## 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 20 Mensuration I (Area of a Trapezium and a Polygon) Ex 20.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1

Other Exercises

Question 1.
A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m² ?
Solution:
Area of floor = 1080 m²
Base of parallelogram shaped tile (b) = 24 cm
and corresponding height (h) = 10 cm
Area of one tile = b x h = 24 x 10 = 240 cm²

Question 2.
A plot is in the form of a rectangle ABCD having semi-circle on BC as shown in Fig. If AB = 60 m and BC = 28 m, find the area of the plot.

Solution:
Length of rectangular portion (l) = 60 m
and breadth (b) = 28 m
Area of the rectangular plot = l x b = 60 x 28 m² = 1680 m²
Radius of semicircular portion (r) = $$\frac { b }{ 2 }$$ = $$\frac { 28 }{ 2 }$$ = 14 m
Area = $$\frac { 1 }{ 2 }$$ πr²
= $$\frac { 1 }{ 2 }$$ x $$\frac { 22 }{ 7 }$$ x 14 x 14 m²
= 308 m²
Total area of the plot = 1680 + 308 = 1988 m²

Question 3.
A playground has the shape of a rectangle, with two semi-circles on its smaller sides as diameters, added to its outside. If the sides of the rectangle are 36 m and 24.5 m, find the area of the playground. (Take π = $$\frac { 22 }{ 7 }$$).
Solution:
Length of rectangular portion (l) = 36 m
and breadth (b) = 24.5 m

= $$\frac { 22 }{ 7 }$$ x 150.0625 m²
= 471.625 m²
Total area of the playground = 471.625 + 882 = 1353.625 m²

Question 4.
A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants of radii 3.5 have been cut. Find the area of the remaining part.
Solution:
Length of rectangular piece (l) = 20 m
breadth (b) = 15 m

Area of rectangular piece = l x b = 20 x 15 = 300 m²
Radius of each quadrant (r) = 3.5 m
Total area of 4 quadrants

Area of the remaining portion = 300 – 38.5 m² = 261.5 m²

Question 5.
The inside perimeter of a running track (shown in Fig.) is 400 m. The length of each of the straight portion is 90 m and the ends are semi-circles. If track is everywhere 14 m wide, find the area of the track. Also, find the length of the outer running track.

Solution:
Inner perimeter = 400 m.
Length (l) = 90 m.
Perimeter of two semicircles = 400 – 2 x 90 = 400 – 180 = 220 m

Question 6.
Find the area of the Figure in square cm, correct to one place of decimal. (Take π = $$\frac { 22 }{ 7 }$$)

Solution:
Length of square (a) = 10 cm.
Area = a² = (10)² = 100 cm²
Base of the right triangle AED = 8 cm
and height = 6 cm

Question 7.
The diameter of a wheel of a bus is 90 cm which makes 315 revolutions per minute. Determine its speed in kilometres per hour. (Take π = $$\frac { 22 }{ 7 }$$)
Solution:
Diameter of the wheel (d) = 90 cm.

Question 8.
The area of a rhombus is 240 cm² and one of the diagonal is 16 cm. Find another diagonal.
Solution:
Area of rhombus = 240 cm²
Length of one diagonal (d1) = 16 cm
Second diagonal (d2)

Question 9.
The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution:
In rhombus, diagonal (d1) = 7.5 cm
and diagonal (d2) = 12 cm

Question 10.
The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.
Solution:
In quadrilateral shaped field ABCD,
diagonal AC = 24 m

and perpendicular BL = 13 m
and perpendicular DM on AC = 8 m
Area of the field ABED = $$\frac { 1 }{ 2 }$$ x AC x (BL + DM)
= $$\frac { 1 }{ 2 }$$ x 24 x (13 + 8) m²
= 12 x 21 = 252 m²

Question 11.
Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
Solution:
Side of rhombus (b) = 6 cm
Altitude (h) = 4 cm

Question 12.
The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m² is Rs 4.
Solution:
Number of rhombus shaped tiles = 300
Diagonals of each tile = 45 cm and 130 cm

Rate of polishing the tiles = Rs 4 per m²
Total cost = 202.5 x 4 = Rs 810

Question 13.
A rectangular grassy plot is 112 m long and 78 broad. It has a gravel path 2.5 m wide all around it on the side. Find the area of the path and the cost of constructing it at Rs 4.50 per square metre.
Solution:
Length of rectangular plot (l) = 112 m
and breadth (b) = 78 m
Width of path = 2.5 m

