RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.1

RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.1

Other Exercises

Question 1.
Construct a quadrilateral ABCD in which AB = 4.4 cm, BC = 4 cm, CD = 6.4 cm, DA = 3.8 cm and BD = 6.6 cm.
Solution:
Steps of construction :
(i) Draw a line segment AB = 4.4 cm.
RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.1 1
(ii) With centre A and radius 3.8 cm and with centre B and radius 6.6 cm, draw arcs intersecting each other at D.
(iii) With centre B and radius 4 cm, and with centre D and radius 6.4 cm, draw arcs intersecting each other at C on the other side of BD.
(iv) Join AD, BD, BC and DC.
The ABCD is the required quadrilateral.

Question 2.
Construct a quadrilateral ABCD such that AB = BC = 5.5 cm, CD = 4 cm, DA = 6.3 cm and AC = 9.4 cm. Measure BD.
Solution:
(i) Draw a line segment AC = 9.4 cm.
(ii) With centre A and C and radius 5.5 cm, draw arcs intersecting each other at B.
(iii) Join AB and CB.
(iv) Again with centre A and radius 6.3 cm, and with centre C and radius 4 cm, draw arcs intersecting each other at D below the line segment AC.
RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.1 2
(v) Join AD and CD.
Then ABCD is the required quadrilateral. On measuring BD, it is 5 cm.

Question 3.
Construct a quadrilateral XYZW in which XY = 5 cm, YZ = 6 cm, ZW = 7 cm, WX = 3 cm and XZ = 9 cm.
Solution:
Steps of construction :
(i) Draw a line segment XZ = 9 cm.
(ii) With centre X and radius 3 cm and with centre Z and radius 7 cm, draw arcs intersecting each other at W.
RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.1 3
(iii) Join XW and ZW.
(iv) Again with centre X and radius 5 cm and with centre Z and radius 6 cm, draw arcs, intersecting each other at Y below the line segment XZ.
(v) Join XY and ZY.
Then XYZW is the required quadrilateral.

Question 4.
Construct a parallelogram PQRS such that PQ = 5.2 cm, PR = 6.8 cm and QS = 8.2 cm.
Solution:
Steps of construction:
In a parallelogram, diagonals bisect each other. Now
(i) Draw a line segment PQ = 5.2 cm.
(ii) With centre P and radius 3.4 cm (\(\frac { 1 }{ 2 }\) of PR) and with centre Q and radius 4.1 cm (\(\frac { 1 }{ 2 }\) of QS) draw arcs intersecting each other at O.
RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.1 4
(iii) Join PQ and QO and produced them to R and S respectively such that PO = OR and QO = OS.
(iv) Join PS, SR and RQ.
Then PQRS is the required parallelogram.

Question 5.
Construct a rhombus with side 6 cm and one diagonal 8 cm. Measure the other diagonal.
Solution:
Steps of construction :
Sides of a rhombus are equal.
(i) Draw a line segment AC = 8 cm.
RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.1 5
(ii) With centres A and C and radius 6 cm, draw two arcs above the line segment AC and two below the line segment AC, intersecting each other at D and B respectively.
(iii) Join AB, AD, BC and CD.
Then ABCD is the required rhombus.
JoinBD.
On measuring BD, it is approximately 9 cm

Question 6.
Construct a kite ABCD in which AB = 4 cm, BC = 4.9 cm and AC = 7.2 cm.
Solution:
Steps of construction :
(i) Draw a line segment AC = 7.2 cm.
RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.1 6
(ii) With centre A and radius 4 cm draw an arc.
(iii) With centre C and radius 4.9 cm, draw another arc which intersects the first arc at B and D.
(iv) Join AB, BC, CD and DA.
Then ABCD is the required kite.

Question 7.
Construct, if possible, a quadrilateral ABCD given, AB = 6 cm BC = 3.7 cm, CD = 5.7 cm, AD = 5.5 cm and BD = 6.1 cm. Give reasons for not being able to construct, if you cannot.
Solution:
Steps of construction :
(i) Draw a line segment BD = 6.1 cm.
(ii) With centre B and radius 6 cm and with centre D and radius 5.5 cm, draw arcs intersecting at A.
(iii) Join AB and AD.
(iv) Again with centre B and radius 3.7 cm and with centre D and radius 5.7 cm, draw two arcs intersecting each other at C below the BD.
RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.1 7
(v) Join BC and DC.
Then ABCD is the required quadrilateral.

Question 8.
Construct, if possible a quadrilateral ABCD in which AB = 6 cm, BC = 7 cm, CD = 3 cm, AD = 5.5. cm and AC = 11 cm. Give reasons for not being able to construct, if you cannot.
Solution:
Steps of construction:
It is not possible to construct this quadrilateral ABCD because
AD + DC = 5.5 cm + 3 cm = 8.5 cm
and AC = 11 cm
AD + DC < AC.
But we know that in a triangle,
Sum of two sides is always greater than its third side.

 

Hope given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5

RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5

Other Exercises

Multiply:

Question 1.
(5x + 3) by (7x + 2)
Solution:
(5x + 3) x (7x + 2)
= 5x (7x + 2) + 3 (7x + 2)
= 35x2 + 10x + 21x + 6
= 35x2 + 31x + 6

Question 2.
(2x + 8) by (x – 3)
Solution:
(2x + 8) x (x – 3)
= 2x (x – 3) + 8 (x – 3)
= 2x2 – 6x + 8x – 24
= 2x2 + 2x – 24

Question 3.
(7x +y) by (x + 5y)
Solution:
(7x + y) x (x + 5y)
= 7x (x + 5y) + y (x + 5y)
= 7x2 + 35xy + xy + 5y2
=7x2 + 36xy + 5y2

Question 4.
(a – 1) by (0.1a2 + 3)
Solution:
(a – 1) x (0.1a2 + 3)
= a (0.1a2 + 3) – 1 (0.1a2+ 3)
= 0.1a3 + 3a-0.1a2-3
= 0.1a3 – 0.1a2 + 3a-3

Question 5.
(3x2 +y2) by (2x2 + 3y2)
Solution:
(3x2+y2) x (2x2 + 3y2)
= 3x2 (2x2 + 3y2) + y2(2x2 + 3y2)
= 6x2 +2 + 9x2y2 + 2x2y2 + 3y2 + 2
= 6x4 + 11 x2y2 + 3y4

Question 6.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 1
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 2

Question 7.
(x6-y6) by (x2+y2)
Solution:
(x6 – y6) x (x2 + y2)
= x6 (x2 + y2) – y6 (x2 + y2)
= x6 x x2 + x6y2 – x2y6 -y6 x y2
= x6 + 2 + x6y2 – x2y6 – y6 +2
= x  + x6y2 – x2y6 – y8

Question 8.
(x2 + y2) by (3a+2b)
Solution:
(x2 + y2) x (3a + 2b)
= x2 (3a + 2b) + y2 (3a + 2b)
= 3x2a + 2x2b + 3y2a + 2y2b
3ax2 + 3av2 + 2bx2 + 2by2

Question 9.
[-3d + (-7ƒ)] by (5d +ƒ)
Solution:
[-3d + (-7ƒ)] x (5d +ƒ)
= -3d x (5d +ƒ) + (-7ƒ) x (5d +ƒ)
= -15d2-3dƒ- 35dƒ- 7ƒ2
= -15d2 – 38dƒ- 7ƒ2

Question 10.
(0.8a – 0.5b) by (1.5a -3b)
Solution:
(0.8a – 0.5b) x (1.5a-3b)
= 0.8a x (1.5a – 36) – 0.56 (1.5a -3b)
= 1.2a2 – 2.4ab – 0.75ab + 1.5b2
= 1.2a2-3.15ab+ 1.5b2

Question 11.
(2x2 y2 – 5xy2) by (x2 -y2)
Solution:
(2x2 y2 – 5xy2) x (x2 -y2)
= 2x2y2 (x2 – y2) – 5x_y2 (x2 – y2)
= 2x2y2 x x2 – 2x2y2 xy2– 5xy2 x x2 + 5x2 xy2
= 2x2 + 2 y2– 2x2 x y2 + 2– 5x1+2 y2+5xy2 + 2
= 2x4y2– 2x2y4 – 5x3y2+ 5xy4

Question 12.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5-q12
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 3

Question 13.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 4
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 5
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 6

Question 14.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 7
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 8

Question 15.
(2x2-1) by (4x3 + 5x2)
Solution:
(2x2-1)x(4x3 + 5x2)
= 2x2 x (4x3 + 5x2) – 1 (4x3 + 5x2)
= 2x2 x 4x3 + 2x2 x 5x2 – 4x3 – 5x2
= 8x2 + 3 + 10x2 + 2-4x3-5x2
= 8x5 + 10x4 – 4x3 – 5x2

Question 16.
(2xy + 3y2) (3y2 – 2)
Solution:
(2xy + 3y2) (3y2 – 2)
= 2xy x (3y2-2) + 3y2 x (3y2-2)
= 2xy x Zy2+ 2xy x (-2) + Zy2 x Zy2 – Zy2 x 2
= 6xy1 + 2– 4xy + 9y2 + 2– 6y2
= 6xy3 – 4xy + 9y4– 6y2
Find the following products and verify the result for x = -1, y = -2 :

Question 17.
(3x-5y)(x+y)
Solution:
(3x-5y)(x+y)
= 3x x (x + y) – 5y x (x + y)
= 3x x x + 3x x y-5y x x-5y x y
= 3x2 + 3xy – 5xy – 5y2
= 3x2 – 2xy – 5y2
Verfification:
x = -1,y = -2
L.H.S. = (3x-5y)(x+y)
= [3 (-1) -5 (-2)] [-1 – 2]
= (-3 + 10) (-3) = 7 x (-3) = -21
R.H.S. = 3x2 – 2xy – 5y2
= 3 (-1)2 – 2 (-1) (-2) -5 (-2)2
=3×1-4-5×4=3-4-20
= 3-24 = -21
∴ L.H.S. = R.H.S.

Question 18.
(x2y-1) (3-2x2y)
Solution:
(x2y-1) (3-2x2y)
= x2y (3 – 2x2y) -1(3-2x2y)
= x2y x 3 – x2y x 2x2y – 1 x 3 + 1 x 2x2y
= 3x2y-2x2 + 2x y1 +1-3 + 2x2y
= 3x2y – 2x4y2– 3 + 2x2y
= 3x2y + 2x2y – 2x4y2 – 3
= 5x2y – 2x4y2 – 3
Verification : (x = -1, y = -2)
L.H.S. = (x2y – 1) (3 – 2x2y)
= [(-1)2 x (-2) -1] [3 – 2 x (-1)2 x (-2)]
= [1 x (-2) -1) [3 – 2 x 1 x (-2)]
= (-2 – 1) (3 + 4) = -3 x 7 = -21
R.H.S. = 5x2y – 2x4y2 – 3
= 5 (-1)2 (-2) -2 (-1)4 (-2)2 -3
5 x 1 (-2) – 2 (1 x 4) -3
= -10-8-3 = -21
∴ L.H.S. = R.H.S

Question 19.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 9
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 10

Simplify :

Question 20.
x2 (x + 2y) (x – 3y)
Solution:
x2 (x + 2y) (x – 3y)
= x2 [x (x – 3y) + 2y (x – 3y)]
= x2 [x2 – 3xy + 2xy – 6y2]
= x2 [x2 – xy – 6y2)
= x2 x x2 – x2 x xy – x26y2
= x4 – x3y – 6x2y2

Question 21.
(x2 – 2y2) (x + 4y)
Solution:
(x2 – 2y2) (x + 4y) x2y2
= [x2 (x + 4y) -2y2 (x + 4y)] x2y2
= (x3 + 4x2y – 2xy2 – 8y3) x2y2
= x2y2 x x3 + x2y2 x 4x2y – 2x2y2 x xy2 – 8x2y2 x y3
= x2 +3 y2 + 4x2 + 2 y2 +1 – 2x2 +1 y2+ 2 – 8x2y2+3
= xy + 44xy3 – 2x3y4 – 8x2y5

