# RD Sharma Solutions for Class 9 Maths Chapter 15 Area of Parallelograms and Triangles

## RD Sharma Solutions Class 9 Maths Chapter 15 – Free PDF Download

RD Sharma Solutions for Class 9 Maths Chapter 15 Areas Of Parallelograms And Triangles is a set of answers to all the exercise questions. In this chapter, students will study about Areas of Parallelograms and Triangles. Parallelogram, in Geometry, is a non-self-intersecting quadrilateral with two pairs of parallel sides. A trapezoid is a quadrilateral with only one pair of parallel sides. In a parallelogram, the diagonals bisect each other. Whereas, a triangle is a polygon with three vertices and three edges. A triangle can be classified into the isosceles triangle, equilateral triangle, and scalene triangle. The diagonal of a parallelogram are of equal lengths and each diagonal of a parallelogram separates it into two congruent triangles.

In this chapter, all the questions on triangles and parallelograms are formulated by expert teachers as per the CBSE 2021-22 syllabus. Students are advised to download the RD Sharma Class 9 Solutions PDF and start practicing to score good marks in the examinations. The RD Sharma Solutions for exercise-wise problems are prepared after vast research is conducted on each topic. The clear format of solutions helps students to solve problems with competence.

## Download PDF of RD Sharma Solutions for Class 9 Maths Chapter 15 Area of Parallelograms and Triangles

### Exercise 15.1 Page No: 15.3

Question 1: Which of the following figures lie on the same base and between the same parallel. In such a case, write the common base and two parallels:

Solution:

(i) Triangle APB and trapezium ABCD are on the common base AB and between the same parallels AB and DC.

So,

Common base = AB

Parallel lines: AB and DC

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

(iii) Consider, parallelogram ABCD and Î”PQR, lies between the same parallels AD and BC. But not sharing common base.

(iv) Î”QRT and parallelogram PQRS are on the same base QR and lies between same parallels QR and PS.

Common base = QR

Parallel lines: QR and PS

(v) Parallelograms PQRS and trapezium SMNR share common base SR, but not between the same parallels.

(vi) Parallelograms: PQRS, AQRD, BCQR are between the same parallels. Also,

Parallelograms: PQRS, BPSC, APSD are between the same parallels.

### Exercise 15.2 Page No: 15.14

Question 1: If figure, ABCD is a parallelogram, AE âŠ¥ DC and CF âŠ¥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

Solution:

In parallelogram ABCD, AB = 16 cm, AE = 8 cm and CF = 10 cm

Since, opposite sides of a parallelogram are equal, then

AB = CD = 16 cm

We know, Area of parallelogram = Base x Corresponding height

Area of parallelogram ABCD:

CD x AE = AD x CF

16 x 18 = AD x 10

Measure of AD = 12.8 cm

Question 2: In Q.No. 1, if AD = 6 cm, CF = 10 cm and AE = 8 cm, find AB.

Solution: Area of a parallelogram ABCD:

From figure:

AD Ã— CF = CD Ã— AE

6 x 10 = CD x 8

CD = 7.5

Since, opposite sides of a parallelogram are equal.

=> AB = DC = 7.5 cm

Question 3: Let ABCD be a parallelogram of area 124 cm2. If E and F are the mid-points of sides AB and CD respectively, then find the area of parallelogram AEFD.

Solution:

ABCD be a parallelogram.

Area of parallelogram = 124 cm2 (Given)

Consider a point P and join AP which is perpendicular to DC.

Now, Area of parallelogram EBCF = FC x AP and

Area of parallelogram AFED = DF x AP

Since F is the mid-point of DC, so DF = FC

From above results, we have

Area of parallelogram AEFD = Area of parallelogram EBCF = 1/2 (Area of parallelogram ABCD)

= 124/2

= 62

Area of parallelogram AEFD is 62 cm2.

Question 4: If ABCD is a parallelogram, then prove that

ar(Î” ABD) = ar(Î” BCD) = ar(Î” ABC)=ar(Î” ACD) = 1/2 ar(||gm ABCD)

Solution:

ABCD is a parallelogram.

When we join the diagonal of parallelogram, it divides it into two quadrilaterals.

