RD SHARMA SOLUTION CHAPTER -14 Areas of Parallelograms and Triangles| CLASS 9TH MATHEMATICS-EDUGROWN

Chapter 14 – Areas of Parallelograms and Triangles Exercise 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 Exercise Ex. 14.2

Question 1In 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 1In 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 cmAD =  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 Exercise 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 = 60^o

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 = 60^o and BDE = 60^o

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, fromQuestion 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