Question 22.
a2b2 (a + 2b) (3a + b)
Solution:
a2b2 (a + 2b) (3a + b)
= a2b2 [a (3a + b).+ 2b (3a + b)]
= a2b2 [3a2 + ab + 6ab + 2b2]
= a2b2 [3a2 + lab + 2b2]
= a2b2 x 3a2 + a2b2 x 7ab + a2b2 x 2b2
= 3a2 + 2b2 + 7a2+1 b2+1+ 2a2b2 + 2
= 3a4b2 + 7a3b3 + 2a2b4

Question 23.
x2 (x-y) y2 (x + 2y)
Solution:
x2 (x -y) y2 (x + 2y)
= [x2 x x – x2 x y] [y2 x x + y2 x 2y]
= (x3 – x2y) (xy2 + 2y3)
= x3 (xy2 + 2y3) – x2y (xy2 + 2y3)
= x3 x xy2 + x3 x 2y3 – x2y x xy2 – x2y x 2y3
= x3 +1 y2 + 2x3y3 – x2 +1 y1+ 2 – 2x2y1 + 3
= x4y2 + 2x3y3 – x3y3 – 2x2y4
= x4y2 + x3y3 – 2x2y4

Question 24.
(x3 – 2x2 + 5x-7) (2x-3)
Solution:
(x3 – 2x2 + 5x – 7) (2x – 3)
= (2x – 3) (x3 – 2x2 + 5x – 7)
= 2x (x3 – 2x2 + 5x – 7) -3 (x3 – 2x2 + 5x – 7)
= 2x x x3 – 2x x 2x2 + 2x x 5x – 2x x 7 -3 x x3 – 3 x (-2x2) – 3 x 5x – 3 x (-7)
= 2x4-4x3 + 10x2– 14x-3x3 + 6x2– 15x + 21
= 2x4 – 4x3 – 3x3 + 10x2 + 6x2– 14x- 15x + 21
= 2x4-7x3 + 16x2-29x+ 21

Question 25.
(5x + 3) (x – 1) (3x – 2)
Solution:
(5x + 3) (x – 1) (3x – 2)
= (5x + 3) [x (3x – 2) -1 (3x – 2)]
= (5x + 3) [3x2 – 2x – 3x + 2]
= (5x + 3) [3x2 – 5x + 2]
= 5x (3x2 – 5x + 2) + 3 (3x2 – 5x + 2)
= (5x x 3x2 – 5x x Sx + 5x x 2)+ [3 x 3x2 + 3 x (-5x) + 3×2]
= 15x3 – 25x2 + 10x + 9x2 – 15x + 6
= 15x3 – 25x2 + 9x2 + 10x – 15x + 6
= 15x3 – 16x2 – 5x + 6

Question 26.
(5-x) (6-5x) (2-x)
Solution:
(5-x) (6-5x) (2-x)
= [5 (6 – 5x) -x (6 – 5x)] (2 – x)
= [30 – 2$x – 6x + 5x2] (2 – x)
= (30 – 3 1x + 5x2) (2-x)
= 2 (30 – 31x + 5x2) – x (30 – 31x + 5x2)
= 60 – 62x + 10x2 – 30x + 3 1x2 – 5x3
= 60 – 62x – 30x + 10x2 + 3 1x2 – 5x3
= 60 – 92x + 41x2 – 5x3

Question 27.
(2x2 + 3x – 5) (3x2 – 5x + 4)
Solution:
(2x2 + 3x – 5) (3x2 – 5x + 4)
= 2x2 (3x2 – 5x + 4) + 3x (3x2 – 5x + 4) -5 (3x2 – 5x + 4)
= 2x2 x 3x2 – 2x2 x 5x + 2x2 x 4 + 3x x 3x2 – 3x x 5x + 3x x 4 – 5 x 3x2 – 5 (-5x) -5×4
= 6x4 – 10x3 + 8x2 + 9x3 – 15x2 + 12x – 15x2 + 25x-20
= 6x4 – 10x3 + 9x3 + 8x2 – 15x2 – 15x2 + 12x + 25x – 20
= 6x4 – x3 – 22x2 + 37x – 20

Question 28.
(3x – 2) (2x – 3) + (5x – 3) (x + 1)
Solution:
(3x – 2) (2x – 3) + (5x – 3) (x + 1)
= 3x (2x – 3) -2 (2x – 3) + 5x (x + 1) – 3 (x + 1)
= 6x2 – 9x – 4x + 6 + 5x2 + 5x – 3x – 3
= 6x2 + 5x2 – 9x – 4x + 5x – 3x + 6 – 3
= 11x2– 11x + 3

Question 29.
(5x – 3) (x + 2) – (2x + 5) (4x – 3)
Solution:
(5x – 3) (x + 2) – (2x + 5) (4x – 3)
= [5x (x + 2) -3 (x + 2)] – [2x (4x – 3) + 5 (4x – 3)]
= [5x2 + 1 0x – 3x – 6] – [8x2 – 6x + 20x -15]
= (5x2 + 7x – 6) – (8x2 + 14x – 15)
= 5x2 + lx – 6 – 8x2 – 14x + 15
= 5x2 – 8x2 + 7x – 14x – 6 + 15
= -3x2 – 7x + 9

Question 30.
(3x + 2y) (4x + 3y) – (2x – y) (7x – 3y)
Solution:
(3x + 2y) (4x + 3y) – (2x – y) (7x – 3y)
= [3x (4x + 3y) + 2y (4x + 3y)]-[2x (7x-3y)-y(7x-3y)]
= (12x2 + 9xy + 8xy + 6y2) – (14x2 – 6xy – 7xy + 3y2)
= (12x2 + 17xy + 6y2) – (14x2 – 13xy + 3y2)
= 12x2 + 17xy + 6y2 – 14x2 + 13xy – 3y2
= 12x2 – 14x2 + 17xy + 13xy + 6y2 – 3y2
= -2x2 + 30xy + 3y2
= -2x2 + 3y2 + 30xy

Question 31.
(x2-3x + 2) (5x- 2) – (3x2 + 4x-5) (2x- 1)
Solution:
(x2-3x + 2) (5x- 2) – (3x2 + 4x-5) (2x- 1)
= [5x (x2 – 3x + 2) -2 (x2 – 3x + 2)] – [2x (3x2 + 4x – 5) -1 (3x2 + 4x – 5)]
= [5x3 – 15x2 + 10x – 2x2 + 6x – 4] – [6x3 + 8x2 – 10x – 3x2 – 4x + 5]
= [5x3 – 15x2 – 2x2 + 10xc + 6x – 4] – [6x3 + 8x2 – 3x2 – 10x – 4x + 5]
= (5x3 – 17x2 + 16x-4) – (6x3 + 5x2 – 14x + 5)
= 5x3 – 17x2 + 16x – 4 – 6x3 – 5x2 + 14x – 5
= 5x3 – 6x3 – 17x2 – 5x2 + 16x + 14x – 4 – 5
= -x3 – 22x2 + 30x – 9

Question 32.
x3 – 2x2 + 3x – 4) (x – 1) – (2x – 3) (x2 – x + 1)
Solution:
(x3 – 2x2 + 3x – 4) (x – 1) – (2x – 3) (x2 – x + 1)
= [x (x3 – 2x2 + 3x – 4) – 1 (x3 – 2x2 + 3x – 4)] – [2x (x2 – x + 1) – 3 (x2 – x + 1)]
= [x4 – 2x3 + 3x2 – 4x – x3 + 2x2 – 3x + 4] [2x3 – 2x2 + 2x – 3x2 + 3x – 3]
= (x4 – 2x3 – x3 + 3x2 + 2x2 – 4x – 3x + 4) (2x3 – 2x2 – 3x2 + 2x + 3x – 3)
= (x4 – 3x3 + 5x2 – 7x + 4) – (2x3 – 5x2 + 5x – 3)
= x4 – 3x3 + 5x2 – 7x + 4 – 2x3 + 5x2 – 5x + 3
= x4 – 3x3 – 2x3 + 5x2 + 5x2 – 7x – 5x + 4 + 3
= x4 – 5x3 + 10x2 – 12x + 7

Hope given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.3

RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III (Special Types of Quadrilaterals) Ex 17.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.3

Other Exercises

Question 1.
Which of the following statements are true for a rectangle ?
(i) It has two pairs of equal sides.
(ii) It has all its sides of equal length.
(iii) Its diagonals are equal.
(iv) Its diagonals bisect each other.
(v) Its diagonals are perpendicular.
(vi) Its diagonals are perpendicular and bisect each other.
(vii) Its diagonals are equal and bisect each other.
(viii) Its diagonals are equal and perpendicular, and bisect each other.
(ix) All rectangles are squares.
(x) All rhombuses are parallelograms.
(xi) All squares are rhombuses and also rectangles.
(xii) All squares are not parallelograms.
Solution:
(i) True.
(ii) False. (Only pair of opposite sides is equal)
(iii) True
(iv) True
(v) False (Diagonals are not perpendicular)
(vi) False (Diagonals are not perpendicular to each other)
(vii) True
(viii) False (Diagonals are equal but not perpendicular)
(ix) False (All rectangles are not square but a special type can be a square)
(x) True
(xi) True
(xii) False (All squares are parallelograms because their opposite sides are parallel and equal)

Question 2.
Which of the following statements are true for a square ?
(i) It is a rectangle.
(ii) It has all its sides of equal length.
(iii) Its diagonals bisect each other at right angle.
(iv) Its diagonals are equal to its sides.
Solution:
(i) True
(ii) True
(iii) True
(iv) False (Each diagonal of a square is greater than its side)

Question 3.
Fill in the blanks in each of the following so as to make the statement true :
(i) A rectangle is a parallelogram in which ……..
(ii) A square is a rhombus in which ……….
(iii) A square is a rectangle in which ………
Solution:
(i) A rectangle is a parallelogram in which one angle is right angle.
(ii) A square is a rhombus in which one angle is right angle.
(iii) A square is a rectangle in which adjacent sides are equal.

Question 4.
A window frame has one diagonal longer than the other. Is the window frame a rectangle ? Why or why not ?
Solution:
No, it is not a rectangle as rectangle has diagonals of equal length.

Question 5.
In a rectangle ABCD, prove that ∆ACB = ∆CAD.
Solution:
In rectangle ABCD, AC is its diagonal.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.3 1
Now in ∆ACB and ∆CAD
AB = CD (Opposite sides of a rectangle)
BC = AD
AC = AC (Common)
∆ACB = ∆CAD (SSS condition)

Question 6.
The sides of a rectangle are in the ratio 2 : 3 and its perimeter is 20 cm. Draw the rectangle.
Solution:
Perimeter of a rectangle = 20 cm
Ratio in the sides = 2 : 3
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.3 2
Let breadth (l) = 2x
Then length (b) = 3x
Perimeter = 2 (l + b)
⇒ 20 = 2 (2x + 3x)
⇒ 4x + 6x = 20
⇒ 10x = 20
⇒ x = \(\frac { 20 }{ 10 }\) = 2
Length = 3x = 3 x 2 = 6
and breadth = 2x = 2 x 2 = 4 cm
Steps of construction:
(i) Draw a line segment AB = 6 cm.
(ii) At A and B draw perpendicular AX and BY.
(iii) Cut off from AX and BY,
AD = BC = 4 cm.
(iv) Join CD.
Then ABCD is the required rectangle.

