left parenthesis triangle ADP right parenthesis space equals 70 space minus space 15 space equals space 55 space cm squared
rightwards double arrow Area space of space triangle APB space equals space Area space of space trapezium space ABCP space minus space Area space of space triangle BPC
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space left parenthesis 55 space minus space 20 right parenthesis space cm squared space space space space left square bracket From space left parenthesis 1 right parenthesis right square bracket
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 35 space cm squared
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Chapter 14 - Areas of Parallelograms and Triangles Excercise 14.61

Question 1

The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 16 cm and 12 cm is

(a) 28 cm2

(b) 48 cm2

(c) 96 cm2

(d) 24 cm2

Solution 1

begin mathsize 12px style AC space equals space 12 space cm space space and space BD space equals space 16 space cm space left parenthesis given right parenthesis
Now comma space consider space triangle ABC.
straight P space and thin space straight Q space are space mid minus points space of space sides space AB space and space BC
So space line space joining space them space will space be space parallel space to space the space third space side space space AC space and space equal space to space 1 half AC.
rightwards double arrow PQ space equals 1 half AC space equals space 6 space cm
Similiary comma space in space triangle ABD comma space PS space parallel to space BD
rightwards double arrow PS space equals space 1 half BD space equals space 8 space cm
Now space we space know space that space by space joining space mid minus points space of space adjacent space sides space of space straight a space Rhombus comma
we space get space straight a space Rectangle space whose space sides space are space PQ space and space PS.
rightwards double arrow Area space equals space PQ space cross times space PS space equals space 6 space cross times space 8 space equals space 48 space cm squared
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 2

A, B, C, D are mid-points of sides of parallelogram PQRS. If ar(PQRS) = 36 cm2, then ar(ABCD) = 

(a) 24 cm2

(b) 18 cm2

(c) 30 cm2

(d) 36 cm2

Solution 2

begin mathsize 12px style straight D space and space straight B space are space joined.
DB space parallel to space PQ space parallel to space RS
Now comma space consider space parallelogram space PQBD space space and space parallelogram space DBRS.
In space PQBD comma space triangle ABD space has space same space base space and space same space height space as space parallelogram space PQBD.
So comma space Area space of space triangle ABD space equals space 1 half cross times Ar left parenthesis PQBD right parenthesis
Similarly comma space Area space of space triangle CDB space equals space 1 half cross times Ar open parentheses RSDB close parentheses
Area space of space open parentheses ABCD close parentheses space equals space Area space of space triangle ABD space plus space Area space of space triangle CDB
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 half left square bracket Ar left parenthesis PQBD right parenthesis space plus space Ar left parenthesis RSDB right parenthesis right square bracket
space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 half Area space of space PQRS space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space
space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 half cross times 36
space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 18 space cm squared
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 3

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is

(a) a rhombus of area 24 cm2

(b) a rectangle of area 24 cm2

(c) a square of area 26 cm2

(d) a trapezium of area 14 cm2

Solution 3

begin mathsize 12px style Figure space obtained space by space joining space the space mid minus points space of space adjacent space sides space of space
rectangle space ABCD space is space straight a space rhombus space PQRS.
AB space equals space 8 space cm comma space AD space equals space 6 space cm
QS space and space PR space are space diagonals space of space Rhombus space PQRS.
QS space equals space AB space equals space 8 space cm
PR space equals space AD space equals space 6 space cm
Ar left parenthesis Rhombus space PQRS right parenthesis space equals space 4 space cross times space Area space of space triangle POR
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 4 space cross times space 1 half cross times space OQ space cross times space OP space space open parentheses triangle POQ space is space straight a space Right space triangle
space OQ space equals space QS over 2 comma space space OP space equals space PR over 2 close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals up diagonal strike 4 squared over 2 cross times 4 cross times 3 space
rightwards double arrow Ar left parenthesis Rhombus space PQRS right parenthesis space equals 24 space cm squared
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 4

