Advertisement Remove all ads

RD Sharma solutions for Mathematics for Class 9 chapter 14 - Areas of Parallelograms and Triangles [Latest edition]

Chapters

Mathematics for Class 9 - Shaalaa.com
Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads

Chapter 14: Areas of Parallelograms and Triangles

Exercise 14.1Exercise 14.2Exercise 14.3Others
Advertisement Remove all ads
Exercise 14.1 [Page 3]

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

Exercise 14.1 | Q 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.

 

 

 

Exercise 14.2 [Page 15]

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

Exercise 14.2 | Q 1 | 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.

Exercise 14.2 | Q 2 | Page 15

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

Exercise 14.2 | Q 3 | Page 15

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.

Exercise 14.2 | Q 4 | Page 15

If ABCD is a parallelogram, then prove that
π‘Žπ‘Ÿ (Δ𝐴𝐡𝐷) = π‘Žπ‘Ÿ (Δ𝐡𝐢𝐷) = π‘Žπ‘Ÿ (Δ𝐴𝐡𝐢) = π‘Žπ‘Ÿ (Δ𝐴𝐢𝐷) = `1/2` π‘Žπ‘Ÿ (||π‘”π‘š 𝐴𝐡𝐢𝐷) .

Exercise 14.3 [Pages 44 - 48]

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

Exercise 14.3 | Q 1 | Page 44

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

Exercise 14.3 | Q 2 | Page 44

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.

Exercise 14.3 | Q 3 | Page 44

Compute the area of trapezium PQRS is Fig. below.

Exercise 14.3 | Q 4 | Page 44

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

Exercise 14.3 | Q 5 | Page 44

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.

Exercise 14.3 | Q 6 | Page 45

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.

Exercise 14.3 | Q 7 | Page 45

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

Exercise 14.3 | Q 8 | Page 45

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

Exercise 14.3 | Q 9 | Page 45

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)

Exercise 14.3 | Q 10 | Page 45

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).

Exercise 14.3 | Q 11 | Page 45

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

Exercise 14.3 | Q 12 | Page 46

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)

Exercise 14.3 | Q 13 | Page 46

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.

Exercise 14.3 | Q 14 | Page 46

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).

Exercise 14.3 | Q 15 | Page 46

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).

Exercise 14.3 | Q 16 | Page 46

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

Exercise 14.3 | Q 17 | Page 46

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.

Exercise 14.3 | Q 18 | Page 46

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.

Exercise 14.3 | Q 19 | Page 46

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) .

Exercise 14.3 | Q 20 | Page 46

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 EFG.

Exercise 14.3 | Q 21 | Page 47

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)

Exercise 14.3 | Q 22 | Page 47

In below fig., PSDA is a parallelogram in which PQ = QR = RS and AP || BQ || CR. Prove
that ar (Δ PQE) = ar (ΔCFD).

Exercise 14.3 | Q 23 | Page 47

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))

Exercise 14.3 | Q 24 | Page 47

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).

Exercise 14.3 | Q 25 | Page 47

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).

Exercise 14.3 | Q 26 | Page 47

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)

Exercise 14.3 | Q 27 | Page 48

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)

Exercise 14.3 | Q 28 | Page 48

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)

Exercise 14.3 | Q 29 | Page 48

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)

Exercise 14.3 | Q 30 | Page 48

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.

[Pages 59 - 60]

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

Q 1 | Page 59

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

Q 2 | Page 59

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

 

Q 3 | Page 59

In the given figure, find the area of ΔGEF.

 
Q 4 | Page 59

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

Q 5 | Page 60

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) 

Q 6 | Page 60

In square ABCDP and Q are mid-point of AB and CD respectively. If AB = 8cm and PQand BD intersect at O, then find area of ΔOPB.

Q 7 | Page 60

ABC is a triangle in which D is the mid-point of BCand F are mid-points of DC and AErespectively. IF area of ΔABC is 16 cm2, find the area of ΔDEF.

Q 8 | Page 60

PQRS is a trapezium having PS and QR as parallel sides. A is any point on PQ and is a point on SR such that AB || QR. If area of ΔPBQ is 17cm2, find the area of ΔASR.

Q 9 | Page 60

ABCD is a parallelogram. P is the mid-point of ABBD and CP intersect at Q such that CQQP = 3.1. If ar (ΔPBQ) = 10cm2, find the area of parallelogram ABCD.

