## 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

### Access Answers to Maths RD Sharma Solutions for Class 9 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.

Common base = *AD*

Parallel lines: *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

AD = 12.8

*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 cm ^{2}. 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 cm^{2} (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 cm ^{2}.*

**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 ll^{gm} ABCD)

Step 2: Let BD be another diagonal

Area (Î”ABD) = Area (Î”BCD) = 1/2( Area of ll^{gm} 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 ll ^{gm} 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,

AB^{2 }+ AD^{2 }= BD^{2}

15^{2} = AB^{2} + 9^{2}

AB^{2 }= 225âˆ’81=144

AB = 12

Area of Î”ABD = 1/2(12Ã—9) cm^{2 }= 54 cm^{2}

In right Î”BCD:

Using Pythagorean Theorem,

CD^{2 }= BD^{2} + BC^{2}

17^{2} = BD^{2} + 8^{2}

BD^{2} = 289 â€“ 64 = 225

or BD = 15

Area of Î”BCD = 1/2(8×17) cm^{2 }= 68 cm^{2}

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

= 54 cm^{2 }+ 68 cm^{2}

*= 112 cm ^{2}*

**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) cm^{2}

= 8cm^{2}

*Area of Î”OTS is 8 cm ^{2}.*

**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,

QR^{2} = QT^{2} + RT^{2}

RT^{2} = QR^{2} âˆ’ QT^{2}

RT^{2} = 17^{2}âˆ’8^{2} = 225

or RT = 15

*Therefore, Area of trapezium = 12Ã—15 cm ^{2} = 180 cm^{2}*

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

**Solution:**

Given: A triangle AOB, with âˆ AOB = 90^{o}, 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

AB^{2} = OB^{2} + OA^{2}

13^{2 }= OB^{2 }+ 12^{2}

OB^{2} = 169 â€“ 144 = 25

or OB = 5 cm

*Now, Area of Î”AOB = Â½(Base x height) cm ^{2} = 1/2(12 x 5) cm^{2 }= 30cm^{2}*

**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

BC^{2} = BM^{2} + MC^{2}

25 = 16 + MC^{2}

MC^{2} = 25 â€“ 16 = 9

or MC = 3

Again,

In right Î” ADL :

Using Pythagorean Theorem, we get

AD^{2} = AL^{2} + DL^{2}

25 = 16 + DL^{2}

DL^{2} = 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 cm ^{2}.*

**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

OD^{2} = OE^{2} + DE^{2}

(10)^{2} = (2âˆš5 )^{2} + DE^{2}

100 â€“ 20 = DE^{2}

DE = 4âˆš5

Now,

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

*Area of rectangle is 40 cm ^{2}.*

**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.

Then, ar(Î”ADC) = ar(Î”BDC) â€¦â€¦(1)

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:

AD = BC

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

Î”ADE â‰… Î”BCF

*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 = 90^{o}

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 cm^{2 }= 48 cm^{2}

*Equation (1) => Area of parallelogram CDEF = 48 cm ^{2}.*

**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 cm^{2}= 48 cm^{2}

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 cm ^{2}) = 24 cm^{2}.*

**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 cm^{2}= 50 cm^{2}

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 cm ^{2}) = 25 cm^{2}.*

**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,

SR^{2} = PR^{2} â€“ PS^{2} = (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 cm ^{2}.*

**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.