If AD is median of ΔABC and P is a point on AC such that

ar(ΔADP) : ar (ΔABC) = 2 : 3, then ar (ΔPDC) : ar(ΔABC) is

(a) 1 : 5

(b) 1 : 5

(c) 1 : 6

(d) 3 : 5

Solution 4

begin mathsize 12px style straight A space median space divides space straight a space triangle space in space two space equal space triangles.
rightwards double arrow Ar left parenthesis ΔABD right parenthesis space equals space Ar left parenthesis ΔADC right parenthesis
Ar left parenthesis triangle PDC right parenthesis space equals space Ar left parenthesis ΔADC right parenthesis space minus space Ar left parenthesis ΔADP right parenthesis
rightwards double arrow Ar left parenthesis ΔPDC right parenthesis space equals space Ar left parenthesis ΔABD right parenthesis space minus space Ar left parenthesis ΔADP right parenthesis space space space space space.... open parentheses 1 close parentheses
Also comma space
Ar left parenthesis ΔABC right parenthesis space equals space 2 space cross times space Ar left parenthesis ΔABD right parenthesis
Dividing space equation space left parenthesis 1 right parenthesis space by space Ar left parenthesis ΔABC right parenthesis comma space we space get
fraction numerator Ar left parenthesis ΔPDC right parenthesis over denominator Ar left parenthesis ΔABC right parenthesis end fraction space equals space fraction numerator Ar left parenthesis ΔABD right parenthesis over denominator Ar left parenthesis ΔABC right parenthesis end fraction space minus space fraction numerator Ar left parenthesis ΔADP right parenthesis over denominator Ar left parenthesis ΔABC right parenthesis end fraction
rightwards double arrow fraction numerator Ar left parenthesis ΔPDC right parenthesis over denominator Ar left parenthesis ΔABC right parenthesis end fraction space equals space fraction numerator Ar left parenthesis up diagonal strike ΔABD right parenthesis over denominator 2 Ar left parenthesis up diagonal strike ΔABD right parenthesis end fraction space minus space fraction numerator Ar left parenthesis ΔADP right parenthesis over denominator 2 Ar left parenthesis ΔABD right parenthesis end fraction space
space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 half minus fraction numerator 1 over denominator up diagonal strike 2 end fraction cross times fraction numerator up diagonal strike 2 over denominator 3 end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 over 6
space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 space colon space 6
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 5

Medians of ΔABC intersect at G. If ar(ΔABC) = 27 cm2, then ar(ΔBGC) =

(a) 6 cm2

(b) 9 cm2

(c) 12 cm2

(d) 18 cm2

Solution 5

begin mathsize 12px style AQ comma space CP space and space RB space are space medians space of space triangle ABC.
Consider space triangle ACP space & space triangle ACQ
Ar open parentheses triangle ACP close parentheses space equals space 27 over 2 cm squared space open curly brackets Median space divides space straight a space triangle space into space two space equal space Area close curly brackets
Ar space left parenthesis triangle ACQ right parenthesis space equals space 27 over 2 cm squared space space open curly brackets Median space divides space straight a space triangle space into space two space equal space Area close curly brackets
rightwards double arrow Ar left parenthesis triangle ACP right parenthesis space equals space Ar open parentheses ACQ close parentheses
Ar left parenthesis triangle AGC right parenthesis space is space common space in space both space the space triangles.
rightwards double arrow Ar left parenthesis triangle CGQ right parenthesis space equals space Ar open parentheses triangle AGP close parentheses space space space space space space space.... open parentheses 1 close parentheses
Similiarly space Ar left parenthesis triangle ABR right parenthesis space equals space 27 over 2 space cm squared space equals space Ar left parenthesis triangle AQB right parenthesis
Ar left parenthesis triangle AGB right parenthesis space is space common space in space both space the space triangels.
rightwards double arrow Ar left parenthesis triangle ARG right parenthesis space equals space Ar left parenthesis triangle GQB right parenthesis space space space space space space space.... open parentheses 2 close parentheses
From space figure space GR comma space GP comma space GQ space are space also space medians space for space triangle AGC comma space triangle AGB space & space triangle CGB space respectively.
rightwards double arrow Ar left parenthesis triangle AGC right parenthesis space plus space Ar left parenthesis triangle AGB right parenthesis space plus space Ar left parenthesis triangle CGB right parenthesis space equals space 27 space cm squared
rightwards double arrow 2 Ar left parenthesis triangle ARG right parenthesis space plus space 2 Ar left parenthesis triangle AGP right parenthesis space plus space Ar left parenthesis triangle BGC right parenthesis space equals 27 space cm squared
rightwards double arrow 2 left parenthesis Ar left parenthesis triangle ARG right parenthesis space plus space Ar left parenthesis triangle AGP right parenthesis right parenthesis space plus space Ar left parenthesis triangle BGC right parenthesis space equals space 27 space cm squared
From space equauations space open parentheses 1 close parentheses space and space open parentheses 2 close parentheses comma
2 left square bracket Ar open parentheses triangle GQB close parentheses plus Ar left parenthesis triangle CGQ right parenthesis right square bracket space plus space Ar left parenthesis triangle BGC right parenthesis space equals space 27 space cm squared
rightwards double arrow 2 left square bracket Ar left parenthesis triangle BGC right parenthesis right square bracket space plus space Ar left parenthesis triangle BGC right parenthesis space equals space 27 space cm squared
rightwards double arrow Ar left parenthesis triangle BGC right parenthesis space equals space 9 space cm squared
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 6