Question 7.
The sides of a rectangle are the ratio 4 : 5. Find its sides if the perimeter is 90 cm.
Solution:
Perimeter of a rectangle = 90 cm.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.3 3
Ratio in sides = 4 : 5
Let first side = 4x
Then second side = 5x
Perimeter = 2 (l + b)
⇒ 2 (4x + 5x) = 90
⇒ 2 x 9x = 90
⇒ 18x = 90
⇒ x = 5
First side = 4x = 4 x 5 = 20 cm
and second side = 5x = 5 x 5 = 25 cm

Question 8.
Find the length of the diagonal of a rectangle whose sides are 12 cm and 5 cm.
Solution:
In rectangle ABCD, AB = 12 cm and AD = 5 cm
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.3 4
BD is its diagonal.
Now, in right angled ∆ABD,
BD² = AB² + AD² (Pythagoras theorem)
= (12)² + (5)² = 144 + 25 = 169 = (13)²
BD = 13 cm
Length of diagonal = 13 cm

Question 9.
Draw a rectangle whose one side measures 8 cm and the length of each of whose diagonals is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8 cm
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.3 5
(ii) At B, draw a perpendicular BX
(iii) With centre A and radius 10 cm, draw an arc which intersects BX at C.
(iv) With centre C and radius equal to AB and with centre A and radius equal to BC, draw arcs which intersect at D.
(v) Join AD, AC, CD and BD.
Then ABCD is the required rectangle.

Question 10.
Draw a square whose each side measures 4.8 cm.
Solution:
Steps of construction :
(i) Draw a line segment AB = 4.8 cm.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.3 6
(ii) At A and B, draw perpendiculars AX and BY.
(iii) Cut off AD = BC = 4.8 cm
(iv) Join CD.
Then ABCD is the required square.

Question 11.
Identify all the quadrilaterals that have:
(i) Four sides of equal length.
(ii) Four right angles.
Solution:
(i) A quadrilateral whose four sides are equal can be a square or a rhombus.
(ii) A quadrilateral whose four angle are right angle each can be a square or a rectangle.

Question 12.
Explain how a square is
(i) a quadrilateral
(ii) a parallelogram
(iii) a rhombus
(iv) a rectangle ?
Solution:
(i) A square is a quadrilateral as it has four sides and four angles.
(ii) A square is a parallelogram, because its opposite sides are parallel and equal.
(iii) A square is a rhombus because it has all sides equal and opposite sides are parallel.
(iv) A square is a rectangle as its opposite sides are equal and each angle is of 90°.

Question 13.
Name the quadrilaterals whose diagonals:
(i) bisect each other
(ii) are perpendicular bisector of each other
(iii) are equal.
Solution:
(i) A quadrilateral whose diagonals bisect each other can be a square, rectangle, rhombus or a parallelogram.
(ii) A quadrilateral whose diagonals are perpendicular bisector of each other can be a square or a rhombus.
(iii) A quadrilateral whose diagonals are equal can be a square or a rectangle.

Question 14.
ABC is a right-angled triangle and O is the mid-point of the side opposite to the right angle. Explain why O is equidistant from A, B and C.
Solution:
In ∆ABC, ∠B = 90°.
O is the mid-point of AC i.e. OA = OC.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.3 7
BO is joined.
Now, we have to prove that OA = OB = OC
Produce BO to D such that OD = OB.
Join DC and DA.
In ∆AOB and ∆COD
OA = OC (O is the mid point of AC)
OB = OD (Construction)
∠AOB = ∠COD (Vertically opposite angles)
∆AOB = ∆COD (SAS condition)
AB = CD (c.p.c.t.) …..(i)
Similarly, we can prove that
∆BOC = ∆AOD
BC = AD …….(ii)
From (i) and (ii)
ABCD is a rectangle.
But diagonals of a rectangle bisect each other and are equal in length.
AC and BD bisect each other at O.
OA = OC = OB.
O is equidistant from A, B and C.

Question 15.
A mason has made a concrete slap. He needs it to be rectangular. In what different ways can he make sure that it is a rectangular ?
Solution:
By definition, a rectangle has each angle of 90° and their diagonals are equal.
The mason will check the slab whether it is a rectangular in shape by measuring that
(i) its each angle is 90°
(ii) its both diagonals are equal.

Hope given RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.3 are helpful to complete your math homework.

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RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2

RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III (Special Types of Quadrilaterals) Ex 17.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2

Other Exercises

Question 1.
Which of the following statements are true for a rhombus ?
(i) It has two pairs of parallel sides.
(ii) It has two pairs of equal sides.
(iii) It has only two pairs of equal sides.
(iv) Two of its angles are at right angles.
(v) Its diagonals bisect each other at right angles.
(vi) Its diagonals are equal and perpendicular.
(vii) it has all its sides of equal lengths.
(viii) It is a parallelogram.
(ix) It is a quadrilateral.
(x) It can be a square.
(xi) It is a square.
Solution:
(i) True
(ii) True
(iii) False (Its all sides are equal)
(iv) False (Opposite angles are equal)
(v) True
(vi) False (Diagonals are not equal)
(vii) True
(viii) True
(ix) True
(x) True (It is a rhombus)
(xi) False

Question 2.
Fill in the blanks, in each of the following, so as to make the statement true:
(i) A rhombus is a parallelogram in which ………..
(ii) A square is a rhombus in which ………..
(iii) A rhombus has all its sides of …….. length.
(iv) The diagonals of a rhombus each ………. other at ………. angles.
(v) If the diagonals of a parallelogram bisect each other at right angles, then it is a ………..
Solution:
(i) A rhombus is a parallelogram in which adjacent sides are equal.
(ii) A square is a rhombus in which one angle is right angle.
(iii) A rhombus has all its sides of equal length.
(iv) The diagonals of a rhombus bisect each other at right angles.
(v) If the diagonals of a parallelogram bisect each other at right angles, then it is a rhombus.

Question 3.
The diagonals of a parallelogram are not perpendicular. Is it a rhombus ? Why or why not ?
Solution:
By definition of a rhombus, its diagonals bisect each other at right angle.
So, the given parallelogram is not a rhombus.

Question 4.
The diagonals of a quadrilateral are perpendicular to each other. Is such a quadrilateral always a rhombus. If your answer is ‘No’, draw a figure to justify your answer.
Solution:
The diagonals of a quadrilateral are perpendicular to each.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 1
It is not always possible to be a rhombus. It can be of the diagonals bisect each other at right angles and if not, then it is not rhombus as shown in the figure given above:

Question 5.
ABCD is a rhombus. If ∠ACB = 40°, find ∠ADB.
Solution:
In rhombus ABCD, diagonals AC and BD intersect each other at O.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 2
∠ACB = 40°, we have to find ∠ADB.
BD || AD and AC.is its transversal..
∠ACB = ∠CAD (Alternate angles)
Now in ∆AOD
∠OAD + ∠AOD + ∠ADO = 180° (Sum of angles of a triangle)
⇒ 40° + 90° + ∠ADO = 180°
⇒ 130° + ∠ADO = 180°
⇒ ∠ADO = 180° – 130° = 50°
∠ADB = 50°

Question 6.
If the diagonals of a rhombus are 12 cm and 16 cm, find the length of each side.
Solution:
In rhombus ABCD, diagonals AC and BD bisect each other at O at right angles.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 3
AC = 16 cm, BD = 12 cm
AO = OC = \(\frac { 16 }{ 2 }\) = 8 cm
BO = OD = \(\frac { 12 }{ 2 }\) = 6 cm.
Now, in right angled ∆AOB,
AB² = AO² + BO² = (8)² + (6)² = 64 + 36 = 100 = (10)²
AB = 10 cm
Each side of rhombus = 10 cm

Question 7.
Construct a rhombus whose diagonals are of length 10 cm and 6 cm.
Solution:
(i) Draw a line segment AC =10 cm.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 4
(ii) Draw its perpendicular bisector and cut off OB = OD = 3 cm (\(\frac { 1 }{ 2 }\) of 6 cm).
(iii) Join AB, BC, CD and DA.
Then ABCD is the required rhombus.

Question 8.
Draw a rhombus, having each side of length 3.5 cm and one of the angles as 40°.
Solution:
Steps of construction
(i) Draw a line segment AB = 3.5 cm.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 5
(ii) Draw a ray BX making an angle of 40° at B and cut off BC = 3.5 cm.
(iii) With centres C and A, and radius 3.5 cm. Draw arcs intersecting each other at D.
(iv) Join AD and CD.
Then ABCD is a required rhombus.

Question 9.
One side of a rhombus is of length 4 cm and the length of an altitude is 3.2 cm. Draw the rhombus.
Solution:
(i) Draw a line segment AB = 4 cm.
(ii) At B, draw a perpendicular BX and cut off BL = 3.2 cm.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 6
(iii) At L, draw a line LY parallel to AB.
(iv) With centres B and radius 4 cm, draw an arc intersecting the line LY at C.
(v) With centre C cut off CD = 4 cm.
(vi) Join BC and AD.
Then ABCD is the required rhombus.

Question 10.
Draw a rhombus ABCD if AB = 6 cm and AC = 5 cm.
Solution:
Steps of construction :
(i) Draw a line segment AB = 6 cm.
(ii) With centre A and radius 5 cm and with centre B and radius 6 cm, draw arcs intersecting each other at C.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 7
(iii) Join AC and BC.
(iv) Again with centres C and A and radius 6 cm, draw arcs intersecting each other at D.
(v) Join AD and CD.
Then ABCD is the required rhombus.

Question 11.
ABCD is a rhombus and its diagonals intersect at O.
(i) Is ∆BOC = ∆DOC ? State the congruence condition used ?
(ii) Also state, if ∠BCO = ∠DCO.
Solution:
In rhombus ABCD, diagonals AC and BD bisect each other at O.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 8
(i) Now in ∆BOC and ∆DOC
OC = OC (Common)
BC = CD (Sides of rhombus)
OB = OC (Diagonals bisect each other at O)
∆BOC = ∆DOC (SSS. condition)
(ii) ∠BCO = ∠DCO

Question 12.
Show that each diagonal of a rhombus bisects the angle through which it passes:
Solution:
In rhombus ABCD, AC is its diagonal we have to prove that AC bisects ∠A and ∠C.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 9
Now, in ∆ABC and ∆ADC
AC = AC (Common)
AB = CD (Sides of a rhombus)
BC = AD
∆ABC = ∆ADC (SSS condition)
∠BAC = ∠DAC and ∠BCA = ∠DCA (c.p.c.t.)
Hence AC bisects the angle A.
Similarly, by joining BD, we can prove that BD bisects ∠B and ∠D.
Hence each diagonal of a rhombus bisects the angle through which it passes.

Question 13.
ABCD is a rhombus whose diagonals intersect at O. If AB = 10 cm, diagonal BD = 16 cm, find the length of diagonal AC.
Solution:
In rhombus ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 10
AB = 10 cm, diagonal BD = 16 cm.
Draw diagonal AC which bisects BD at O at right angle.
BO = OD = 8 cm and AO = OC.
Now in ∆AOB,
AB² = AO² + BO²
⇒ (10)² = AO² + (8)²
⇒ 100 = AO² – 64
⇒ AO² = 100 – 64 = 36 = (6)²
AO = 6.
AC = 2AO = 2 x 6 = 12 cm

Question 14.
The diagonals of a quadrilateral are of lengths 6 cm and 8 cm. If the diagonals bisect each other at right angles, what is the length of each side of the quadrilateral ?
Solution:
In a quadrilateral ABCD, diagonals AC and BD bisect each other at right angles.
AC = 8 cm and BD = 6 cm.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 11
AO = OC = 4 cm and BO = OD = 3 cm
Now, in right ∆AOB,
AB² = AO² + BO² = (4)² + (3)² = 16 + 9 = 25 = (5)²
AB = 5 cm
Hence each side of quadrilateral will be 5 cm.

