#### Chapters

Chapter 2: Exponents of Real Numbers

Chapter 3: Rationalisation

Chapter 4: Algebraic Identities

Chapter 5: Factorisation of Algebraic Expressions

Chapter 6: Factorisation of Polynomials

Chapter 7: Linear Equations in Two Variables

Chapter 8: Co-ordinate Geometry

Chapter 9: Introduction to Euclidβs Geometry

Chapter 10: Lines and Angles

Chapter 11: Triangle and its Angles

Chapter 12: Congruent Triangles

Chapter 13: Quadrilaterals

Chapter 14: Areas of Parallelograms and Triangles

Chapter 15: Circles

Chapter 16: Constructions

Chapter 17: Heronβs Formula

Chapter 18: Surface Areas and Volume of a Cuboid and Cube

Chapter 19: Surface Areas and Volume of a Circular Cylinder

Chapter 20: Surface Areas and Volume of A Right Circular Cone

Chapter 21: Surface Areas and Volume of a Sphere

Chapter 22: Tabular Representation of Statistical Data

Chapter 23: Graphical Representation of Statistical Data

Chapter 24: Measures of Central Tendency

Chapter 25: Probability

## Chapter 14: Areas of Parallelograms and Triangles

### RD Sharma solutions for Mathematics for Class 9 Chapter 14 Areas of Parallelograms and Triangles Exercise 14.1 [Page 3]

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

### RD Sharma solutions for Mathematics for Class 9 Chapter 14 Areas of Parallelograms and Triangles Exercise 14.2 [Page 15]

In fig below, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8

cm and CF = 10 cm, find AD.

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

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.

If ABCD is a parallelogram, then prove that

ππ (Δπ΄π΅π·) = ππ (Δπ΅πΆπ·) = ππ (Δπ΄π΅πΆ) = ππ (Δπ΄πΆπ·) = `1/2` ππ (||^{ππ} π΄π΅πΆπ·) .

### RD Sharma solutions for Mathematics for Class 9 Chapter 14 Areas of Parallelograms and Triangles Exercise 14.3 [Pages 44 - 48]

In the below figure, compute the area of quadrilateral ABCD.

In the below 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.

Compute the area of trapezium PQRS is Fig. below.

In the below fig. ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of

ΔAOB.

In the below fig. ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm,

and distance between AB and DC is 4cm. Find the value of x and area of trapezium ABCD.

In the below fig. OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If

OE = 2√5, find the area of the rectangle.

In the below fig. ABCD is a trapezium in which AB || DC. Prove that ar (ΔAOD) =

ar(ΔBOC).

In the given below fig. ABCD, ABFE and CDEF are parallelograms. Prove that ar (ΔADE)

= ar (ΔBCF)

In the below Fig, ABC and ABD are two triangles on the base AB. If line segment CD is

bisected by AB at O, show that ar (Δ ABC) = ar (Δ ABD)

If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in

area. If G is the mid-point of median AD, prove that ar (Δ BGC) = 2 ar (Δ AGC).

A point D is taken on the side BC of a ΔABC such that BD = 2DC. Prove that ar(Δ ABD) =

2ar (ΔADC).

ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove

that: (1) ar (ΔADO) = ar (ΔCDO) (2) ar (ΔABP) = ar (ΔCBP)

ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects

CD at F.

(i) Prove that ar (ΔADF) = ar (ΔECF)

(ii) If the area of ΔDFB = 3 cm2, find the area of ||^{gm} ABCD.

ABCD is a parallelogram whose diagonals AC and BD intersect at O. A line through O

intersects AB at P and DC at Q. Prove that ar (Δ POA) = ar (Δ QOC).

In the below fig. D and E are two points on BC such that BD = DE = EC. Show that ar

(ΔABD) = ar (ΔADE) = ar (ΔAEC).

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:

ar(ΔAPB) × ar(ΔCPD) = ar(ΔAPD) × ar (ΔBPC)

If P is any point in the interior of a parallelogram ABCD, then prove that area of the

triangle APB is less than half the area of parallelogram.

ABCD is a parallelogram. E is a point on BA such that BE = 2 EA and F is a point on DC

such that DF = 2 FC. Prove that AE CF is a parallelogram whose area is one third of the

area of parallelogram ABCD.

In a ΔABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point

of AP. Prove that :

(1) ar (Δ PBQ) = ar (Δ ARC)

(2) ar (Δ PRQ) =`1/2`ar (Δ ARC)

(3) ar (Δ RQC) =`3/8` ar (Δ ABC) .

ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, E is a point of DC

such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:

(1) ar ( ADEG) = ar (GBCD)

(2) ar (ΔEGB) = `1/6` ar (ABCD)

(3) ar (ΔEFC) = `1/2` ar (ΔEBF)

(4) ar (ΔEBG) = ar (ΔEFC)

^{(5)}ΔFind what portion of the area of parallelogram is the area of οEFG.

