## 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 diagonals of a parallelogram are of equal length 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** 2023-24 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.

## 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 ΔPQR, lies between the same parallels AD and BC. But not sharing common base.

**(iv)** ΔQRT 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 ⊥ DC and CF ⊥ 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 × CF = CD × 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(Δ ABD) = ar(Δ BCD) = ar(Δ ABC)=ar(Δ 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 (ΔABC) = Area (ΔACD) = 1/2(Area of ll^{gm} ABCD)

Step 2: Let BD be another diagonal

Area (ΔABD) = Area (ΔBCD) = 1/2( Area of ll^{gm} ABCD)

Now,

From Step 1 and step 2, we have

*Area (ΔABC) = Area (ΔACD) = Area (ΔABD) = Area (ΔBCD) = 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 ΔABD,

Using Pythagorean Theorem,

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

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

AB^{2 }= 225−81=144

AB = 12

Area of ΔABD = 1/2(12×9) cm^{2 }= 54 cm^{2}

In right ΔBCD:

Using Pythagorean Theorem,

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

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

BD^{2} = 289 – 64 = 225

or BD = 15

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

Now, area of quadrilateral ABCD = Area of ΔABD + Area of ΔBCD

= 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 ΔOTS if PQ = 8 cm.**

**Solution:**

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

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

In ΔPQS ,

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 ΔOTS = 1/2(TO×TS)

= 1/2(4×4) cm^{2}

= 8cm^{2}

*Area of ΔOTS 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 ΔQRT

= PT × RT + 1/2 (QT×RT)

= 8 × RT + 1/2(8×RT)

= 12 RT

In right ΔQRT,

Using Pythagorean Theorem,

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

RT^{2} = QR^{2} − QT^{2}

RT^{2} = 17^{2}−8^{2} = 225

or RT = 15

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

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

**Solution:**

Given: A triangle AOB, with ∠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 ΔOAB:

Using Pythagorean Theorem, we get

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

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

OB^{2} = 169 – 144 = 25

or OB = 5 cm

*Now, Area of ΔAOB = ½(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 ΔBMC :

Using Pythagorean Theorem, we get

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

25 = 16 + MC^{2}

MC^{2} = 25 – 16 = 9

or MC = 3

Again,

In right Δ ADL :

Using Pythagorean Theorem, we get

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

25 = 16 + DL^{2}

DL^{2} = 25 – 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 = 2√5 cm, find the area of the rectangle.**

**Solution**:

From given:

Radius = OD = 10 cm and OE = 2√5 cm

In right ΔDEO,

By Pythagoras theorem

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

(10)^{2} = (2√5 )^{2} + DE^{2}

100 – 20 = DE^{2}

DE = 4√5

Now,

Area of rectangle OCDE = Length x Breadth = OE x DE = 2√5 x 4√5 = 40

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

**Question 7: In figure, ABCD is a trapezium in which AB || DC. Prove that ar(ΔAOD) = ar(ΔBOC)**

**Solution:**

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

To Prove: ar(ΔAOD) = ar(ΔBOC)

Proof:

From figure, we can observe that ΔADC and ΔBDC are sharing common base i.e. DC and between same parallels AB and DC.

Then, ar(ΔADC) = ar(ΔBDC) ……(1)

ΔADC is the combination of triangles, ΔAOD and ΔDOC. Similarly, ΔBDC is the combination of ΔBOC and ΔDOC triangles.

Equation (1) => ar(ΔAOD) + ar(ΔDOC) = ar(ΔBOC) + ar(ΔDOC)

or *ar(ΔAOD) = ar(ΔBOC)*

Hence Proved.

**Question 8: In figure, ABCD, ABFE and CDEF are parallelograms. Prove that ar(ΔADE) = ar(ΔBCF).**

**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

ΔADE ≅ ΔBCF

*Since both the triangles are congruent, then ar(ΔADE) = ar(ΔBCF).*

Hence Proved,

**Question 9: Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that: ar(ΔAPB) x ar(ΔCPD) = ar(ΔAPD) x ar(ΔBPC).**

**Solution:**

Consider: BQ and DR are two perpendiculars on AC.

To prove: ar(ΔAPB) x ar(ΔCPD) = ar(ΔAPD) x ar(ΔBPC).

Now,

L.H.S. = ar(ΔAPB) x ar(ΔCDP)

= (1/2 x AP × BQ) × (1/2 × PC × DR)

= (1/2 x PC × BQ) × (1/2 × AP × DR)

= *ar(ΔAPD) x ar(ΔBPC)*

= 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(ΔABC) = ar(ΔABD).**

**Solution:**

Draw two perpendiculars CP and DQ on AB.

Now,

ar(ΔABC) = 1/2×AB×CP ⋅⋅⋅⋅⋅⋅⋅(1)

ar(ΔABD) = 1/2×AB×DQ ⋅⋅⋅⋅⋅⋅⋅(2)

To prove the result, ar(ΔABC) = ar(ΔABD), we have to show that CP = DQ.

In right angled triangles, ΔCPO and ΔDQO

∠CPO = ∠DQO = 90^{o}

CO = OD (Given)

∠COP = ∠DOQ (Vertically opposite angles)

By AAS condition: ΔCP0 ≅ ΔDQO

So, CP = DQ …………..(3)

(By CPCT)

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

*ar(ΔABC) = ar(ΔABD)*

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(△ABC) : ar(△BDE).**

**Solution: **

Given: ABC and BDE are two equilateral triangles.

We know, area of an equilateral triangle = √3/4 (side)^{2}

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

Find ar(△ABC):

ar(△ABC) = √3/4 (a)^{ 2}

Find ar(△BDE):

ar(△BDE) = √3/4 (a/2)^{ 2}

(D is the mid-point of BC)

Now,

ar(△ABC) : ar(△BDE)

or √3/4 (a)^{ 2} : √3/4 (a/2)^{ 2}

or 1 : 1/4

or 4:1

*This implies, ar(△ABC) : ar(△BDE) = 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 ΔGEF.**

**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 = ½(Area of parallelogram)

In this case,

*Area of a triangle EFG = ½(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 ΔEFG.**

**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 = ½(Area of parallelogram)

In this case,

*Area of a triangle EFG = ½(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(△RAS).**

**Solution: **

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

In △PSR:

Using Pythagoras theorem,

SR^{2} = PR^{2} – PS^{2} = (13)^{ 2} – (5)^{ 2} = 169 – 25 = 114

or SR = 12

Now,

Area of △RAS = 1/2 x SR x PS

= 1/2 x 12 x 5

= 30

*Therefore, Area of △RAS 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:

- Introduction to Area of Parallelograms and Triangles
- 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 build a good knowledge of the basics as well as advanced mathematical concepts. It also helps students in retaining and quickly retrieving the concepts.
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- These solutions are the best resource materials for students who aspire to gain proficiency in the subject and score good marks in academics.

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