Hope given RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1

RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III (Special Types of Quadrilaterals) Ex 17.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1

Other Exercises

Question 1.
Given below is a parallelogram ABCD. Complete each statement along with the definition or property used:
(i) AD = ………
(ii) ∠DCB = ……….
(iii) OC’ = …….
(iv) ∠DAB + ∠CDA = …….
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 1
Solution:
(i) AD = BC
(ii) ∠DCB = ∠ADC
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 2
(iii) OC = OA
(iv) ∠DAB + ∠CDA = 180°

Question 2.
The following figures are parallelograms. Find the degree values of the unknowns x,y, z.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 3
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 4
Solution:
In a parallelogram, opposite angles are equal and sum of adjacent angle is 180°.
(i) In parallelogram ABCD,
∠B = 100°
∠A = ∠C = 180° (Co-interior angles)
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 5
⇒ x + 100° = 180°
⇒ x = 180° – 100°
⇒ x = 80°
But ∠A = ∠C and ∠B = ∠D (Opposite angles)
A = x
⇒ z = x
⇒ z = 80°
and ∠D = ∠B
⇒ y = 100°
x = 80°, y = 100° and z = 80°
(ii) In parallelogram PQRS, side PQ is produced to T.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 6
∠S = 50°
∠PQR = ∠S (Opposite angles)
w = 50°
But ∠P + ∠PQR = 180° (Sum of adjacent angles)
⇒ x + 50° = 180°
⇒ x = 180° – 50° = 130°
⇒ But ∠P = ∠R (Opposite angles)
x = y
⇒ y = 130°
But w + z = 180° (A linear pair)
⇒ 50° + z = 180°
⇒ z = 180° – 50° = 130°
⇒ x = 130°, y = 130°, ∠ = 130
(iii) In parallelogram LMNP, PM is its diagonal ∠NPM = 30°, ∠PMN = 90°
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 7
PN || LM (Opposite sides of a parallelogram) and PM is its transversal
∠NPM = ∠PML
⇒ 30° = x
x = 30°
In ∆PMN,
∠P = 30°, ∠M = 90°
But ∠P + ∠M + ∠N = 180° (Sum of angles of a triangle)
⇒ 30° + 90° + z = 180°
⇒ 120° + z = 180°
⇒ z = 180° – 120° = 60°
But ∠L = ∠N (Opposite angles of a parallelogram)
y = z
⇒ y = 60°
Hence x = 30°, y = 60° and ∠ = 60°
(iv) In rhombus ABCD, diagonals AC and BD bisect each other at right angles.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 8
x = 90°
In ∆OCD,
∠O + ∠C + ∠D = 180°
⇒ 90° + 30° + y = 180°
⇒ 120° + y = 180°
⇒ y = 180° – 120° = 60°
y = 60°
CD || AB, BD is the transversal
y = z (Alternate angles)
z = 60°
Hence x = 90°, y = 60° and z = 60°
(v) In parallelogram PQRS, side QR is produced to T
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 9
∠Q = 80°
∠P + ∠Q = 180° (Sum of adjacent angles)
⇒ x + 80° = 180°
⇒ x = 180° – 80° = 100°
∠Q = ∠S (Opposite angles of a parallelogram)
⇒ 80° = y
⇒ y = 80°
PQ || SR and QRT is transversal
∠TRS = ∠RQP (Corresponding angles)
⇒ ∠ = 80°
Hence x = 100°, y = 80° and z = 80°
(vi) In parallelogram TUVW, UW is its diagonal
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 10
∠TUW = 40° and ∠V = 112°
∠T = ∠V (Opposite angles)
y = 112°
In ∆TUW,
∠T + ∠V + ∠W = 180° (Sum of angles of a parallelogram)
⇒ y + 40° + x = 180°
⇒ 112° + 40° + x = 180°
⇒ 152° + x = 180°
⇒ x = 180° – 152° = 28°
UV || TW and UW is its transversal
∠WUV = ∠TWU (Alternate angles)
⇒ z = x
⇒ z = 28°
Hence x = 28°, y = 112°, z = 28°

Question 3.
Can the following figures be parallelograms. Justify your answer.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 11
Solution:
(i) In quadrilateral PLEH
∠H = 100°, ∠L = 80°
But there are opposite angles
∠H ≠ ∠L
PLEH is not a parallelogram.
(ii) In quadrilateral GNIR,
RI = 8 cm, GN = 8 cm, RG = 5 cm and IN = 5 cm
PI = GH and RG = IN
But there are opposite sides of the quadrilateral.
GNIR is a parallelogram.
(iii) In quadrilateral BEST,
BS and ET are its diagonals
But these diagonal do not bisect each other.
BEST is not a parallelogram

Question 4.
In the adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the geometrical truths you use to find them.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 12
Solution:
In the figure, HOPE is a parallelogram in which HG is produced and HP is the diagonal
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 13
∠EHP = 40° and ∠POQ = 70°
But ∠POQ + ∠POH = 180° (Linear pair)
⇒ 70° + w = 180°
⇒ w = 180° – 70° = 110°
But ∠E = ∠POE (Opposite angles of a parallelogram)
x = 110°
HE || OP and HP is its transversal.
∠EHP = ∠HPO (Alternate angles)
⇒40° = y
⇒ y = 40°
In ∆PHO,
Ext. ∠POQ = ∠PHO + ∠HPO
⇒ 70° = z + y
⇒ 70° = z + 40°
⇒ z = 70° – 40° = 30°
Hence x = 110°, y = 40°, z = 30°

Question 5.
In the following figures GUNS and RUNS are parallelograms. Find x and y.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 14
Solution:
(i) In parallelogram GUNS,
Opposite sides are parallel and equal
3x = 18
⇒ x = 6
and 3y – 1 = 26
⇒ 3y = 26 + 1 = 27
⇒ y = 9
x = 6, y = 9
(ii) Diagonals of a parallelogram bisect each other.
y – 7 = 20
⇒ y = 20 + 7 = 27
and x – 27 = 16
⇒ x – 27 = 16
⇒ x = 16 + 27 = 43
x = 43, y = 27

Question 6.
In the following figure RISK and CLUE are parallelograms. Find the measure of x.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 15
Solution:
In the figure, RISK and CLUE are parallelograms
∠K = 120° and ∠L = 70°
In parallelogram RISK
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 16
RK || IS
∠RKS = ∠ISU (Corresponding angles)
⇒ ∠ISU = 120°
In parallelogram CLUE,
∠E = ∠L (Opposite angles of a parallelogram)
∠E = 70° (∠L = 70°)
Now in ∆EOS,
Ext. ∠ISU = x + ∠E
⇒ 120° = x + 70°
⇒ x = 120° – 70° = 50°
x = 50°

Question 7.
Two opposite angles of a parallelogram are (3x – 2)° and (50 – x)°. Find the measure of each angle of the parallelogram.
Solution:
In parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 17
∠A = (3x – 2)° and ∠C = (50 – x)°
∠A = ∠C (Opposite angles of a parallelogram)
⇒ 3x – 2° = 50° – x
⇒ 3x + x = 50° + 2°
⇒ 4x = 52°
x = 13°

Question 8.
If an angle of a parallelogram is two- third of its adjacent angle, find the angles of the parallelogram.
Solution:
Let the parallelogram is ABCD and ∠A of a parallelogram ABCD be x
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 18
Hence other angles will be
∠C = ∠A = 108° (Opposite angles)
and ∠D = ∠B = 72° (Opposite angles)
Hence ∠A = 108°, ∠B = 72°, ∠C = 108° and ∠D = 72°

Question 9.
The measure of one angle of a parallelogram is 70°. What are the measures of the remaining angles ?
Solution:
Let in parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 19
∠A = 70°
But ∠A + ∠B= 180° (Sum of adjacent angles)
⇒ 70° + ∠B = 180°
⇒ ∠B = 180° – 70°
⇒ ∠B = 110°
But ∠C = ∠A and ∠D = ∠B (Opposite angles)
∠C = 70° and ∠D = 110°
Hence ∠B = 110°, ∠C = 70° and ∠D = 110°

Question 10.
Two adjacent angles of a parallelogram are as 1 : 2. Find the measures of all the angles of the parallelogram.
Solution:
Let in parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 20
∠A : ∠B = 1 : 2
Let ∠A = x, then ∠B = 2x
But ∠A + ∠B = 180° (Sum of adjacent angles)
⇒ x + 2x = 180°
⇒ 3x = 180°
∠A = x = 60°
and ∠B = 2x = 2 x 60°= 120°
But ∠C = ∠A and ∠D = ∠B (Opposite angles)
∠C = 60° and ∠D = 120°
Hence ∠A = 60°, ∠B = 120°, ∠C = 60° and ∠D = 120°

Question 11.
In a parallelogram ABCD, ∠D = 135°, determine the measure of ∠A and ∠B.
Solution:
In parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 21
∠D = 135°
But ∠A + ∠D= 180° (Sum of adjacent angles)
∠A + 135° = 180°
∠A = 180° – 135° = 45°
But ∠B = ∠D (Opposite angles)
∠B = 135°
Hence ∠A = 45° and ∠B = 135°

Question 12.
ABCD is a parallelogram in which ∠A = 70°, compute ∠B, ∠C and ∠D.
Solution:
In parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 22
∠A = 70°
But ∠A + ∠B = 180° (Sum of adjacent angles)
⇒ 70° + ∠B = 180°
⇒ ∠B = 180° – 70° = 110°
But ∠C = ∠A and ∠D = ∠B (Opposite angles)
∠C = 70° and ∠D = 110°
Hence ∠B = 110°, ∠C = 70° and ∠D = 110°

Question 13.
The sum of two opposite angles of a parallelogram is 130°. Find all the angles of the parallelogram.
Solution:
Let in parallelogram
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 23
∠A + ∠C = 130°
But ∠A = ∠C (Opposite angles)
⇒ ∠C = \(\frac { 130 }{ 2 }\) = 65°
∠B + ∠D = 180° (Sum of adjacent angles)
⇒ 65° + ∠B = 180°
⇒ ∠B = 180° – 65° = 115°
⇒ ∠B = 115°
But ∠D = ∠B (Opposite angles)
∠D = 115°
Hence ∠A = 65°, ∠B = 115°, ∠C = 65° and ∠D = 115°

Question 14.
All the angles of a quadrilateral are equal to each other. Find the measure of each. Is the quadrilateral a parallelogram ? What special type of parallelogram is it ?
Solution:
All the angles of a quadrilateral are equal and sum of the four angles = 360°
Each angle will be = \(\frac { 360 }{ 4 }\) = 90°
Let in quadrilateral ABCD,
∠A = ∠B = ∠C = ∠D = 90°
It is a parallelogram as opposite angles are equal
i.e., ∠A = ∠C and ∠B = ∠D.
Each angle is of 90°
This parallelogram is a rectangle.

Question 15.
Two adjacent sides of a parallelogram are 4 cm and 3 cm respectively. Find its perimeter.
Solution:
Length of two adjacent sides = 4 cm and 3 cm
i.e., l = 4 cm and b = 3 cm
Perimeter = 2 (l + b) = 2 (4 + 3) cm = 2 x 7 = 14 cm

Question 16.
The perimeter of a parallelogram is 150 cm. One of its sides is greater ii.au the other by 25 cm. Find the length of the sides of the parallelogram.
Solution:
Perimeter of a parallelogram = 150 cm
Let l he the longer side and b be the shorter side
l = b + 25 cm.
⇒ 2 (l + b) = 150
⇒ l + b = 75
⇒ b + 25 + b = 75
⇒ 2b = 75 – 25 = 50
⇒ b = 25
l = b + 25 = 25 + 25 = 50
Sides are 50 cm, 25 cm

Question 17.
The shorter side of a parallelogram is 4.8 cm and the longer side is half as much again as the shorter side. Find the perimeter of the parallelogram.
Solution:
In a parallelogram shorter side (b) = 4.8 cm.
longer side (l) = 4.8 + \(\frac { 1 }{ 2 }\) x 4.8
= 4.8 + 2.4 = 7.2 cm
Perimeter = 2 (l + b) = 2 (7.2 + 4.8) cm = 2 x 12.0 = 24 cm

Question 18.
Two adjacent angles of a parallelogram are (3x – 4)° and (3x + 10)°. Find the angles of the parallelogram.
Solution:
Let in parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 24
∠A = (3x + 4)° and ∠B = (3x + 10)°
But ∠A + ∠B = 180° (Sum of adj adjacent angles)
⇒ 3x – 4 + 3x + 10 = 180°
⇒ 6x + 6° = 180°
⇒ 6x = 180° – 6° = 174°
⇒ x = 29°
∠A = 3x – 4 = 3 x 29 – 4 = 87° – 4° = 83°
∠B = 3x + 10 = 3 x 29° + 10° = 87° + 10° = 97°
But ∠C = ∠A and ∠D = ∠B (Opposite angles)
∠C = 83° and ∠D = 97°
Hence ∠A = 83°, ∠B = 97°, ∠C = 83° and ∠D = 97°