In a ΔABC if D and E are mid-points of BC and AD respectively such that ar(ΔAEC) = 4cm2, then ar(ΔBEC) =

(a) 4 cm2

(b) 6 cm2

(c) 8 cm2

(d) 12 cm2

Solution 6

E is the mid-point of AD and CE is median of ΔACD.

Hence Ar(ΔAEC) = Ar(ΔCED) = 4 cm2    ....(1)

(Median divides a Δ in two two equal Areas)

Also AD is median of ΔABC and and ED is median of ΔBEC.

So Ar(ΔBED) = Ar(CED) = 4 cm2  [From eq (1)]

So Ar(ΔBEC) = Ar(ΔBED) + Ar(ΔCED) = 4 + 4 = 8 cm2

Hence, correct option is (c).

Question 7

In figure, ABCD is a parallelogram. If AB = 12 cm, AE = 7.5 cm, CF = 15 cm, then AD = 

(a) 3 cm

(b) 6 cm

(c) 8 cm

(d) 10.5 cm

Solution 7

Area of parallelogram = AD × FC = AB × AE

Thus,

AD × 15 = 12 × 7.5

AD = 6 cm

Hence, correct option is (b).

Question 8

In figure, PQRS is a parallelogram. If X and Y are mid-points of PQ and SR respectively and diagonal SQ is joined. The ratio ar(||gm XQRY) : ar (ΔQSR) =

(a) 1 : 4

(b) 2 : 1

(c) 1 : 2

(d) 1 : 1

Solution 8

begin mathsize 12px style Diagonal space SQ space divides space parallel to to the power of gm space in space two space equal space areas.
Hence space Ar left parenthesis triangle QSR right parenthesis space equals 1 half space Ar left parenthesis PQRS right parenthesis
Also space XY space divides space the space parallel to to the power of gm space into space two space equal space parts.
Hence comma space Area left parenthesis parallel to to the power of gm XQRY right parenthesis space equals space 1 half space Ar space left parenthesis PQRS right parenthesis
Thus comma space Ratio space of space Ar left parenthesis parallel to to the power of gm XQRY right parenthesis space colon space Ar left parenthesis triangle QSR right parenthesis space equals space 1 space colon space 1
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 9

Diagonal AC and BD of trapezium ABCD, in which AB || DC, intersect each other at O. The triangle which is equal in area of ΔAOD is

(a) ΔAOB

(b) ΔBOC

(c) ΔDOC

(d) ΔADC

Solution 9

ΔABD & ΔABC have same base AB and are between same parallels.