In Fig. below, CD || AE and CY || BA.

(i) Name a triangle equal in area of ΔCBX

(ii) Prove that ar (Δ ZDE) = ar (ΔCZA)

(iii) Prove that ar (BCZY) = ar (Δ EDZ)

In below fig., PSDA is a parallelogram in which PQ = QR = RS and AP || BQ || CR. Prove

that ar (Δ PQE) = ar (ΔCFD).

In the below 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))

D is the mid-point of side BC of ΔABC and E is the mid-point of BD. if O is the mid-point

of AE, prove that ar (ΔBOE) = `1/8` ar (Δ ABC).

In the below 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 (ΔACQ).

In the below fig. ABCD and AEFD are two parallelograms. Prove that

(1) PE = FQ

(2) ar (Δ APE) : ar (ΔPFA) = ar Δ(QFD) : ar (Δ PFD)

(3) ar (ΔPEA) = ar (ΔQFD)

In the below figure, ABCD is parallelogram. O is any point on AC. PQ || AB and LM ||

AD. Prove that ar (||gm DLOP) = ar (||^{gm} BMOQ)

In a ΔABC, if L and M are points on AB and AC respectively such that LM || BC. Prove

that:

(1) ar (ΔLCM ) = ar (ΔLBM )

(2) ar (ΔLBC) = ar (ΔMBC)

(3) ar (ΔABM) ar (ΔACL)

(4) ar (ΔLOB) ar (ΔMOC)

In Fig. below, ABC and BDE are two equilateral triangles such that D is the mid-point of

BC. AE intersects BC in F. Prove that

(1) ar (Δ BDE) = `1/2` ar (ΔABC)

(2) Area ( ΔBDE) `= 1/2 ` ar (ΔBAE)

(3) ar (BEF) = ar (ΔAFD)

(4) area (Δ ABC) = 2 area (ΔBEC)

(5) ar (ΔFED) `= 1/8` ar (ΔAFC)

(6) ar (Δ BFE) = 2 ar (ΔEFD)

In the following figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:-

(i) ΔMBC ≅ ΔABD

(ii) ar (BYXD) = 2 ar(MBC)

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

(iv) ΔFCB ≅ ΔACE

(v) ar(CYXE) = 2 ar(FCB)

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

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

Note : Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.

### RD Sharma solutions for Mathematics for Class 9 Chapter 14 Areas of Parallelograms and Triangles [Pages 59 - 60]

If *ABC* and *BDE* are two equilateral triangles such that *D* is the mid-point of *BC*, then find ar (Δ*ABC*) : ar (Δ*BDE*).

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

In the given figure, find the area of ΔGEF.

In the given figure, *ABCD* is a rectangle with sides *AB* = 10 cm and *AD* = 5 cm. Find the area of ΔEFG.

*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 (ΔRAS) *

In square *ABCD*, *P* and *Q* are mid-point of *AB* and *CD* respectively. If *AB* = 8cm and *PQ*and *BD* intersect at *O*, then find area of Δ*OPB*.

*ABC* is a triangle in which *D* is the mid-point of *BC*. *E *and *F* are mid-points of *DC* and *AE*respectively. IF area of Δ*ABC* is 16 cm^{2}, find the area of Δ*DEF*.

*PQRS* is a trapezium having *PS* and *QR* as parallel sides. *A* is any point on *PQ* and *B *is a point on *SR* such that *AB* || *QR*. If area of Δ*PBQ* is 17cm^{2}, find the area of Δ*ASR*.

*ABCD* is a parallelogram. *P* is the mid-point of *AB*. *BD* and *CP* intersect at *Q* such that *CQ*: *QP* = 3.1. If ar (Δ*PBQ*) = 10cm^{2}, find the area of parallelogram *ABCD*.

*P* is any point on base *BC* of Δ*ABC** *and *D* is the mid-point of* BC*. *DE* is drawn parallel to*PA* to meet *AC* at *E*. If ar (Δ*ABC*) = 12 cm^{2}, then find area of Δ*EPC*.

### RD Sharma solutions for Mathematics for Class 9 Chapter 14 Areas of Parallelograms and Triangles [Pages 60 - 62]

Two parallelograms are on the same base and between the same parallels. The ratio of their areas is

1 : 2

2 : 1

1 : 1

3 : 1

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

1 : 1

1 : 2

2 : 1

1 : 3

Let *ABC* be a triangle of area 24 sq. units and *PQR* be the triangle formed by the mid-points of the sides of Δ *ABC*. Then the area of Δ*PQR* is

12 sq. units

6 sq. units

4 sq. units

3 sq. units

The median of a triangle divides it into two

congruent triangle

isosceles triangles

right triangles

triangles of equal areas

In a Δ*ABC*, *D*, *E*, *F* are the mid-points of sides *BC*, *CA* and *AB* respectively. If ar (Δ*ABC*) = 16cm^{2}, then ar (trapezium *FBCE*) =