Question 19.
In a parallelogram ABCD, the diagonals bisect each other at O. If ∠ABC = 30°, ∠BDC = 10° and ∠CAB = 70°.
Find : ∠DAB, ∠ADC, ∠BCD, ∠AOD, ∠DOC, ∠BOC, ∠AOB, ∠ACD, ∠CAB, ∠ADB, ∠ACB, ∠DBC and ∠DBA.
Solution:
In parallelogram ABCD, diagonal AC and ED bisect each other at O.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 25
∠ABC = 30°, ∠CAB = 70° and ∠BDC = 10°
∠ADC = ∠ABC = 30° (Opposite angles)
and ∠ADB = ∠ADC – ∠BDC = 30° – 10° = 20°
AB || DC and AC is the transversal
∠ACD = ∠CAB = 70°(Altemate angles)
AB || DC and BD is transversal
∠CDB = ∠ABD = 10°(Altemate angles)
In ∆ABC
∠CAB + ∠ABC + ∠BCA = 180° (Sum of anlges of a triangle)
⇒ 70° + 30° + ∠BCA = 180°
⇒ 100° + ∠BCA = 180°
∠BCA = 180° – 100° = 80°
∠BCD = ∠BCA + ∠ACD = 80° + 70° = 150°
∠BCD = ∠DAB (Opposite angles)
⇒ ∠DAB = 150° and ∠CAD = 150° – 70° = 80°
In ∆OCD,
∠ODC + ∠OCD + ∠COD = 180° (Angles of a triangle)
⇒ ∠ACD + ∠ACD + ∠COD = 180°
⇒ 70° + 10° + ∠COD = 180°
⇒ 80° + ∠COD = 180°
⇒ ∠COD = 180° – 80° = 100°
∠COD = 100°
But ∠AOD + ∠COD = 180° (Linear pair)
⇒ ∠AOD + 100° = 180°
⇒ ∠AOD = 180° – 100° = 80°
⇒ ∠AOD = 80°
But ∠AOB = ∠COD
and ∠BOC = ∠AOD (Vertically opposite angles)
∠AOB = 100° and ∠BOC = 80°
Hence ∠DAB = 150°, ∠ADC = 30°, ∠BCD = 150°
∠AOD = 80° ∠DOC = 100°
∠BOC = 80° ∠AOB = 100°
∠ACD = 70°, ∠CAB = 70°, ∠AD = 20°, ∠ACB = 80°, ∠DBC = 20°, ∠DBA = 10

Question 20.
Find the angles marked with a question mark shown in the figure.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 26
Solution:
In parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 27
CE ⊥ AB and CF ⊥ AD
∠BCE = 40°
In ∆BCE,
∠BCE + ∠CEB + ∠EBC = 180° (Sum of angles of a triangle)
⇒ 40° + 90° + ∠EBC = 180°
⇒ 130° + ∠EBC = 180°
⇒ ∠EBC = 180° – 130° = 50°
or ∠B = 50°
But ∠D = ∠B (Opposite angles)
∠D = 50° or ∠ADC = 50°
Similarly in ∆DCF,
∠DCF + ∠CFD + ∠FDC = 180°
⇒ ∠DCF + 90° + 50° = 180°
⇒ ∠DCF + 140° = 180°
⇒ ∠DCF = 180° – 140° = 40°
But ∠C + ∠B= 180° (Sum of adjacent angles)
⇒ ∠BCE + ∠ECF + ∠DCF + ∠B = 180°
⇒ 40° + ∠ECF + 40° + 50° = 180°
⇒ ∠ECF + 130° = 180°
⇒ 40° + ∠ECF + 40° + 50° = 180°
⇒ ∠ECF + 130° = 180°
⇒ ∠ECF = 180° – 130° = 50°
∠ECF = 50°

Question 21.
The angle between the altitudes of a parallelogram through the same vertex of an obtuse angle of the parallelogram is 60°. Find the angles of the parallelogram.
Solution:
In parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 28
∠A is an obtused angle
AE ⊥ BC and AF ⊥ DC
∠EAF = 60°
In quadrilateral AECF,
∠EAF + ∠F + ∠C + ∠E = 360° (Sum of angles of a quadrilateral)
⇒ 60° + 90° + ∠C + 90° = 360°
⇒ 240° + ∠C = 360°
⇒ ∠C = 360° – 240° = 120°
∠C = 120°
But ∠A = ∠C (Opposite angles)
∠A = 120°
But ∠A + ∠B = 180° (Sum of adjacent angles)
⇒ 120° + ∠B = 180°
⇒ ∠B = 180° – 120° = 60°
∠B = 60°
But ∠D = ∠B (Opposite angles)
∠D = 60°
Hence ∠A = 120°, ∠B = 60°, ∠C = 120° and ∠D = 60°

Question 22.
In the figure, ABCD and AEFG are parallelograms. If ∠C = 55°, what is the measure of ∠F ?
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 29
Solution:
In the parallelogram ABCD,
∠A = ∠C …..(i) (Opposite angles of a parallelogram)
Similarly in parallelogram AEFG,
∠A = ∠F …(ii)
From (i) and (ii),
∠C = ∠F = 55° (∠C = 55°)
Hence ∠F = 55°

Question 23.
In the figure, BDEF and DCEF are each a parallelogram. It is true that BD = DC ? Why or why not ?
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 30
Solution:
In parallelogram BDEF,
BD = EF ……(i) (Opposite sides of a parallelogram)
Similarly, in parallelogram DCEF
DC = EF
From (i) and (ii),
BD = DC
Hence it is true that BD = DC.

Question 24.
In the figure, suppose it is known that DE = DF. Then is ∆ABC isosceles ? Why or why not ?
Solution:
In parallelogram BDEF,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 31
∠B = ∠E ……(i) (Opposite angles of a parallelogram)
Similarly, in parallelogram DCEF,
∠C = ∠F ……(ii)
But DE = DF (Given)
In ∆DEF
∠E = ∠F
From (i) and (ii),
∠B = ∠C
AC = AB (Sides opposite to equal angles)
∆ABC is an isosceles triangle.

Question 25.
Diagonals of parallelogram ABCD intersect at O as shown in the figure. XY contains O, and X, Y are points on opposite sides of the parallelogram. Give reasons for each of the following:
(i) OB = OD
(ii) ∠OBY = ∠ODX
(iii) ∠BOY = ∠DOX
(iv) ∆BOY = ∆DOX
Now, state if XY is bisected at O.
Solution:
In parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 32
diagonals AC and BP intersect each other at O
O is the mid-point of AC and BD.
Through O, XY is draw such that X lies on AD and Y, on BC.
(i) OB = OD (O is mid-point of BD)
(ii) AD || BC and BD is transversal
∠OBY = ∠ODX (Alternate angles)
(iii) ∠BOY = ∠DOX (Vertically opposite angles)
(iv) Now in ∆BOY and ∆DOX,
OB = OD
∠OBY = ∠ODX
∠BOY = ∠DOX
∆BOY = ∆DOX (ASA axiom)
OY = OX (c.p.c.t.)
Hence XY is bisected at O.

Question 26.
In the figure ABCD is a parallelogram, CE bisects ∠C and AF bisects ∠A. In each of the following, if the statement is true, give a reason for the same.
(i) ∠A = ∠C
(ii) ∠FAB = \(\frac { 1 }{ 2 }\) ∠A
(iii) ∠DCE = \(\frac { 1 }{ 2 }\) ∠C
(iv) ∠CEB = ∠FAB
(v) CE || AF.
Solution:
In parallelogram ABCD,
CE is the bisector of ∠C and and AF is the bisector of ∠A.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 33
(i) ∠A = ∠C (Opposite angles of a parallelogram)
(ii) AF is the bisector of ∠A
∠FAB = \(\frac { 1 }{ 2 }\) ∠A
(iii) CE is the bisector of ∠C
∠DCE = \(\frac { 1 }{ 2 }\) ∠C
(iv) From (i), (ii) and (iii)
∠FAB = ∠DCE
(v) ∠FAB = ∠DCE
But these are opposite angles of quadrilateral AECF
AB or AE || DC or FC
AECF is a parallelogram
CE || AF
Hence proved

Question 27.
Diagonals of a parallelogram ABCD intersect at O. AL and CM are drawn perpendiculars to BD such that L and M lie on BD. Is AL = CM ? Why or why not ?
Solution:
In parallelogram ABCD, diagonals AC and BD intersect each other at O.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 34
O is the mid-point of AC and BD.
AL ⊥ BD and CM ⊥ BD.
In ∆ALO and ∆CMO
∠L = ∠M (Each 90°)
∠AOL = ∠COM (Vertically opposite angles)
AO = CO (O is mid-point of AC)
∆ALO = ∆CMO (AAS axiom)
AL = CM (c.p.c.t.)
Hence proved

Question 28.
Points E and F lie on diagonal AC of a parallelogram ABCD such that AE = CF. What type of quadrilateral is BFDE ?
Solution:
In parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 35
AC is its diagonal. E and F are points on AC such that AE = CF
Join EB, BF, FD and DE .
Join also diagonal BD which intersects AC at O
O is the mid-point of AC and BD
AO = OC
But AE = CF
⇒ AO – AE = CO – CF
⇒ EO = OF
But BO = OD (O is mid-point of BD)
Diagonals EF and BD of quadrilateral bisect each other at O.
BFDE is a parallelogram.

Question 29.
In a parallelogram ABCD, AB = 10 cm, AD = 6 cm. The bisector of ∠A meets DC in E. AE and BC produced meet at F. Find the length of CF.
Solution:
AB || DC
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 36
∠DEA = ∠EAB (Alternate angles)
= ∠DAE (EA is bisector of ∠A)
In ∆DAE,
∠DEA = ∠DAE
AD = DE = 6 cm
But DE = AB = 10 cm.
EC = DC – DE = 10 – 6 = 4 cm
AD || BC or BF and AF is transversal
∠DAE = ∠EFC (Alternate angle)
But ∠DAE = ∠DEA Prove
= ∠FEC (DEA = FEC vertically opposite angles)
In ∆ECF,
CE = CF = 4 cm (CE = 4 cm)

 

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RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.1

RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.1

Other Exercises

Question 1.
Identify the terms, their co-efficients for each of the following expressions.
(i) 7x2yz – 5xy
(ii) x2 + x + 1
(iii)3x2y2 – 5x2y2 + z2 + z2
(iv) 9 – ab + be-ca
(v) \(\frac { a }{ 2 } +\frac { b }{ 2 }\)-ab
(vi)2x – 0.3xy + 0.5y
Solution:
(i) Co-efficient of 7x2yz = 7
co-efficient of -5xy = -5
(ii) Co-efficient of x1 = 1
co-efficient of x = 1
co-efficient of 1 = 1
(iii) Co-efficient of 3x2_y2 = 3
co-efficient of -5x2y2z2 = -5
co-efficient of z2 – 1
(iv) Co-efficient of 9 = 9
co-efficient of -ab = -1
co-efficient of be = 1
co-efficient of -ca = -1
(v) Co-efficient of \(\frac { a }{ 2 } =\frac { 1 }{ 2 }\)
Co-efficient of \(\frac { b }{ 2 } =\frac { 1 }{ 2 }\)
co-efficient of -ab = -1
(vi) co-efficient of 0.2x = 0.2
co-efficient of-0.3xy = -0.3
co-efficient of 0.5y = 0.5

Question 2.
Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any category ?
(i) x+y
(ii) 1000
(iii) x + x2 + x3 + x4
(iv) 7 + a + 5b
(v) 2b – 3 b2    
(vi) 2y – 3y2 +4y3
(vii) 5x – 4y + 3x
(viii) 4a – 15a2
(ix) xy+yz + zt + tx

(x)   pqr
(xi) p2q + pq2       
(xii)  2p + 2 q
Solution:
Monomials are (ii), (x)
Binomials are (i), (v), (viii), (xi), (xii)
Trinomials are (iv), (vi) and (vii)
None of these are (iii) and (ix)

 

Hope given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.1 are helpful to complete your math homework.