Then,

Ar(ΔABD) = Ar(ΔABC)

But Ar(ΔAOB) is common in both.

Thus, Ar(ΔAOD) = Ar(ΔBOC)

Hence, correct option is (b).

Question 10

ABCD is a trapezium in which AB || DC. If ar(ΔABD)= 24 cm2 and AB = 8 cm, then height of ΔABC is

(a) 3 cm

(b) 4 cm

(c) 6 cm

(d) 8 cm

Solution 10

begin mathsize 12px style triangle ABD space & space triangle ABC space are space on space same space base space AB space and space are space between space same space parallels.
rightwards double arrow Ar left parenthesis triangle ABD right parenthesis space equals space Ar open parentheses triangle ABC close parentheses
Ar open parentheses triangle ABD close parentheses space equals space 1 half cross times 8 cross times straight h space space space equals space 24 space cm squared
rightwards double arrow straight h space equals space 6 space cm
Now space Ar left parenthesis triangle ABC right parenthesis space equals space 1 half cross times 8 cross times straight h space space space equals space 24 space cm squared
rightwards double arrow straight h space equals space 6 space cm
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Chapter 14 - Areas of Parallelograms and Triangles Excercise 14.62

Question 1

begin mathsize 12px style The space mid minus points space of space the space sides space of space straight a space triangle space ABC space along space with space any space of space the space vertices space as space the space fourth
point space make space straight a space parallelogram space of space straight a space area space equal space to
left parenthesis straight a right parenthesis space ar left parenthesis triangle ABC right parenthesis
left parenthesis straight b right parenthesis space 1 half ar open parentheses triangle ABC close parentheses
left parenthesis straight c right parenthesis space 1 third ar space open parentheses triangle ABC close parentheses
left parenthesis straight d right parenthesis space 1 fourth ar left parenthesis triangle ABC right parenthesis end style

Solution 1

begin mathsize 12px style AQRP space is space straight a space required space parallelogram space by space by space joining space the space mid minus points.
All space 4 space triangles space formed space are space congruent space and space space are space equal space in space area.
So space area space of space any space one space triangle space equals space 1 fourth Ar open parentheses triangle ABC close parentheses
Ar left parenthesis triangle APQ right parenthesis space plus space Ar left parenthesis triangle PQR right parenthesis space equals space 1 half Ar open parentheses triangle ABC close parentheses
rightwards double arrow Ar left parenthesis AQRP right parenthesis space equals space 1 half Ar left parenthesis triangle ABC right parenthesis

Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 2

In figure, ABCD and FECG are parallelograms equal in area. If ar(ΔAQE) = 12 cm2, then ar(||gm FGBQ) =

(a) 12 cm2

(b) 20 cm2

(c) 24 cm2

(d) 36 cm2

Solution 2

begin mathsize 12px style Ar left parenthesis parallel to to the power of gm space ABCD right parenthesis space equals space Ar open parentheses parallel to to the power of gm space FECG close parentheses
Ar space of space open parentheses parallel to to the power of gm space QBCE close parentheses space is space common space in space both comma
rightwards double arrow Ar open parentheses parallel to to the power of gm space AQED close parentheses space equals space Ar open parentheses parallel to to the power of gm space FGBQ close parentheses space space space space space space.... open parentheses 1 close parentheses
Now space AE space is space diagonal space of space AQED.
rightwards double arrow Ar open parentheses triangle AQE close parentheses space equals space 1 half space Ar open parentheses parallel to to the power of gm space AQED close parentheses space
rightwards double arrow 12 space cm squared space equals space 1 half Ar open parentheses parallel to to the power of gm space AQED close parentheses space
rightwards double arrow Ar open parentheses parallel to to the power of gm space AQED close parentheses equals space 2 space cross times space 12 space cm space equals space 24 space cm squared
rightwards double arrow space Ar open parentheses parallel to to the power of gm space FGBQ close parentheses space equals space 24 space cm squared space space space space space space space left square bracket From space left parenthesis 1 right parenthesis right square bracket
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 3