Inner length = 112 – 2 x 2.5 = 112 – 5 = 107 m
and inner breadth = 78 – 2 x 2.5 = 78 – 5 = 73 m
Area of path = outer area – inner area
= (112 x 78 – 107 x 73) m² = 8736 – 7811 = 925 m²
Rate of constructing = Rs 4.50 per m²
Total cost = 925 x Rs 4.50 = Rs 4162.50

Question 14.
Find the area of a rhombus, each side of which measures 20 cm and one of whose diagonals is 24 cm.
Solution:
Side of rhombus = 20 cm.
One diagonal (d1) = 24 cm

Diagonals of a rhombus bisect each other at right angle
AB = 20 cm
and OA = $$\frac { 1 }{ 2 }$$ AC = $$\frac { 1 }{ 2 }$$ x 24 cm = 12 cm
In right-angled ∆AOB,
AB² = AO² + BO² (Pythagoras theorem)
⇒ (20)² = (12)² + BO²
⇒ 400 = 144 + BO²
⇒ BO² = 400 – 144 = 256 = (16)²
⇒ BO = 16 cm
and diagonal BD = 2 x BO = 2 x 16 = 32 cm
Now area of rhombus ABCD

Question 15.
The length of a side of a square field is 4 m. What will be the altitude of the rhombus if the area of the rhombus is equal to the square field and one of its diagonal is L m ?
Solution:
Side of square = 4 m
Area of square = (a)² = 4 x 4 =16 m²

Diagonals of a rhombus bisect each other at right angles.
In right ∆AOB
AB² = QA² + BO² (Pythagoras theorem)
= (8)² + (1)² = 64 + 1 = 65
AB = √65 m.
Now, length of perpendicular AL (h)

Question 16.
Find the area of the field in the form of a rhombus, if the length of each side be 14 cm and the altitude be 16 cm.
Solution:
Length of each side of rhombus = 14 cm.
Length of altitude = 16 cm
Area = Base x altitude = 14 x 16 cm² = 224 cm²

Question 17.
The cost of fencing a square field at 60 paise per metre is Rs 1,200. Find the cost of reaping the field at the rate of 50 paise per 100 sq. metres.
Solution:
Cost of fencing the square field = Rs 1,200
Rate = 60 paise per m.

Question 18.
In exchange of a square plot one of whose sides is 84 m, a man wants to buy a rectangular plot 144 m long and of the same area as of the square plot. Find the width of the rectangular plot.
Solution:
Side of a square plot = 84 m
Area = (a)² = (84)² = 84 x 84 m² = 7056 m²
Area of rectangular field = 7056 m²
Length (l) = 144 m

Question 19.
The area of a rhombus is 84 m². If its perimeter is 40 m, then find its altitude.
Solution:
Area of rhombus = 84 m²
Perimeter = 40 m

Question 20.
A garden is in the form of a rhombus whose side is 30 metres and the corresponding altitude is 16 m. Find the cost of levelling the garden at the rate of Rs 2 per m².
Solution:
Side of rhombus garden (b) = 30 m.
Altitude (h) = 16 m
Area = Base x altitude = 30 x 16 = 480 m²
Rate of levelling the garden = Rs 2 per m²
Total cost = Rs 480 x 2 = Rs 960

Question 21.
A field in the form of a rhombus has each side of length 64 m and altitude 16 m. What is the side of a square field which has the same area as that of a rhombus ?
Solution:
Length of side of rhombus (b) = 64 m
and altitude (h) = 16 m
Area = b x h = 64 x 16 m² = 1024 m²
Now area of square = 1024 m²
Side of the square = √Area = √1024 m = 32 m

Question 22.
The area of a rhombus is equal to the area of a triangle whose base and the corresponding altitude are 24.8 cm and 16.5 cm respectively. If one of the diagonals of the rhombus is 22 cm, find the length of the other diagonal.
Solution:
Base of triangle (b) = 24.8 cm
and altitude (h) = 16.5 cm
Area of triangle = $$\frac { 1 }{ 2 }$$ x base x height
= $$\frac { 1 }{ 2 }$$ x bh= $$\frac { 1 }{ 2 }$$ x 24.8 x 16.5 cm² = 204.6 cm²
Area of rhombus = 204.6 cm²
Length of one diagonal (d1 = 22 cm
Second diagonal (d2)