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RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3

RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3

Other Exercises

Question 1.
The list price of a refrigerator is Rs 9700. If a value added tax of 6% is to be charged on it, how much one has to pay to buy the refrigerator ?
Solution:
List price of refrigerator = Rs 9700
Rate of VAT = 6%
Amount of VAT = Rs. \(\frac { 9700 x 6 }{ 100 }\) = Rs 582
Total price to be paid = Rs 9700 + 582 = Rs 10282

Question 2.
Vikram bought a watch for Rs 825. If this amount includes 10% VAT on the list price, what was the list price of the watch ?
Solution:
Price of watch including VAT = Rs 825
Rate of VAT = 10%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 1

Question 3.
Aman bought a shirt for Rs 374.50 which includes 7% VAT. Find the list price of the shirt.
Solution:
Cost price of the shirt = Rs 374.50
Rate of VAT = 7%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 2

Question 4.
Rani purchases a pair of shoes whose sale price is Rs 175. If he pays a VAT at the rate of 7%, how much amount does he pays as VAT. Find the net value of the pair of shoes.
Solution:
Sale price of a pair of shoes = Rs 175
Rate of VAT = 7%
Total VAT paid = Rs. \(\frac { 175 x 7 }{ 100 }\) = Rs 12.25
and total amount paid = Rs 175 + Rs 12.25 = Rs 187.25

Question 5.
Swarna paid Rs 20 as VAT on a pair of shoes worth Rs 250. Find the rate of VAT.
Solution:
Amount of VAT paid = Rs 20
Price of the pair of shoes = Rs 250
Rate of VAT = \(\frac { 20 x 100 }{ 250 }\) = 8%

Question 6.
Sarita buys goods worth Rs 5,500. She gets a rebate of 5% on it. After getting the rebate if VAT at the rate of 5% is charged, find the amount she will have to pay for the goods.
Solution:
Price of goods = Rs 5,500
Rate of rebate = 5%
Sales price after rebate
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 3
= Rs 5,225
Rate of VAT = 5%
Amount of VAT
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 4
Amount to be paid after paying VAT = Rs 5,225 + 261.25 = Rs 5486.25

Question 7.
The cost of furniture inclusive of VAT is Rs 7,150. If the rate of VAT is 10%, find the original cost of the furniture.
Solution:
Cost of furniture including VAT = Rs 7,150
Rate of VAT = 10%
Original cost of furniture
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 5

Question 8.
A refrigerator is available for Rs 13,750 including VAT. If the rate of VAT is 10%, find the original cost of refrigerator.
Solution:
Cost of refrigerator including VAT = Rs 13,750
Rate of VAT = 10%
Actual price of refrigerator
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 6

Question 9.
A colour TV is available for Rs 13,440 inclusive of VAT. If the original cost of TV is Rs 12,000, find the rate of VAT.
Solution:
Cost of TV including VAT = Rs 13,440
Actual cost = Rs 12,000
Amount of VAT = Rs 13,440 – Rs 12,000 = Rs 1,440
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 7

Question 10.
Reena goes to a shop to buy a radio, costing Rs 2,568. The rate of VAT is 7%. She tells the shopkeeper to reduce the price of the radio such that she has to pay Rs 2,568 inclusive of VAT. Find the reduction needed in the price of radio.
Solution:
Price of radio inclusive of VAT = Rs 2,568
Rate of VAT = 7%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 8
But the cost of radio in the beginning = Rs 2,568
Reduction = Rs 2568 – Rs 2,400 = Rs 168

Question 11.
Rajat goes to a departmental store and buys the following articles :
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 9
Calculate the total amount he has to pay to the store.
Solution:
Price of 2 pairs of shoes @ Rs 800 = Rs800 x 2 = Rs 1,600
Rate of VAT = 5%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 10
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 11
Total amount to be paid = Rs 1,680 + Rs 1,590 + Rs 1,352 = Rs 4,622

Question 12.
Ajit buys a motorcycle for Rs 17,600 including value added tax. If the rate of VAT is 10%, what is the sale price of the motorcycle ?
Solution:
Cost price of motorcycle (including VAT) = Rs 17,600
Rate of VAT = 10%
Sale price of motorcycle
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 12

Question 13.
Manoj buys a leather coat costing Rs 900 at Rs 990 after paying the VAT. Calculate the VAT charged on the cost.
Solution:
Cost price of coat = Rs 900
And sale price including VAT = Rs 990
Amount of VAT = Rs 990 – Rs 900 = Rs 90
Rate of VAT = \(\frac { 90 x 100 }{ 900 }\) = 10%

Question 14.
Rakesh goes to a departmental store and purchases the following articles:
(i) biscuits and bakery products costing Rs 50, VAT @ 5%.
(ii) medicines costing Rs 90. VAT @ 10%,
(iii) clothes costing Rs 400, VAT @ 1% and
(iv) cosmetics costing Rs 150, VAT @ 10% Calculate the total amount to be paid by Rakesh to the store.
Solution:
(i) Cost of biscuits and bakery product = Rs 50
VAT = 5%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 13
Total amount of the bill = Rs 52.50 + Rs 99 + Rs 404 + Rs 165 = Rs 720.50

Question 15.
Rajeeta purchased a set of cosmetics. She paid Rs 165 for it including VAT. If the rate of VAT is 10%, find the sale price of the set.
Solution:
Total price of set including VAT = Rs 165
Rate of VAT = 10%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 14

Question 16.
Sunita purchases a bicycle for Rs 660. She has paid a VAT of 10%. Find the list price of bicycle.
Solution:
Cost price of bicycle including VAT = Rs. 660
Rate of VAT = 10%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 15

Question 17.
The sales price of a television inclusive of VAT is Rs 13,500. If VAT is charged at the rate of 8% of the list price, find the list price of the television.
Solution:
Sale price of television including VAT = Rs 13,500
Rate of VAT = 8%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 16
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 17

Question 18.
Shikha purchased a car with a marked price of Rs 2,10,000 at a discount of 5%. If VAT is charged at the rate of 10%, find the amount Shikha had paid for purchasing the car.
Solution:
Marked price of car = Rs 2,10,000
Rate of discount = 5%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 18

Question 19.
Sharuti bought a set of cosmetic items for Rs 345 including 15% value added tax and a purse for Rs 110 including 10% VAT. What percent is the VAT charged on the whole transactions ?
Solution:
Cost price of set = Rs 345
VAT = 15%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 19
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 20

Question 20.
List price of a cooler is Rs 2,563. The rate of VAT is 10%. The customer requests the shopkeeper to allow a discount in the price of the cooler to such an extent that the price remains Rs 2,563 inclusive of VAT. Find the discount in the price of the cooler.
Solution:
List price of cooler = Rs. 2,563
On request the price of cooler is paid Rs 2,563 including VAT
VAT = 10%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 21
Amount of discount = Rs 2,563 – Rs 2,330 = Rs 233

Question 21.
List price of a washing machine is Rs 9,000. If the dealer allows a discount of 5% on the cash payment, how much money will a customer pay to the dealer in cash, if the rate of VAT is 10%.
Solution:
List price of washing machine = Rs 9,000
Rate of discount = 5%
Amount after giving discount
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 22

Hope given RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax Ex 13.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4

RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4

Other Exercises

Question 1.
The present population of a town is 28,000. If it increases at the rate of 5% per annum, what will be its population after 2 years ?
Solution:
Present population = 28000
Rate of increase (R) = 5% p.a.
Period (n) = 2 years
Population after 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 1

Question 2.
The population of a city is 125000. If the annual birth rate and death rate are 5.5% and 3.5% respectively, calculate the population of city after 3 years.
Solution:
Present population = 125000
Rate of birth = 5.5%
and rate of death = 3.5%
Increase = 5.5 – 3.5 = 2% p.a.
Period = 3 years.
Population after 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 2
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 3

Question 3.
The present population of a town is 25000. It grows at 4%, 5% and 8% during first year, second year and third year respectively. Find its population after 3 years.
Solution:
Present population = 25000
Increase in first year = 4%
in second year= 5% and
in third year = 8%
Population after 3 years =
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 4

Question 4.
Three years ago, the population of a town was 50000. If the annual increase during three successive years be at the rate of 4%, 5% and 3% respectively, find the present population.
Solution:
Three years ago,
Population of a town = 50000
Annual increase in population in first year = 4%
in second year = 5%
and in third year = 3%
Present population
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 5

Question 5.
There is a continuous growth in population of a village at the rate of 5% per annum. If its present population is 9261, what it was 3 years ago ?
Solution:
Let 3 years ago, population = P
Present population = 9261
Rate of increase (R) = 5% p.a.
Period (n) = 3 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 6

Question 6.
In a factory, the production of scooters rose to 46305 from 40000 in 3 years. Find the annual rate of growth of the production of scooters.
Solution:
Production of scooters 3 years ago (P) = 40000
Present production (A) = 46305
Period (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 7
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 8

Question 7.
The annual rate of growth in population of a certain city is 8%. If its present population is 196830, what it was 3 years ago ?
Solution:
Let 3 years ago, the population of a city = P
Rate of growth (R) = 8% p.a.
Present population = 196830
Period (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 9

Question 8.
The population of a town increases at the rate of 50 per thousand. Its population after 2 years will be 22050. Find its present population.
Solution:
Population after 2 years = 22050
Rate of increase = 50 per thousand
Period (n) = 2 years
Let present population = P, then
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 10

Question 9.
The count of bacteria in a culture grows by 10% in the first hour, decreases by 8% in the second hour and again increases by 12% in the third hour. If the count of bacteria in the sample is 13125000, what will be the count of bacteria after 3 hours ?
Solution:
Present count of bacteria = 13125000
In first hour increase = 10%
decrease in second hour = 8%
increase in third hour = 12%
Count of bacteria after 3 hours
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 11

Question 10.
The population of a certain city was 72000 on the last day of the year 1998. During next year it increased by 7% but due to an epidemic it decreased by 10% in the following year. What was its population at the end of the year 2000 ?
Solution:
On the last day of 1998,
Population of a town = 72000
In the first year, increase = 7%
In the second year, decrease = 10%
Population in the last day of 2000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 12

Question 11.
6400 workers were employed to construct a river bridge in four years. At the end of the first year, 25% workers were retrenched. At the end of the second year, 25% of those working at that time were retrenched. However, to complete the project in time, the number of workers was increased by 25% at the end of the third year. How many workers were working during the fourth year ?
Solution:
Number of workers at the beginning = 6400
Period = 4 years.
At the end of 1st year, workers retrenched = 25%
At the end of second year, workers retrenched = 25%
At the end of third year, workers increased = 25%
Total number of workers during the 4 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 13

Question 12.
Aman started a factory with an initial investment of Rs 1,00,000. In the first year, he incurred a loss of 5%. However, during the second year, he earned a profit of 10% which in the third year rose to 12%. Calculate his net profit for the entire period of three years.
Solution:
Initial investment = Rs 100000
Loss in the first year = 5%
Profit in the second year = 10%
Profit in the third year = 12%
Investment at the end of 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 14

Question 13.
The population of a town increases at the rate of 40 per thousand annually. If the present population be 175760, what was the population three years ago ?
Solution:
Present population (A) = 175760
Increase rate = 40 per 1000
Period = 3 years
Let 3 years ago,
population was = P
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 15

Question 14.
The production of a mixi company in 1996 was 8000 mixies. Due to increase in demand it increases its production by 15% in the next two years and after two years its demand decreases by 5%. What will be its production after 3 years ?
Solution:
Production of Mixi in 1996 = 8000
Increase in next 2 years = 15%
Decrease in the third year = 5%
Production after 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 16

Question 15.
The population of a city increases each year by 4% of what it had been at the beginning of each year. If the population in 1999 had been 6760000, find the population of the city in (i) 2001 (ii) 1997.
Solution:
Population of a city in 1999 = 6760000
Increase = 4%
(i) Population in 2001 is after 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 17
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 18

Question 16.
Jitendra set up a factory by investing Rs 25,00,000. During the first two successive years his profits were 5% and f 10% respectively. If each year the profit was on previous year’s capital, compute his total profit.
Solution:
Investment in the beginning = Rs 25,00,000
Profit during the first 2 years = 5% and 10% respectively
Investment after 2 years will be
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 19
= Rs 28,87,500
Amount of profit = Rs 28,87,500 – Rs 25,00,000 = Rs 3,87,500

Hope given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 are helpful to complete your math homework.