Step 1: Let AC is the diagonal, then, Area (Î”ABC) = Area (Î”ACD) = 1/2(Area of llgm ABCD)

Step 2: Let BD be another diagonal

Area (Î”ABD) = Area (Î”BCD) = 1/2( Area of llgm ABCD)

Now,

From Step 1 and step 2, we have

Area (Î”ABC) = Area (Î”ACD) = Area (Î”ABD) = Area (Î”BCD) = 1/2(Area of llgm ABCD)

Hence Proved.

### Exercise 15.3 Page No: 15.40

Question 1: In figure, compute the area of quadrilateral ABCD.

Solution:

A quadrilateral ABCD with DC = 17 cm, AD = 9 cm and BC = 8 cm (Given)

In right Î”ABD,

Using Pythagorean Theorem,

152 = AB2 + 92

AB2 = 225âˆ’81=144

AB = 12

Area of Î”ABD = 1/2(12Ã—9) cm2 = 54 cm2

In right Î”BCD:

Using Pythagorean Theorem,

CD2 = BD2 + BC2

172 = BD2 + 82

BD2 = 289 â€“ 64 = 225

or BD = 15

Area of Î”BCD = 1/2(8×17) cm2 = 68 cm2

Now, area of quadrilateral ABCD = Area of Î”ABD + Area of Î”BCD

= 54 cm2 + 68 cm2

= 112 cm2

Question 2: In figure, PQRS is a square and T and U are, respectively, the mid-points of PS and QR . Find the area of Î”OTS if PQ = 8 cm.

Solution:

T and U are mid points of PS and QR respectively (Given)

Therefore, TU||PQ => TO||PQ

In Î”PQS ,

T is the mid-point of PS and TO||PQ

So, TO = 1/2 PQ = 4 cm

(PQ = 8 cm given)

Also, TS = 1/2 PS = 4 cm

[PQ = PS, As PQRS is a square)

Now,

Area of Î”OTS = 1/2(TOÃ—TS)

= 1/2(4Ã—4) cm2

= 8cm2

Area of Î”OTS is 8 cm2.

Question 3: Compute the area of trapezium PQRS in figure.

Solution:

From figure,

Area of trapezium PQRS = Area of rectangle PSRT + Area of Î”QRT

= PT Ã— RT + 1/2 (QTÃ—RT)

= 8 Ã— RT + 1/2(8Ã—RT)

= 12 RT

In right Î”QRT,

Using Pythagorean Theorem,

QR2 = QT2 + RT2

RT2 = QR2 âˆ’ QT2

RT2 = 172âˆ’82 = 225

or RT = 15

Therefore, Area of trapezium = 12Ã—15 cm2 = 180 cm2

Question 4: In figure, âˆ AOB = 90o, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of Î”AOB.

Solution:

Given: A triangle AOB, with âˆ AOB = 90o, AC = BC, OA = 12 cm and OC = 6.5 cm

As we know, the midpoint of the hypotenuse of a right triangle is equidistant from the vertices.

So, CB = CA = OC = 6.5 cm

AB = 2 CB = 2 x 6.5 cm = 13 cm

In right Î”OAB:

Using Pythagorean Theorem, we get

AB2 = OB2 + OA2

132 = OB2 + 122

OB2 = 169 â€“ 144 = 25

or OB = 5 cm

Now, Area of Î”AOB = Â½(Base x height) cm2 = 1/2(12 x 5) cm2 = 30cm2

Question 5: In figure, 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:

Given: ABCD is a trapezium, where AB = 7 cm, AD = BC = 5 cm, DC = x cm, and

Distance between AB and DC = 4 cm

Consider AL and BM are perpendiculars on DC, then

AL = BM = 4 cm and LM = 7 cm.

In right Î”BMC :

Using Pythagorean Theorem, we get

BC2 = BM2 + MC2

25 = 16 + MC2

MC2 = 25 â€“ 16 = 9

or MC = 3

Again,

Using Pythagorean Theorem, we get

25 = 16 + DL2

DL2 = 25 â€“ 16 = 9

or DL = 3

Therefore, x = DC = DL + LM + MC = 3 + 7 + 3 = 13

=> x = 13 cm

Now,

Area of trapezium ABCD = 1/2(AB + CD) AL

= 1/2(7+13)4

= 40

Area of trapezium ABCD is 40 cm2.

Question 6: In figure, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If OE = 2âˆš5 cm, find the area of the rectangle.

