CBSE Class 8

RD Sharma Class 8 Solutions Chapter 15 Understanding Shapes I (Polygons) Ex 15.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 15 Understanding Shapes I Ex 15.1

Question 1.
Draw rough diagrams to illustrate the following:
(i) Open curve
(ii) Closed curve
Solution:

Question 2.
Classify the following curves as open or closed.

Solution:
Open curves : (i), (iv) and (v) are open curves.
(ii) , (iii), and (vi) are closed curves.

Question 3.
Draw a polygon and shade its interior. Also draw its diagonals, if any.
Solution:
In the given polygon, the shaded portion is its interior region AC and BD are the diagonals of polygon ABCD.

Question 4.
Illustrate, if possible, each one of the following with a rough diagram:
(i) A closed curve that is not a polygon.
(ii) An open curve made up entirely of line segments.
(iii) A polygon with two sides.
Solution:
(i) Close curve but not a polygon.

(ii) An open curve made up entirely of line segments.

(iii) A polygon with two sides. It is not possible. At least three sides are necessary

Question 5.
Following are some figures : Classify each of these figures on the basis of the following:


(i) Simple curve
(ii) Simple closed curve
(iii) Polygon
(iv) Convex polygon
(v) Concave polygon
(vi) Not a curve
Solution:
(i) It is a simple closed curve and a concave polygon.
(ii) It is a simple closed curve and convex polygon.
(iii) It is neither a curve nor polygon.
(iv) it is neither a curve not a polygon.
(v) It is a simple closed curve but not a polygon.
(vi) It is a simple closed curve but not a polygon.
(vii) It is a simple closed curve but not a polygon.
(viii) It is a simple closed curve but not a polygon.

Question 6.
How many diagonals does each of the following have ?
(i) A convex quadrilateral
(ii) A regular hexagon
(iii) A triangle.
Solution:
(i) A convex quadrilateral
Here n = 4

Question 7.
What is a regular polygon ? State the name of a regular polygon of:
(i) 3 sides
(ii) 4 sides
(iii) 6 sides.
Solution:
A regular polygon is a polygon which has all its sides equal and so all angles are equal,
(i) 3 sides : It is an equilateral triangle.
(ii) 4 sides : It is a square.
(iii) 6 sides : It is a hexagon.

Hope given RD Sharma Class 8 Solutions Chapter 15 Understanding Shapes I Ex 15.1 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 14 Compound Interest Ex 14.5

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5

Other Exercises

• RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1
• RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2
• RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3
• RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4
• RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5

Question 1.
Mr. Cherian purchased a boat for Rs 16,000. If the total cost of the boat is depreciating at the rate of 5% per annum, calculate its value after 2 years.
Solution:
Cost of boat = Rs 16,000
Rate of depreciating = 5% p.a.
Period = 2 years
Value of boat after 2 years

Question 2.
The value of a machine depreciates at the rate of 10% per annum. What will be its value 2 years hence, if the present value is Rs 1,0,000 ? Also, find the total depreciation during this period.
Solution:
Present value of machine = Rs 1,00,000
Rate of depreciation = 10% p.a.
Period (n) = 2 years
Value of machine after 2 years

Question 3.
Pritam bought a plot of land for Rs 6,40,000. Its value is increasing by 5% of its previous value after every six months. What will be the value of the plot after 2 years ?
Solution:
Present value of plot = Rs 6,40,000
Increase = 5% per half year
Period (n) = 2 years or 4 half years

Question 4.
Mohan purchased a house for Rs 30,000 and its value is depreciating at the rate of 25% per year. Find the value of the house after 3 years.
Solution:
Present value of the house (P) = Rs 30,000
Rate of depreciation = 25% p.a.
Period (n) = 3 years
Value of house after 3 years

Question 5.
The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs 43,740, find its purchased price.
Solution:
Let the purchase price of machine = Rs P
Rate of depreciation = 10% p.a.
Period (n) = 3 years.
and present value = Rs 43,740

Question 6.
The value of a refrigerator which was purchased 2 years ago, depreciates at 12% per annum. If its present value is Rs 9,680, for how much was it purchased ?
Solution:
Let the refrigerator was purchased for = Rs P
Rate of depreciation (R) = 12% p.a.
Period (n) = 2 years
and present value (A) = Rs 9,680

Question 7.
The cost of a TV set was quoted Rs 17,000 at the beginning of 1999. In the beginning of2000, the price was hiked by 5%. Because of decrease in demand the cost was reduced by 4% in the beginning of 2001. What is the cost of the TV set in 2001 ?
Solution:
List price of TV set in 1999 = Rs 17,000
Rate of hike in 2000 = 5%
Rate of decrease in 2001 = 4%
Price of TV set in 2001

Question 8.
Ashish started the business with an initial investment of Rs 5,00,000. In the first year, he incurred a loss of 4%. However, during the second year he earned a profit of 5% which in third year, rose to 10%. Calculate the net profit for the entire period of 3 years.
Solution:
Initial investment = Rs 5,00,000
In the first year, rate of loss = 4%
In the second year, rate of gain = 5%
and in the third year, rate of gain = 10%
Investment after 3 years

Hope given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 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 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

• RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.1
• RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.2
• RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.3
• RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.4
• RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.5

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.

(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.

(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.

(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.

(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.

(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.

(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.

(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

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

• 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 6.2
• 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 6.4
• 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 6.6
• RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7

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)
= 6x^{2 +2} + 9x²y² + 2x²y² + 3y^{2 + 2}
= 6x⁴ + 11 x²y² + 3y⁴

Question 6.

Solution:

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
= x^{6 + 2} + x⁶y² – x²y⁶ – y^{6 +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
= 2x^{2 + 2} y²– 2x² x y^{2 + 2}– 5x¹⁺² y²+5xy^{2 + 2}
= 2x⁴y²– 2x²y⁴ – 5x³y²+ 5xy⁴

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

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
= 8x^{2 + 3} + 10x^{2 + 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
= 6xy^{1 + 2}– 4xy + 9y^{2 + 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-2x^{2 + 2}x y^{1 +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.

Solution:

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
= x^{2 +3} y² + 4x^{2 + 2} y^{2 +1} – 2x^{2 +1} y^{2+ 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
= 3a^{2 + 2}b² + 7a²⁺¹ b²⁺¹+ 2a²b^{2 + 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
= x^{3 +1} y² + 2x³y³ – x^{2 +1} y^{1+ 2} – 2x²y^{1 + 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 (7_x-3y)-y(7_x-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 (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

• RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1
• RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2
• RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.3

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.

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

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.

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

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

(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.

(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.

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.

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