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RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3

RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3

Other Exercises

Find each of the following products (1-8)
Question 1.
5x2 x 4x3
Solution:
5x2 x 4x3 = 5 x 4 x x2 x x3
= 20x2 + 3 = 20xs

Question 2.
3a2 x 4b4
Solution:
-3a2 x 4b4 = -3 x 4 x a2b4
= -12a2b4

Question 3.
(-5xy) x (-3x2yz)
Solution:
(-5xy) x (-3x2yz)
= (-5) x (-3)xy x x2yz
= 15x1 + 2xy1+ 1z= 15x3y2z

Question 4.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 1
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 2
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 3

Question 5.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 4
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 5

Question 6.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 6
Solution:

Question 7.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 7
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 8
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 9

Question 8.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 10
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 11

Find each of the following products : (9-17)

Question 9.
(7ab) x (-5ab2c) x (6abc2)
Solution:
(7ab) x (-5ab2c) x (6abc2)
= 7 x (-5) x 6 x a x a x a x b x b2 x b x c x c2
=-210 x a1+1+1 x b1+2+1x c1+2
=-210 x a3b4c3

Question 10.
(-5a) x (-10a2) x (-2a3)
Solution:
(-5a) x (-10a2) x (-2a3)
= (-5) (-10) (-2) x a x a2 x a3
= -100a1 + 2 + 3 = -100a6

Question 11.
(-4x2) x (-6xy2) x (-3yz2)
Solution:
(-4x2) x (-6xy2) x (-3yz2)
= (-4) x (-6) x (-3) x2 x x x y2 x y xz2
= -72x2+1 x y2+1 x z2
= 72x3y3z3

Question 12.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 12
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 13

Question 13.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 14
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 15

Question 14.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 16
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 17

Question 15.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 18
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 19

Question 16.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 20
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 21
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 22

Question 17.
(2.3xy) x (0.1x) x (0.16)
Solution:
(2.3xy) x (0.1x) x (0.16)
= 2.3 x 0.1 x 0.16 x x x x x y
= 0.0368x1 +1 x y = 0.0368x2y

Express each of the following products as a monomials and verify the result in each case for x = 1 : (18 -26)

Question 18.
(3x) x (4x) x (-5x)
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 23

Question 19.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 24
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 25
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 26

Question 20.
(5x4) x (x2)3 x (2x)2
Solution:
(5x4) x (x2)3 x (2x)2
= 5x4 x x2 x 3 x 2x x 2x
= 5x4 * x6 x 4x2 = 5 x 4 x x4 + 6 + 2
= 20x12
Verification:
L.H.S. = (5x4) x (x2)3 x (2x)2
= 5 x (1)4 x [(1)2]3 x (2 x 1)2
= 5 x 1 x (1)2 x 3x (2)2
= 5 x 16 x 22 = 5 x 1 x 4 = 20
R.H.S. = 20x12 = 20 (1)12 = 20 x 1 = 20
∴ L.H.S. = R.H.S.

Question 21.
(x2)3 x (2x) x (-4x) x 5
Solution:
(x2)3 x (2x) x (-4x) x (5)
= x2 x 3 X 2x X (-4x) X 5
= x6 x 2x x (-4x) x 5 = 2 x (-4) x 5x6+1 +1
= -40x8
Verification
L.H.S. = (x2)3 x (2x) x (-4x) x (5)
= (12)3 x (2 x 1) x (-4 x 1) x 5
= 1x 2 x (- 4) x 5 = 16 x 2 x (-4) x 5
= 1 x 2 x (-4) x 5 = -40
R.H.S. = -40x8 = -40 x (1)8
= -40 x 1 = -40
∴ L.H.S. = R.H.S.

Question 22.
Write down the product of -8x2y6 and – 20xy Verify the product for x = 2.5, y = 1.
Solution:
Product of -8x2y6 and -20xy
= (-8x2y6) x (-20xy)
= -8 x (-20) x2 x x x y6 x y = 160x2 + 1 x y6 + 1
= 160x3y3
Verification.
L.H.S. = (-8x2y6) x (-20xy)
= -8 x (2.5)2 x (1) x (-20 x 2.5 x 1)
= -8 x 6.25 x 1 x -20 x 2.5
= (-50) x (-50) = 2500
R.H.S. = 160 x = 160 (2.5)3 x (1)7
= 160 x 15.625 x 1 =2500
∴ L.H.S. = R.H.S.

Question 23.
Evaluate : (3.2x6y3) x (2.1x2y2) when x = 1 and y = 0.5.
Solution:
3.2x6y3 x 2.1x2y2
= 3.2 x 2.1 x x6+2 x y3+2
= 6.72x8y5 = 6.72 x (1)8 x (0.5)5
= 6.72 x 1 x 0.03125
= 0.21

Question 24.
Find the value of (5x6) x (-1.5x2y3) x (-12xy2) when x = 1, y = 0.5.
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 27

Question 25.
Evaluate : (2.3a5b2) x (1.2a2b2) when a = 1, b = 0.5.
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 28

Question 26.
Evaluate : (-8x2y6) x (-20xy) for x = 2.5 and y = 1.
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 29

Express each of the following products as a monomials and verify the result for x = 1,y = 2: (27-31)

Question 27.
(-xy3) x (yx3) x (xy)
Solution:
(-xy3) x (yx3) x (xy)
= -x x xx x x yx y x y = -x1 + 3 + 1 x y3 + 1 + = -x5y5
Verification:
L.H.S. = (-xy3) x (yx3) x (xy)
= (-1 x 23) x [2 x (1)3] x (1 X 2)
= (-1 x 8) x (2 x 1) x (1 x 2)
= -8 x 2 x 2 = -32
R.H.S. =-x5y5  = -(1)5 (2)5
= -1 x 32 =-32
∴ L.H.S. = R.H.S.

Question 28.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 30
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 31
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 32

Question 29.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 33
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 34

Question 30.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 35
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 36

Question 31.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 37
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 38
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 39

Question 32.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 40
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 41
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 42

Question 33.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 43
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 44

Hope given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3

RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3

Other Exercises

Question 1.
On what sum will the compound interest at 5% p.a. annum for 2 years compounded annually be Rs 164 ?
Solution:
Let Principal (P) = Rs 100
Rate (R) = 5% p.a.
Period (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 1
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 2

Question 2.
Find the principal of the interest compounded annually at the rate of 10% for two years is Rs 210.
Solution:
Let principal (P) = Rs 100
Rate (R) = 10% p.a.
Period (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 3
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 4

Question 3.
A sum amounts to Rs 756.25 at 10% per annum in 2 years, compounded annually. Find the sum.
Solution:
Amount (A) = Rs 756.25
Rate (R) = 10% p.a.
Period (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 5

Question 4.
What sum will amount to Rs 4913 in 18 months, if the rate of interest is 12\(\frac { 1 }{ 2 }\) % per annum, compounded half-yearly.
Solution:
Amount (A) = Rs 4,913
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 6

Question 5.
The difference between the compound interest and simple interest on a certain sum at 15% per annum for 3 years is Rs 283.50. Find the sum.
Solution:
Let sum (P) = Rs 100
Rate (R) = 15% p.a.
Period (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 7
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 8
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 9

Question 6.
Rachna borrowed a certain sum at the rate of 15% per annum. If she paid at the end of two years Rs 1,290 as interest compounded annually, find the sum she borrowed.
Solution:
C.I. = Rs 1,290
Rate (R) = 15% p.a.
Period (n) = 2 years
Let sum (P) = Rs 100
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 10

Question 7.
The interest on a sum of Rs 2,000 is being compounded annually at the rate of 4% per annum. Find the period for which the compound interest is Rs 163.20.
Solution:
Sum (P) = Rs 2,000
C.I. = Rs 163.20
Amount (A) = P + C.I. = Rs 2000 + Rs 163.20 = Rs 2163.20
Rate (R) = 4%
Let period = n years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 11
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 12

Question 8.
In how much time would Rs 5,000 amount to Rs 6,655 at 10% per annum compound interest ?
Solution:
Principal (P) = Rs 5,000
Amount (A) = Rs 6,655
Rate (R) = 10%
Let period = n years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 13

Question 9.
In what time will Rs 4,400 becomes Rs 4,576 at 8% per annum interest compounded half-yearly ?
Solution:
Principal (P) = Rs 4,400
Amount (A) = Rs 4,576
Rate (R) = 8% or 4% half-yearly
Let period = n half-years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 14
n = 1
Period = 1 half year

Question 10.
The difference between the S.I. and C.I. on a certain sum of money for 2 years at 4% per annum is Rs 20. Find the sum.
Solution:
Difference between C.I. and S.I. = Rs 20
Rate (R) = 4% p.a.
Period (n) = 2 years
Let principal (P) = Rs 100
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 15
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 16

Question 11.
In what time will Rs 1,000 amount to Rs 1,331 at 10% per annum compound interest.
Solution:
Principal (P) = Rs 1,000
Amount (A) = Rs 1,331
Rate (R) = 10% p.a.
Let period = n year
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 17

Question 12.
At what rate percent compound interest per annum will Rs 640 amount to Rs 774.40 in 2 years ?
Solution:
Principal (P) = Rs 640
Amount (A) = Rs 774.40
Period (n) = 2 years.
Let R be the rate of interest p.a.
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 18
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 19

Question 13.
Find the rate percent per annum if Rs 2000 amount to Rs 2,662 in 1\(\frac { 1 }{ 2 }\) years, interest being compounded half-yearly ?
Solution:
Principal (P) = Rs 2,000
Amount (A) = Rs 2,662
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 20

Question 14.
Kamala borrowed from Ratan a certain rate for two years simple interest. She lent this sum at the same rate to Hari for two years compound interest. At the end of two years, she received Rs 210 as compound interest, but paid Rs 200 only as simple interest. Find the sum and the rate of interest.
Solution:
Simple interest = Rs 200
and compound interest = Rs 210.
Period = 2 years
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 21

Question 15.
Find the rate percent per annum, if Rs 2,000 amount to Rs 2,315.25, in an year and a half, interest being compounded six monthly.
Solution:
Principal (P) = Rs 2,000
Amount (A) = Rs 2315.25
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 22
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 23

Question 16.
Find the rate at which a sum of money will double itself in 3 years, if the interest is compounded annually.
Solution:
Let Principal (P) = Rs 100
then Amount (A) = Rs 200
Period (n) = 3 years
Let R be the rate % p.a.
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 24

Question 17.
Find the rate at which a sum of money will become four times the original amount in 2 years, if the interest is compounded half- yearly.
Solution:
Let Principal (P) = Rs 100
Then Amount (A) = Rs 400
Period (n) = 2 years or 4 half years
Let R be the rate % half-yearly, then
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 25
Rate % = 41.42% half yearly and 82.84% p.a.