Hope given RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 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

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 6 Algebraic Expressions and Identities Ex 6.6

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 6.6

Other Exercises

Question 1.
Write the following squares of bionomials as trinomials :

Solution:
Using the formulas
(a + b)2 = a2 + 2ab + b2 and (a – b)2 = a2 – 2ab + b2
(i) (a + 2)2 = (a)2 + 2 x a x 2 + (2)2
{(a + b)2 = a2 + 2ab + b2}
= a2 + 4a + 4
(ii) (8a + 3b)2 = (8a)2 + 2 x 8a * 3b + (3b)2 = 642 + 48ab + 9 b2
(iii) (2m+ 1)2 = (2m)2 + 2 x 2m x1 + (1)2
= 4m2 + 4m + 1

Question 2.
Find the product of the following binomials :

Solution:

Question 3.
Using the formula for squaring a binomial, evaluate the following :
(i) (102)2
(ii) (99)2
(iii) (1001)2

(iv) (999)2
(v) (703)
2
Solution:
(i) (102)2 = (100 + 2)2
= (100)2 + 2 x 100 x 2 + (2)2
{(a + b)2 = a2 + 2ab + b2}
= 10000 + 400 + 4 = 10404
(ii) (99)2 = (100 – 1)2
= (100)2 – 2 x 100 X 1 +(1)2
{(a – b)2 = a2 – 2ab + b2}
= 10000 -200+1
= 10001 -200 =9801
(iii) (1001 )2 = (1000 + 1)2
{(a + b)2 = a2 + 2ab + b2}
= (1000)2 + 2 x 1000 x 1 + (1)2
= 1000000 + 2000 + 1 = 1002001
(iv) (999)2 = (1000 – 1)2
{(a – b)2 = a2 – 2ab + b2}
= (1000)2 – 2 x 1000 x 1 + (1)2
= 1000000 – 2000 + 1
= 1000001 -2000 = 998001

Question 4.
Simplify the following using the formula:
(a – b) (a + b) = a2 – b2 :
(i) (82)2 (18)2
(ii) (467)2 (33)2
(iii) (79)2 (69)2
(iv) 197 x 203
(v) 113 x 87
(vi) 95 x 105
(vii) 1.8 x 2.2
(viii) 9.8 x 10.2
Solution:
(i) (82)2 – (18)2 = (82 + 18) (82 – 18)
{(a + b)(a- b) = a2 – b2} = 100 x 64 = 6400
(ii) (467)2 – (33)2 = (467 + 33) (467 – 33)
= 500 x 434 = 217000
(ii) (79)2 – (69)2 = (79 + 69) (79 – 69)
148 x 10= 1480
(iv) 197 x 203 = (200 – 3) (200 + 3)
= (200)2 – (3)2
= 40000-9 = 39991
(v) 113 x 87 = (100 + 13) (100- 13)
= (100)2 – (13)2
= 10000- 169 = 9831
(vi) 95 x 105 = (100 – 5) (100 + 5)
= (100)2 – (5)2
= 10000 – 25 = 9975
(vii) 8 x 2.2 = (2.0 – 0.2) (2.0 + 0.2)
= (2.0)2 – (0.2)2
= 4.00 – 0.04 = 3.96
(viii)9.8 x 10.2 = (10.0 – 0.2) (10.0 + 0.2)
(10.0)2 – (0.2)2
= 100.00 – 0.04 = 99.96

Question 5.
Simplify the following using the identities :

Solution:

Question 6.
Find the value of x, if
(i)  4x = (52)2 – (48)2
(ii) 14x = (47)2 – (33)2
(iii)  5x = (50)2 – (40)2
Solution:
(i) 4x = (52)2 – (48)2
4x = (52 + 48) (52 – 48)

Question 7.
If x + $$\frac { 1 }{ x }$$= 20, find the value of x2+ $$\frac { 1 }{ { x }^{ 2 } }$$

Solution:

Question 8.
If x – $$\frac { 1 }{ x }$$ = 3, find the values of x2 + $$\frac { 1 }{ { x }^{ 2 } }$$ and x4 + $$\frac { 1 }{ { x }^{ 4 } }$$

Solution:

Question 9.
If x2 – $$\frac { 1 }{ { x }^{ 2 } }$$= 18, find the values of x+ $$\frac { 1 }{ x }$$  and x– $$\frac { 1 }{ x }$$
Solution:

Question 10.
Ifx+y = 4 and xy = 2, find the value of x2+y2.
Solution:
x + y = 4
Squaring on both sides,
(x + y)2 = (4)2
⇒ x2 +y2 + 2xy = 16
⇒ x2+y2 + 2 x 2 = 16                       (∵ xy = 2)
⇒ x2 + y2 + 4 = 16
⇒ x2+y2 = 16 – 4= 12           ‘
∴ x2+y2 = 12

Question 11.
If x-y = 7 and xy = 9, find the value of x2+y2.
Solution:
x-y = 7
Squaring on both sides,
(x-y)2 = (7)2
⇒ x2+y2-2xy = 49
⇒ x2 + y2 – 2 x 9 = 49                    (∵ xy = 9)
⇒ x2 +y2 – 18 = 49
⇒ x2 + y2 = 49 + 18 = 67
∴ x2+y2 = 67

Question 12.
If 3x + 5y = 11 and xy = 2, find the value of 9x2 + 25y2
Solution:
3 x + 5y = 11, xy = 2
Squaring on both sides,
(3x + 5y)2 = (11)2
⇒ (3x)2 + (5y)2 + 2 x 3x x 5y = 121
⇒ 9x2 + 25y2 + 30 x 7 = 121
⇒ 9x2 + 25y2+ 30 x 2 = 121           (∵ xy = 2)
⇒ 9x2 + 25y2 + 60 = 121
⇒ 9x2 + 25y2 = 121 – 60 = 61
∴ 9x2 + 25y2 = 61

Question 13.
Find the values of the following expressions :
(i)16x2 + 24x + 9, when X’ = $$\frac { 7 }{ 45 }$$
(ii) 64x2 + 81y2 + 144xy when x = 11 and y = $$\frac { 4 }{ 3 }$$
(iii) 81x2 + 16y2-72xy, whenx= $$\frac { 2 }{ 3 }$$ andy= $$\frac { 3 }{ 4 }$$
Solution:

Question 14.
If x + $$\frac { 1 }{ x }$$ = 9, find the values of x+ $$\frac { 1 }{ { x }^{ 4 } }$$.
Solution:

Question 15.
If x + $$\frac { 1 }{ x }$$ = 12, find the values of x–  $$\frac { 1 }{ x }$$.
Solution:

Question 16.
If 2x + 3y = 14 and 2x – 3y = 2, find the value of xy.
Solution:
2x + 3y = 14, 2x – 3y= 2
We know that
(a + b)2 – (a – b)2 = 4ab
∴ (2x + 3y)2 – (2x – 3y)2 = 4 x 2x x 3y = 24xy
⇒ (14)2 – (2)2 = 24xy
⇒ 24xj= 196-4= 192
⇒ xy = $$\frac { 192 }{ 24 }$$ = 8
∴  xy = 8

Question 17.
If x2 + y2 = 29 and xy = 2, find the value of
(i) x+y
(ii) x-y
(iii) x4 +y4
Solution:
x2 + y2 = 29, xy = 2
(i) (x + y)2 = x2 + y2 + 2xy
= 29 + 2×2 = 29+ 4 = 33
∴  x + y= ±√33
(ii) (x – y)2 = x2 + y2 – 2xy
= 29- 2×2 = 29- 4 = 25
∴ x-y= ±√25= ±5
(iii) x2 + y2 = 29
Squaring on both sides,
(x2 + y2)2 = (29)2
⇒ (x2)2 + (y2)2 + 2x2y2 = 841
⇒ x4 +y + 2 (xy)2 = 841
⇒ x4 + y + 2 (2)2 = 841          (∵ xy = 2)
⇒ x4 + y + 2×4 = 841
⇒ x4 + y + 8 = 841
⇒ x4 + y = 841 – 8 = 833
∴ x4 +y = 833

Question 18.
What must be added to each of the following expressions to make it a whole square ?’
(i) 4x2 – 12x + 7
(ii) 4x2 – 20x + 20
Solution:
(i) 4x2 – 12x + 7 = (2x)2 – 2x 2x x 3 + 7
In order to complete the square,
we have to add  32 – 7 = 9 – 7 = 2
∴ (2x)2 – 2 x 2x x 3 + (3)2
= (2x-3)2
∴ Number to be added = 2
(ii) 4x2 – 20x + 20
⇒ (2x)2 – 2 x 2x x 5 + 20
In order to complete the square,
we have to add (5)2 – 20 = 25 – 20 = 5
∴ (2x)2 – 2 x 2x x 5 + (5)2
= (2x – 5)2
∴ Number to be added = 5

