RD SHARMA Solutions for Class 9 Maths Chapter 14 - Areas of Parallelograms and Triangles

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:

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

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 1
In the given figure, ABCD is parallelogram, AE Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles DC and CF Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles AD. If AB = 16 cm. AE = 8 cm and CF = 10 cm, find AD. 
 
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
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 = Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles 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

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 3

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 3

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 4

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 4

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

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

Question 1

In fig., compute the area of quadrilateral ABCD.

 

 

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 1

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

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.

 

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 2

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 3

Compute the area of trapezium PQRS in fig.

 

 

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 3

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 4

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

 

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 4

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

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.

 

 

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 5

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 6

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 6

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 7

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

 

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 7

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 8

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 8

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

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).

 

 

 

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 9

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 10

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 10

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 11

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 11

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 12

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 12

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 13

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 13




Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 14

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 14

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

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).

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 15

Draw a line l through A parallel to BC.

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

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

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 16

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 17

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 17

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 18

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 18

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 19

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 19

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
(i) 
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

(ii) 
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
(iii)
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Question 20

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 20

 Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
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)

 

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 21

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

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).

 

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 22

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

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)

 


Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 23

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Question 24

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 24

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

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)

 

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 25

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

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)

 

 

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 26

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

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).

 

 

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles
Solution 27

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 28

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 28

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

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 (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBDE) = Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles(ABC)

(ii) ar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBDE) = Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBAE)

(iii) ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBFE) = ar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAFD)

(iv) ar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC) = 2 ar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBEC)

(v) ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesFED) = Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar(AFC)

(vi) ar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBFE) = 2 ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesEFD)

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

 

Solution 29

Given, ABC and BDE are two equilateral triangles.

Let AB = BC = CA = x. Then, BD = Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles= DE = BE

(i) We have,

ar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC) = Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesx2

ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBDE) = Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBDE) = Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC)

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

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesACB = Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDBE = 60o

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBE||AC(Since, alternate angles are equal)

Triangles BAE and BEC are on the same base BE and between the same parallels BE and AC.

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBAE) = ar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBEC)

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBAE) =2 ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBDE)

[Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles ED is a median of Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesEBC Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBEC) = 2ar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBDE)]

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBDE) = Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBAE)

(iii) Since Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC and Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBDE are equilateral triangles.

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesRd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC = 60o and Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBDE = 60o

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesRd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC = Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBDE

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAB||DE(Since, alternate angles are equal)

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

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBED) = ar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAED)

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBED) ar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesEFD) = ar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAED) ar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesEFD)

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBEF) = ar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAFD)

(iv) Since ED is a median of Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBEC

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBEC) = 2 ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBDE)

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBEC) = Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC)[From (i), ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBDE) = Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC)]

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBEC) = Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC)

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC) = 2 ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBEC)

(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(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBDE) = Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC)

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

From part (iii),

ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBFE) = ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAFD)

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

(vi) ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAFC) = ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAFD) + ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesADC)

= ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBFE) + Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC)

(Using part (iii); and AD is the median of Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC)

= ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBFE) + Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles4 ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBDE)(Using part (i))

= ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBFE) + 2 ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBDE) (2)

Now, from part (v),

ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBFE) = 2ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesFED) (3)

ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBDE) = ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesBFE) + ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesFED)

= 2 ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesFED) + ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesFED)

= 3 ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesFED) (4)

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

ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesAFC) = 2ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesFED) + 2 Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles3 ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesFED) = 8 ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesFED)

Hence, ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesFED) = Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar(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 Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesDE meets BC at Y. Show that

(i) Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesMBC Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesRd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABD

(ii) ar (BYXD) = 2ar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesMBC)

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

(iv) Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesFCB Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesRd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesACE

(v) ar(CYXE) = 2ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesFCB)

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

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

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 30

(i) In Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesMBC and Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABD, we have

MB = AB

BC = BD

And Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesMBC = Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABD

[Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesRd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesMBC and Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABD are obtained by adding Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABC to a right angle]

So, by SAS congruence criterion, we have

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesMBC Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesRd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABD

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesMBC) = ar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABD) (1)

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

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABD) = Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar (rect. BYXD)

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar (rect. BYXD) = 2 ar(Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABD)

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar (rect. BYXD) = 2 ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesMBC)...(2)

[Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesABD) = ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesMBC), from (1)]

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

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles2 ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesMBC) = 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, Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesFCB = Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesACE

[Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesRd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesFCB and Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesACE are obtained by adding Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesACB to a right angle]

So, by SAS congruence criterion, we have

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesFCB Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesRd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesACE

(v) We have,

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesFCB Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesRd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesACE

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Trianglesar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesFCB) = ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesACE)

Clearly, Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesACE and rectangle CYXE are on the same base CE ad between the same parallels CE and AX.

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles2 ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesACE) = ar (CYXE)

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles2 ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesFCB) = ar (CYXE) (4)

(vi) Clearly, Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesFCB and rectangle FCAG are on the same base FC and between the same parallels FC and BG.

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles2ar (Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesFCB) = ar(FCAG) (5)

From (4) and (5), we get

ar(CYXE) = ar (ACFG)

(vii) Applying Pythagoras theorem in Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And TrianglesACB, we have

BC2 = AB2 + AC2

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Chapter 14 - Areas of Parallelograms and Triangles Exercise 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

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

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

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

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

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

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

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

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

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Chapter 14 - Areas of Parallelograms and Triangles Exercise 14.61

Question 7

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 7

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 8

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 8

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 9

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 9

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 10

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 10

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 11

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 11

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 12

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 12

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

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 13

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

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 13

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Area of parallelogram = AD × FC = AB × AE

Thus,

AD × 15 = 12 × 7.5

AD = 6 cm

Hence, correct option is (b).

Question 14

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

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 14

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 15

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 15

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

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 16

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 16

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Chapter 14 - Areas of Parallelograms and Triangles Exercise 14.62

Question 19

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 19

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 20

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

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Solution 20

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 17

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 17

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Question 18

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 18

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Areas Of Parallelograms And Triangles