Question 18.
A certain sum amounts to Rs 5,832 in 2 years at 8% compounded interest. Find the sum.
Solution:
Amount (A) = Rs 5,832
Let P be the sum
Rate (R) = 8% p.a.
Period (n) = 2 years
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 26

Question 19.
The difference between the compound interest and simple interest on a certain sum for 2 years at 7.5% per annum is Rs 360. Find the sum.
Solution:
Let sum (P) = Rs 100
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 27
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 28

Question 20.
The difference in simple interest and compound interest on a certain sum of money at 6\(\frac { 2 }{ 3 }\) % per annum for 3 years is Rs 46. Determine the sum.
Solution:
Let sum (P) = Rs 100
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 29
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 30
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 31

Question 21.
Ishita invested a sum of Rs 12,000 at 5% per annum compound interest. She received an amount of Rs 13,230 after n years, Find the value of n.
Solution:
Principal (P) = Rs 12,000
Amount (A) = Rs 13,230
Rate (R) = 5% p.a.
Period = n years
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 32

Question 22.
At what rate percent per annum will a sum of Rs 4,000 yield compound interest of Rs 410 in 2 years ?
Solution:
Principal (P) = Rs 4,060
C.I. = Rs 410
Amount (A) = Rs 4,000 + 410 = Rs 4,410
Let rate = R % p.a.
Period (n) = 2 years
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 33

Question 23.
A sum of money deposited at 2% per annum compounded annually becomes Rs 10404 at the end of 2 years. Find the sum deposited.
Solution:
Amount (A) = Rs 10,404
Rate (R) = 2% p.a.
Period (n) = 2 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 18
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 19

Question 24.
In how much time will a sum of Rs 1,600 amount to Rs 1852.20 at 5% per annum compound interest ?
Solution:
Principal (P) = Rs 1,600
Amount (A) = Rs 1852.20
Rate (R) = 5% p.a.
Let n be the time
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 20

Question 25.
At what rate percent will a sum of Rs 1,000 amount to Rs 1102.50 in 2 years at compound interest ?
Solution:
Principal (P) = Rs 1,000
Amount (A) = Rs 1102.50 .
Period (n) = 2 years
Let R be the rate of interest p.a.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 21

Question 26.
The compound interest on Rs 1,800 at 10% per annum for a certain period of time is Rs 378. Find the time in years.
Solution:
Principal (P) = Rs 1,800
C.I. = Rs 378
Amount (A) = P + C.I. = Rs 1,800 + 378 = Rs 2,178
Rate (R) = 10% p.a.
Let n be the period in years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 22
Comparing, we get:
n = 2
Period = 2 years

Question 27.
What sum of money will amount to Rs 45582.25 at 6\(\frac { 3 }{ 4 }\) % per annum in two years, interest being compounded annually ?
Solution:
Amount (A) = Rs 45582.25
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 23

Question 28.
Sum of money amounts to Rs 4,53,690 in 2 years at 6.5% per annum compounded annually. Find the sum.
Solution:
Amount (A) = Rs 4,53,690
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 24
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 25
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 26

Hope given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2

RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2

Other Exercises

Question 1.
Compute the amount and the compound interest in each of the following by using the formula when :
(i) Principal = Rs 3,000, Rate = 5%, Time = 2 years
(ii) Principal = Rs 3,000 Rate = 18%, Time = 2 years
(iii) Principal = Rs 5,000 Rate = 10 paise per rupee per annum, Time = 2 years
(iv) Principal = Rs 2,000, Rate = 4 paise per rupee per annum, Time = 3 years.
(v) Principal = Rs 12,800, Rate = 7\(\frac { 1 }{ 2 }\) %, Time = 3 years
(vi) Principal = Rs 10,000, Rate = 20% per annum compounded half-yearly, time = 2 years
(vii) Principal = Rs 1,60,000, Rate = 10 paise per rupee per annum compounded half- yearly, Time = 2 years.
Solution:
(i) Principal (P) = Rs 3,000
Rate (R) = 5% p.a.
Time (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 1
and compound interest (C.I) = A – P = Rs 3307.50 – Rs 3,000 = Rs 307.50
(ii) Principal (P) = Rs 3,000
Rate (R) = 18% p.a.
Time (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 2
and compound interest (C.I.) = A – P = Rs 4177.20 – Rs 3,000 = Rs 1177.20
(iii) Principal (P) = Rs 5,000
Rate (R) =10 paise per rupee or 10% p.a.
Time (n) = 2 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 3
C.I. = A – P = Rs 6,050 – Rs 5,000 = Rs 1,050
(iv) Principal (P) = Rs 2,000
Rate (R) = 4 paise per rupee or 4% p.a.
Time (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 4
C.I. = A – P = Rs 2249.73 – Rs 2,000 = Rs 249.73
(v) Principal (P) = Rs 12,800
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 5
C.I. = A – P = Rs 15901.40 – Rs 12,800 = Rs 3101.40
(vi) Principal (P) = Rs 10,000
Rate (R) = 20% p.a. or 10% half-yearly
Time = 2 years or 4 half-years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 6
C.I. = A – P = Rs 14,641 – Rs 10,000 = Rs 4,641
(vii) Principal (P) = Rs 1,60,000
Rate (R) = 10 paise per rupee or 10% p.a. or 5% half-yearly
Time (n) = 2 years or 4 half-years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 7
C.I. = A – P = Rs 1,94,481 – Rs 1,60,000 = Rs 34,481

Question 2.
Find the amount of Rs 2,400 after 3 years, when the interest is compounded annually at the rate of 20% per annum.
Solution:
Principal (P) = Rs 2,400
Rate (R) = 20%
Time (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 8

Question 3.
Rahman lent Rs 16,000 to Rasheed at the rate of 12\(\frac { 1 }{ 2 }\) % per annum compound interest. Find the amount payable by Rasheed to Rahman after 3 years.
Solution:
Principal (P) = Rs 16,000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 9

Question 4.
Meera borrowed a sum of Rs 1,000 from Sita for two years. If the rate of interest is 10% compounded annually find the amount that Meera has to pay back.
Solution:
Amount of loan (P) = Rs 1,000
Rate (R) = 10% p.a.
Period (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 10

Question 5.
Find the difference between the compound interest and simple interest. On a sum of Rs 50,000 at 10% per annum for 2 years.
Solution:
Principal (P) = Rs 50,000
Rate (R) = 10% p.a.
Period (n) = 2 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 11
Difference between C.I. and S.I. = Rs 10,500 – Rs 10,000 = Rs 500

Question 6.
Amit borrowed Rs 16,000 at 17\(\frac { 1 }{ 2 }\) % per annum simple interest on the same day, he lent it to Ashu at the same rate but compounded annually. What does he gain at the end of 2 years ?
Solution:
Amount of loan (P) = Rs 16,000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 12
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 13
C.I. = A – P = Rs 22,090 – Rs 16,000 = Rs 6,090
Now gain = C.I. – S.I = Rs 6,090 – 5,600 = Rs 490

Question 7.
Find the amount of Rs 4,096 for 18 months at 12\(\frac { 1 }{ 2 }\) % per annum, interest being compounded semi-annually ?
Solution:
Principal (P) = Rs 4,096
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 14

Question 8.
Find the amount and the compound interest on Rs 8,000 for 1\(\frac { 1 }{ 2 }\) years at 10% per annum, compounded half-yearly.
Solution:
Principal (P) = Rs 8,000
Rate (R) = 10% p.a. or 5% half yearly
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 15
and C.I. = A – P = Rs 9,261 – Rs 8,000 = Rs 1,261

Question 9.
Kamal borrowed Rs 57,600 from LIC against her policy at 12\(\frac { 1 }{ 2 }\) % per annum to build a house. Find the amount that she pays LIC after 1\(\frac { 1 }{ 2 }\) years if the interest is calculated half-yearly.
Solution:
Amount of loan (P) = Rs 57,600
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 16

Question 10.
Abha purchased a house from Avas Parishad on credit. If the cost of the house is Rs 64,000 and the rate of interest is 5% per annum compounded half-yearly, find the interest paid by Abha after one year and a half.
Solution:
Price of house (P) = Rs 64,000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 17
Compound interest (C.I.) = A – P = Rs 68,921 – Rs 64,000 = Rs 4,921

Question 11.
Rakesh lent out Rs 10,000 for 2 years at 20% per annum compounded annually. How much more he could earn if the interest be compounded half-yearly ?
Solution:
Principal (P) = Rs 10,000
Rate (R) = 20% p.a. or 10% half-yearly
Period (n) = 2 years or 4 half-years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 18
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 19
C.I. = Rs 14,400 – Rs 10,000 = Rs 4,400
Now difference in C.I. = Rs 4,641 – Rs 4,400 = Rs 241

Question 12.
Romesh borrowed a sum of Rs 2,45,760 at 12.5% per annum compounded annually. On the same day, he lent out his money to Ramu at the same rate of interest but compounded semi-annually. Find his gain after 2 years.
Solution:
In first case,
Principal (P) = Rs 2,45,760
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 20
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 21
C.I. = A – P = Rs 313203.75 – Rs 2,45,760 = Rs 67443.75
Gain = 67443.75 – Rs 65,280 = Rs2163.75

Question 13.
Find the amount that David would receive if he invests Rs 8,192 for 18 months at 12\(\frac { 1 }{ 2 }\) % per annum, the interest being compounded half-yearly.
Solution:
Principal (P) = Rs 8,192
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 22
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 23

Question 14.
Find the compound interest on Rs 15,625 for 9 months at 16% per annum, compounded quarterly.
Solution:
Principal (P) = Rs 15,625
Rate (R) = 16% p.a. or 4% quarterly
Period (n) = 9 months or 3 quarters
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 24
Compound interest = A – P = Rs 17,576 – Rs 15,625 = Rs 1,951

Question 15.
Rekha deposited Rs 16,000 in a foreign bank which pays interest at the rate of 20% per annum compounded quarterly, find the interest received by Rekha after one year.
Solution:
Principal (P) = Rs 16,000
Rate (R) = 20% p.a. or 5% quarterly
Period (n) = one year or 4 quarters
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 25
C.I. = A – P = Rs 19448.10 – Rs 16,000 = Rs 3448.10

Question 16.
Find the amount of Rs 12,500 for 2 years compounded annually, the rate of interest being 15% for the first year and 16% for the second year.
Solution:
Principal (P) = Rs 12,500
Rate (R1) = 15% p.a. for first year
R2 = 16% p.a. for second year
Period = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 26

Question 17.
Ramu borrowed Rs 15,625 from a finance company to buy a scooter. If the rate of interest be 16% per annum compounded annually, what payment he will have to make after 2\(\frac { 1 }{ 4 }\) years ?
Solution:
Principal (P) = Rs 15,625
Rate (R) = 16%
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 27
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 28

Question 18.
What will Rs 1,25,000 amount to at the rate of 6% if interest is calculated after every 4 months for one year ?
Solution:
Principal (P) = Rs 1,25,000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 29

Question 19.
Find the compound interest at the rate of 5% for 3 years on that principal which in 3 years at the rate of 5% per annum gives Rs 12,000 as simple interest.
Solution:
In first case,
S.I. = Rs 12,000
Rate (R) = 5% p.a.
Period (T) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 30
= Rs 80,000
In second case,
Principal (P) = Rs 80,000
Rate (R) = 5% p.a.
Period (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 31
C.I. = A – P = Rs 92,610 – 80,000 = Rs 12,610

Question 20.
A sum of money was lent for 2 years at 20% compounded annually. If the interest is payable half-yearly instead of yearly, then the interest is Rs 482 more. Find the sum.
Solution:
Let Sum (P) = Rs x
Rate (R) = 20% p.a. or 10% half-yearly
Period (n) = 2 years or 4 half years
In first case,
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 32
Interests = A – P = Rs 146.41 – Rs 100 = Rs 46.41
Now difference in interests = Rs 46.41 – Rs 44.00 = Rs 2.41
If difference is 2.41 then sum is 100 If difference is Rs 482, then sum
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 33

Question 21.
Simple interest on a sum of money for 2 years at 6\(\frac { 1 }{ 2 }\) % per annum is Rs 5,200. What will be the compound interest on the sum at the same rate for the same period ?
Solution:
In first case,
S.I. = Rs 5,200
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 34
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 35
Compound interest = A – P = Rs 45,369 – Rs 40,000 = Rs 5,369

Question 22.
Find the compound interest at the rate of 5% per annum for 3 years on that principal which in 3 years at the rate of 5% per annum gives Rs 1,200 as simple interest.
Solution:
In first case,
S.I. = Rs 1,200
Rate (R) = 5% p.a.
Period (T) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 36
In second case,
Principal (P) = Rs 8,000
Rate (R) = 5% p.a.
Period (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 37
= Rs 9,261
C.I. = A – P = Rs 9,261 – Rs 8000 = Rs 1,261

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