4 cm

^{2}8 cm

^{2}12 cm

^{2}10 cm

^{2}

*ABCD* is a parallelogram. *P* is any point on *CD*. If ar (Δ*DPA*) = 15 cm2 and ar (Δ*APC*) = 20 cm^{2}, then ar (Δ*APB*) =

15 cm

^{2}20 cm

^{2}35 cm

^{2}30 cm

^{2}

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

28 cm

^{2}48 cm

^{2}96 cm

^{2}24 cm

^{2}

A, B, C, D are mid-points of sides of parallelogram *PQRS*. If ar (*PQRS*) = 36 cm^{2}, then ar (*ABCD*) =

24 cm

^{2}18cm

^{2}30 cm

^{2}36 cm

^{2}

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is

a rhombus of area 24 cm

^{2}a rectangle of area 24 cm

^{2}a square of area 26 cm

^{2}a trapezium of area 14 cm

^{2}

If *AD* is median of Δ*ABC* and *P *is a point on *AC* such that

ar (Δ*ADP*) : ar (Δ*ABD*) = 2 : 3, then ar (Δ *PDC*) : ar (Δ *ABC*)

1 : 5

1 : 5

1 : 6

3 : 5

Medians of Δ*ABC* intersect at G. If ar (Δ*ABC*) = 27 cm^{2}, then ar (Δ*BGC*) =

6 cm

^{2}9 cm

^{2}12 cm

^{2}18 cm

^{2}

In a Δ*ABC* if *D* and *E* are mid-points of *BC* and *AD* respectively such that ar (Δ*AEC*) = 4cm^{2}, then ar (Δ*BEC*) =

4 cm

^{2}6 cm

^{2}8 cm

^{2}12 cm

^{2}

In the given figure, *ABCD* is a parallelogram. If *AB* = 12 cm, *AE* = 7.5 cm, *CF* = 15 cm, then *AD* =

In the given figure, *PQRS* is a parallelogram. If *X* and *Y* are mid-points of *PQ* and *SR*respectively and diagonal *Q* is joined. The ratio ar (||^{gm} *XQRY*) : ar (Δ*QSR*) =

1 : 4

2 : 1

1 : 2

1 : 1

Diagonal *AC* and *BD* of trapezium *ABCD*, in which *AB* || *DC*, intersect each other at *O*. The triangle which is equal in area of Δ*AOD* is

Δ

*AOB*Δ

*BOC*Δ

*DOC*Δ

*ADC*

*ABCD* is a trapezium in which *AB* || *DC*. If ar (Δ*ABD*) = 24 cm^{2} and *AB* = 8 cm, then height of Δ*ABC* is

3 cm

4 cm

6 cm

8 cm

*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 areas of quadrilaterals *ABFE*and *EFCD* is

*a*:*b*(

*a*+ 3*b*): (3*a*+*b*)(3

*a*+*b*) : (*a*+ 3*b*)(2

*a*+*b*) : (3*a*+*b*)

*ABCD* is a rectangle with *O* as any point in its interior. If ar (Δ*AOD*) = 3 cm^{2} and ar (Δ*ABOC*) = 6 cm^{2}, then area of rectangle *ABCD* is

9 cm

^{2}12 cm

^{2}15 cm

^{2}18 cm

^{2}

The mid-points of the sides of a triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to

ar (Δ

*ABC*)- \[\frac{1}{2}\] ar (Δ
*ABC*) - \[\frac{1}{3}\] ar (Δ
*ABC*) - \[\frac{1}{4}\] ar (Δ
*ABC*)

In the given figure, *ABCD* and *FECG* are parallelograms equal in area. If ar (Δ*AQE*) = 12 cm^{2}, then ar (||^{gm} *FGBQ*) =

12 cm

^{2}20 cm

^{2}24 cm

^{2}36 cm

^{2}

## Chapter 14: Areas of Parallelograms and Triangles

## RD Sharma solutions for Mathematics for Class 9 chapter 14 - Areas of Parallelograms and Triangles

RD Sharma solutions for Mathematics for Class 9 chapter 14 (Areas of Parallelograms and Triangles) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Mathematics for Class 9 solutions in a manner that help students grasp basic concepts better and faster.

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Concepts covered in Mathematics for Class 9 chapter 14 Areas of Parallelograms and Triangles are Corollary: Triangles on the same base and between the same parallels are equal in area., Corollary: A rectangle and a parallelogram on the same base and between the same parallels are equal in area., Theorem: Parallelograms on the Same Base and Between the Same Parallels., Concept of Area.

Using RD Sharma Class 9 solutions Areas of Parallelograms and Triangles exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in RD Sharma Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 9 prefer RD Sharma Textbook Solutions to score more in exam.

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