Solution:

From given:

Radius = OD = 10 cm and OE = 2âˆš5 cm

In right Î”DEO,

By Pythagoras theorem

OD2 = OE2 + DE2

(10)2 = (2âˆš5 )2 + DE2

100 â€“ 20 = DE2

DE = 4âˆš5

Now,

Area of rectangle OCDE = Length x Breadth = OE x DE = 2âˆš5 x 4âˆš5 = 40

Area of rectangle is 40 cm2.

Question 7: In figure, ABCD is a trapezium in which AB || DC. Prove that ar(Î”AOD) = ar(Î”BOC)

Solution:

ABCD is a trapezium in which AB || DC (Given)

To Prove: ar(Î”AOD) = ar(Î”BOC)

Proof:

From figure, we can observe that Î”ADC and Î”BDC are sharing common base i.e. DC and between same parallels AB and DC.

Î”ADC is the combination of triangles, Î”AOD and Î”DOC. Similarly, Î”BDC is the combination of Î”BOC and Î”DOC triangles.

Equation (1) => ar(Î”AOD) + ar(Î”DOC) = ar(Î”BOC) + ar(Î”DOC)

or ar(Î”AOD) = ar(Î”BOC)

Hence Proved.

Question 8: In figure, ABCD, ABFE and CDEF are parallelograms. Prove that ar(Î”ADE) = ar(Î”BCF).

Solution:

Here, ABCD, CDEF and ABFE are parallelograms:

Which implies:

DE = CF and

AE = BF

Again, from triangles ADE and BCF:

AD = BC, DE = CF and AE = BF

By SSS criterion of congruence, we have

Since both the triangles are congruent, then ar(Î”ADE) = ar(Î”BCF).

Hence Proved,

Question 9: Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that: ar(Î”APB) x ar(Î”CPD) = ar(Î”APD) x ar(Î”BPC).

Solution:

Consider: BQ and DR are two perpendiculars on AC.

To prove: ar(Î”APB) x ar(Î”CPD) = ar(Î”APD) x ar(Î”BPC).

Now,

L.H.S. = ar(Î”APB) x ar(Î”CDP)

= (1/2 x AP Ã— BQ) Ã— (1/2 Ã— PC Ã— DR)

= (1/2 x PC Ã— BQ) Ã— (1/2 Ã— AP Ã— DR)

= ar(Î”APD) x ar(Î”BPC)

= R.H.S.

Hence proved.

Question 10: In figure, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that ar(Î”ABC) = ar(Î”ABD).

Solution:

Draw two perpendiculars CP and DQ on AB.

Now,

ar(Î”ABC) = 1/2Ã—ABÃ—CP â‹…â‹…â‹…â‹…â‹…â‹…â‹…(1)

ar(Î”ABD) = 1/2Ã—ABÃ—DQ â‹…â‹…â‹…â‹…â‹…â‹…â‹…(2)

To prove the result, ar(Î”ABC) = ar(Î”ABD), we have to show that CP = DQ.

In right angled triangles, Î”CPO and Î”DQO

âˆ CPO = âˆ DQO = 90o

CO = OD (Given)

âˆ COP = âˆ DOQ (Vertically opposite angles)

By AAS condition: Î”CP0 â‰… Î”DQO

So, CP = DQ â€¦â€¦â€¦â€¦..(3)

(By CPCT)

From equations (1), (2) and (3), we have

ar(Î”ABC) = ar(Î”ABD)

Hence proved.

### Exercise VSAQs Page No: 15.55

Question 1: If ABC and BDE are two equilateral triangles such that D is the mid-point of BC, then find ar(â–³ABC) : ar(â–³BDE).

Solution:

Given: ABC and BDE are two equilateral triangles.

We know, area of an equilateral triangle = âˆš3/4 (side)2

Let “a” be the side measure of given triangle.

Find ar(â–³ABC):

ar(â–³ABC) = âˆš3/4 (a) 2

Find ar(â–³BDE):

ar(â–³BDE) = âˆš3/4 (a/2) 2

(D is the mid-point of BC)

Now,

ar(â–³ABC) : ar(â–³BDE)

or âˆš3/4 (a) 2 : âˆš3/4 (a/2) 2

or 1 : 1/4

or 4:1

This implies, ar(â–³ABC) : ar(â–³BDE) = 4:1

Question 2: In figure, ABCD is a rectangle in which CD = 6 cm, AD = 8 cm. Find the area of parallelogram CDEF.

Solution:

ABCD is a rectangle, where CD = 6 cm and AD = 8 cm (Given)

From figure: Parallelogram CDEF and rectangle ABCD on the same base and between the same parallels, which means both have equal areas.