ABCD is a trapezium with parallel sides AB = a and DC = b. If E and F are mid-points of non-parallel sides AD and BC respectively, then the ratio of area of quadrilaterals ABFE and EFCD is

(a) a : b

(b) (a + 3b) : (3a + b)

(c) (3a + b) : (a + 3b)

(d) (2a + b) : (3a + b)

Solution 3

begin mathsize 12px style AP space is space drawn space parallel space to space BC. space
ABCP space is space straight a space parallelogram.
AB space equals space PC space equals space straight a
DP space equals space DC space minus space PC space equals space straight b space minus space straight a
Area space left parenthesis ABFE right parenthesis space equals space Ar open parentheses triangle AQE close parentheses space plus space Ar open parentheses parallel to to the power of gm space ABFQ close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 fourth open parentheses Ar open parentheses triangle ADP close parentheses close parentheses space plus space 1 half space Ar space open parentheses parallel to to the power of gm space ABCP close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space equals 1 fourth cross times 1 half cross times open parentheses straight b space minus space straight a close parentheses straight h space plus space 1 half cross times space straight a space cross times space straight h
space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator open parentheses 3 straight a space plus space straight b close parentheses straight h over denominator 8 end fraction
Siimiliarly comma space Area space of space trapazium space EFCD
equals space Ar left parenthesis EQPD right parenthesis space plus space Ar open parentheses parallel to to the power of gm space QFCP close parentheses
equals space 3 over 4 Ar open parentheses triangle ADP close parentheses space plus space 1 half open parentheses parallel to to the power of gm space ABCP close parentheses
equals space 3 over 4 cross times 1 half cross times open parentheses straight b space minus space straight a close parentheses space cross times space straight h space plus space 1 half cross times straight a cross times straight h
equals space fraction numerator open parentheses 3 straight b plus straight a close parentheses straight h over denominator 8 end fraction
rightwards double arrow Ratio space of space Ar space left parenthesis Quad space ABFE right parenthesis space colon space Ar left parenthesis Quad space EFCD right parenthesis space equals space fraction numerator open parentheses 3 straight a plus straight b close parentheses straight h over denominator 8 end fraction space space colon space fraction numerator open parentheses 3 straight b plus straight a close parentheses straight h over denominator 8 end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open parentheses 3 straight a plus straight b close parentheses space colon space open parentheses straight a plus 3 straight b close parentheses space space
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 4

ABCD is a rectangle with O as any point in its interior. If ar (ΔAOD) = 3 cm2 and ar (ΔBOC) = 6 cm2, then area of rectangle ABCD is

(a) 9 cm2

(b) 12 cm2

(c) 15 cm2

(d) 18 cm2

Solution 4

begin mathsize 12px style straight A space line space PQ space is space drawn space from space AB space parallel space to space AD space & space BC.
Now comma space triangle AOD space has space height space equals space AP
And comma space triangle BOC space has space height space equals space BP
Area space of space triangle AOD space equals space 1 half cross times AD cross times AP space equals 3 space cm squared
rightwards double arrow AD space cross times space AP space equals space 6 space cm squared space space space.... open parentheses 1 close parentheses
Ar open parentheses triangle BOC close parentheses space equals space 1 half cross times BC cross times BP space equals 6 space cm squared
rightwards double arrow BC space cross times space BP space equals space 12 space cm squared space space space space space.... open parentheses 2 close parentheses
Adding space equations space open parentheses 1 close parentheses space space and space open parentheses 2 close parentheses comma space we space get
AD space cross times space AP space plus space BC space cross times space BP space equals space 18 space cm squared
rightwards double arrow AD space cross times space AP space plus space AD space cross times space BP space equals space 18 space cm squared space space space space left parenthesis AD space equals space BC right parenthesis
rightwards double arrow AD left parenthesis AP space plus space BP right parenthesis space equals space 18 space cm squared
rightwards double arrow AD cross times AB space equals space 18 space cm squared space equals space Area space of space rectangle space ABCD
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

CBSE Class 9 Maths Homework Help

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