Q 10 | Page 60

P is any point on base BC of ΔABC and D is the mid-point of BCDE is drawn parallel toPA to meet AC at E. If ar (ΔABC) = 12 cm2, then find area of ΔEPC

[Pages 60 - 62]

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

Q 1 | Page 60

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

Q 2 | Page 60

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

Q 3 | Page 60

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 ΔPQR is

  • 12 sq. units

  •  6 sq. units

  • 4 sq. units

  • 3 sq. units

Q 4 | Page 60

The median of a triangle divides it into two 

  • congruent triangle

  •  isosceles triangles

  •  right triangles

  •  triangles of equal areas

Q 5 | Page 60

In a ΔABCDEF are the mid-points of sides BCCA and AB respectively. If ar (ΔABC) = 16cm2, then ar (trapezium FBCE) =

  •  4 cm2

  •  8 cm2

  • 12 cm2

  • 10 cm2

Q 6 | Page 60

ABCD is a parallelogram. P is any point on CD. If ar (ΔDPA) = 15 cm2 and ar (ΔAPC) = 20 cm2, then ar (ΔAPB) =

  • 15 cm2

  • 20 cm2

  • 35 cm2

  • 30 cm2

Q 7 | Page 61

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 cm2

  •  48 cm2

  • 96 cm2

  • 24 cm2

Q 8 | Page 61

A, B, C, D are mid-points of sides of parallelogram PQRS. If ar (PQRS) = 36 cm2, then ar (ABCD) =

  •  24 cm2

  •  18cm2

  •  30 cm2

  •  36 cm2

Q 9 | Page 61

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 cm2

  • a rectangle of area 24 cm2

  • a square of area 26 cm2

  •  a trapezium of area 14 cm2

Q 10 | Page 61

If AD is median of ΔABC and is a point on AC such that
ar (ΔADP) : ar (ΔABD) = 2 : 3, then ar (Δ PDC) : ar (Δ ABC)

  • 1 : 5

  • 1 : 5

  • 1 : 6

  • 3 : 5

Q 11 | Page 61

Medians of ΔABC intersect at G. If ar (ΔABC) = 27 cm2, then ar (ΔBGC) =

  • 6 cm2

  •  9 cm2

  • 12 cm2

  •  18 cm2

Q 12 | Page 61

In a ΔABC if D and E are mid-points of BC and AD respectively such that ar (ΔAEC) = 4cm2, then ar (ΔBEC) =

  • 4 cm2

  • 6 cm2

  •  8 cm2

  • 12 cm2

Q 13 | Page 61

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

Q 14 | Page 61

In the given figure, PQRS is a parallelogram. If X and Y are mid-points of PQ and SRrespectively and diagonal Q is joined. The ratio ar (||gm XQRY) : ar (ΔQSR) =

  •  1 : 4

  •  2 : 1

  •  1 : 2

  • 1 : 1

Q 15 | Page 61

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

  •  ΔAOB

  •  ΔBOC

  • ΔDOC

  • ΔADC

Q 16 | Page 61

ABCD is a trapezium in which AB || DC. If ar (ΔABD) = 24 cm2 and AB = 8 cm, then height of ΔABC is

  •  3 cm

  • 4 cm

  •  6 cm

  •  8 cm

Q 17 | Page 62

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 ABFEand EFCD is

  • a : b

  • (a + 3b): (3a + b)

  • (3a + b) : (a + 3b)

  •  (2a + b) : (3a + b)

Q 18 | Page 62

ABCD is a rectangle with O as any point in its interior. If ar (ΔAOD) = 3 cm2 and ar (ΔABOC) = 6 cm2, then area of rectangle ABCD is

  • 9 cm2

  • 12 cm2

  • 15 cm2

  • 18 cm2

Q 19 | Page 62

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 (ΔABC)

  • \[\frac{1}{2}\] ar (ΔABC)

     

  • \[\frac{1}{3}\] ar (ΔABC)

     

  • \[\frac{1}{4}\]  ar (ΔABC)

     

Q 20 | Page 62

In the given figure, ABCD and FECG are parallelograms equal in area. If ar (ΔAQE) = 12 cm2, then ar (||gm FGBQ) = 

  •  12 cm2

  •  20 cm2

  • 24 cm2

  •  36 cm2

Chapter 14: Areas of Parallelograms and Triangles

Exercise 14.1Exercise 14.2Exercise 14.3Others
Mathematics for Class 9 - Shaalaa.com

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.

Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. RD Sharma textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.

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.

Get the free view of chapter 14 Areas of Parallelograms and Triangles Class 9 extra questions for Mathematics for Class 9 and can use Shaalaa.com to keep it handy for your exam preparation

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×