Area of parallelogram CDEF = Area of rectangle ABCD ….(1)

Area of rectangle ABCD = CD x AD = 6 x 8 cm2 = 48 cm2

Equation (1) => Area of parallelogram CDEF = 48 cm2.

Question 3: In figure, find the area of Î”GEF.

Solution:

From figure:

Parallelogram CDEF and rectangle ABCD on the same base and between the same parallels, which means both have equal areas.

Area of CDEF = Area of ABCD = 8 x 6 cm2= 48 cm2

Again,

Parallelogram CDEF and triangle EFG are on the same base and between the same parallels, then

Area of a triangle = Â½(Area of parallelogram)

In this case,

Area of a triangle EFG = Â½(Area of parallelogram CDEF) = 1/2(48 cm2) = 24 cm2.

Question 4: In figure, ABCD is a rectangle with sides AB = 10 cm and AD = 5 cm. Find the area of Î”EFG.

Solution:

From figure:

Parallelogram ABEF and rectangle ABCD on the same base and between the same parallels, which means both have equal areas.

Area of ABEF = Area of ABCD = 10 x 5 cm2= 50 cm2

Again,

Parallelogram ABEF and triangle EFG are on the same base and between the same parallels, then

Area of a triangle = Â½(Area of parallelogram)

In this case,

Area of a triangle EFG = Â½(Area of parallelogram ABEF) = 1/2(50 cm2) = 25 cm2.

Question 5: PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then find ar(â–³RAS).

Solution:

PQRS is a rectangle with PS = 5 cm and PR = 13 cm (Given)

In â–³PSR:

Using Pythagoras theorem,

SR2 = PR2 â€“ PS2 = (13) 2 â€“ (5) 2 = 169 â€“ 25 = 114

or SR = 12

Now,

Area of â–³RAS = 1/2 x SR x PS

= 1/2 x 12 x 5

= 30

Therefore, Area of â–³RAS is 30 cm2.

## RD Sharma Solutions for Class 9 Maths Chapter 15 Area of Parallelograms and Triangles

In the 15th Chapter of Class 9,Â Â RD Sharma Solutions students will study important concepts. Some are listed below:

• Area of Parallelograms and Triangles Introduction.
• Figures on the same base and between the same parallels
• Geometric figures Regions
• Area Axioms
• Parallelogram on the same base and between the same parallels

### Key benefits of studying RD Sharma books

• RD Sharma books provide several questions to obtain confidence in solving complex problems with ease. This develops critical thinking and strengthens the mental ability of the students.
• These solutions facilitate the students to have a good knowledge of basic as well as advanced mathematical concepts. It also helps students in retaining and quickly retrieving the concepts.
• The illustrations provided in RD Sharma books are offered in a step by step approach for a comfortable and better understanding of concepts. The accurate answers also boost to grasp the methods of solving the problems effortlessly.
• These solutions are the best resource materials for students who aspire to gain proficiency in the subject and score good marks in academics.

## Frequently Asked Questions on RD Sharma Solutions for Class 9 Maths Chapter 15

### Why RD Sharma Solutions for Class 9 Maths Chapter 15 is important from an exam point of view?

RD Sharma Solutions for Class 9 Maths Chapter 15 Areas of Parallelogram and Triangles consist of important questions which help to secure high marks. The concepts present in this chapter are essential not only for board exams but also for other competitive exams as well. Subject matter experts explained the concepts in a simple and easy manner so that students understand them more effectively.

### Is it necessary to practice all the questions provided in RD Sharma Solutions for Class 9 Maths Chapter 15?

Yes, practising all the questions provided in RD Sharma Solutions for Class 9 Maths Chapter 15 boosts skills required to secure more marks in exams. These solutions are provided in simple language so that students solve the problems without any difficulty and grasp the concepts in-depth. Subject experts designed the solutions in a descriptive manner as per the current CBSE board.

### How does RD Sharma Solutions for Class 9 Maths Chapter 15 Areas of Parallelogram and Triangles boost exam preparation among students?

RD Sharma Solutions for Class 9 Maths Chapter 15 are solved in a detailed manner with illustrations to help students in effective exam preparation. The solutions prepared by experts clear the doubts immediately and boost conceptual knowledge among students. By practising these solutions on a regular basis, students not only grasp the concepts thoroughly but also procure more marks in board exams.