## NCERT Solutions Class 9 Maths Chapter 15 – Free PDF Download

**RD Sharma Solutions for Class 9 Maths Chapter 15Â Areas Of Parallelograms And Triangles** is a set of answers to all the exercise questions. In this chapter, students will study about area 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.

We will be seeing questions on triangles and parallelograms which are covered in every exercise given below. Students are advised to download the RD Sharma Class 9 Solutions pdf and start practicing to score good marks in the examinations.Â The **RD Sharma Solutions** for exercise wise problems are prepared after vast research is conducted on each topic. The step-wise solutions prepared by subject experts help 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, 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 offer several questions for practice at the end of each chapter. 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.
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**RD Sharma books**are offered in a step by step approach for a comfortable and better understanding of concepts. - These are the one of the best resource materials currently available for students who aspire to score well in Maths.