Question 19.
Simplify :
(i) (x-y) (x + y) (x2 + y2) (x4 + y4)
(ii) (2x – 1) (2x + 1) (4x2 + 1) (16x4 + 1)
(iii) (7m – 8m)2 + (7m + 8m)2
(iv) (2.5p -5q)2 – (1.5p – 2.5q)2
(v) (m2 – n2m)2 + 2m3n2

Solution:
(i) (x – y) (x + y) (x2 + y2) (x4 +y)
= (x2 – y2) (x2 + y) (x4 + y4)
= [(x2)2 – (y2)2] (x4+y4)
= (x4-y4) (x4+y4)
= (x4)2 – (y4)2 = x8 – y8
(ii) (2x – 1) (2x + 1) (4x2 + 1) (16x4 + 1)
= [(2x)2 – (1)2] (4x2 + 1) (16x4 + 1)
= (4x2 – 1) (4x2 + 1) (16x4 + 1)
= [(4x2)2-(1)2] (16x4+ 1)
= (16x4-1) (16x4+ 1)
= (16x4)2– (1)2 = 256x8 – 1
(iii) (7m – 8m)2 + (7m + 8n)2
= (7m)2 + (8n)2 – 2 x 7m x 8n + (7m)2 + (8n)2 + 2 x 7m x 8n
= 49m2 + 64m2 – 112mn + 49m2 + 64m2 + 112mn
= 98 m2 + 128n2
(iv) (2.5p – 1.5q)2 – (1.5p – 2.5q)2
= (2.5p)2 + (1.5q)2 – 2 x 2.5p x 1.5q
= [(1.5p)2 + (1.5q)2 – 2 x 1.5 p x 2.5q]
= (6.25p2 + 2.25q2 – 7.5 pq) – (2.25p2 + 6.25q2-7.5pq)
= 6.25p2 + 2.25q2 – 7.5pq – 2.25p2 – 6.25q2 + 7.5pq
= 6.25p2 – 2.25p2 + 2.25g2 – 6.25q2
= 4.00P2 – 4.00q2
= 4p2 – 4q2 = 4 (p2 – q2)
(v) (m2 – n2m)2 + 2m3M2
= (m2)2 + (n2m)2 -2 x m2 x n2m + 2;m3m2
= m4 + n4m2 – 2m3n2 + 2m3n2
= m4 + n4m2 = m4 + m2n4

Question 20.
Show that :

Solution:

Hope given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 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 19 Visualising Shapes Ex 19.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.2

Other Exercises

Question 1.
Which among the following are nets for a cube ?

Solution:
Nets for a cube are (ii), (iv) and (vi)

Question 2.
Name the polyhedron that can be made by folding each net:

Solution:
(i) This net is for a square

(ii) This net is for triangular prism.

(iii) This net is for triangular prism.

(iv) This net is for hexagonal prism.

(v) This net is for hexagon pyramid.

(vi) This net is for cuboid.

Question 3.
Dice are cubes where the numbers on the opposite faces must total 7. Which of the following are dice ?

Solution:
Figure (i) shows the net of cube or dice.

Question 4.
Draw nets for each of the following polyhedrons:

Solution:
(i) Net for cube is given below :

(ii) Net of a triangular prism is as under :

(iii) Net of hexagonal prism is as under :

(iv) The net for pentagonal pyramid is as under:

Question 5.
Match the following figures:

Solution:
(a) (iv)
(b) (i)
(c) (ii)
(d) (iii)

Hope given RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.2 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 19 Visualising Shapes Ex 19.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.1

Other Exercises

Question 1.
What is the least number of planes that can enclose a solid ? What is the name of the solid ?
Solution:
The least number of planes that can enclose a solid is called a Tetrahedron.

Question 2.
Can a polyhedron have for its faces :
(i) three triangles ?
(ii) four triangles ?
(iii) a square and four triangles ?
Solution:
(i) No, polyhedron has three faces.
(ii) Yes, tetrahedron has four triangles as its faces.
(iii) Yes, a square pyramid has a square as its base and four triangles as its faces.

Question 3.
Is it possible to have a polyhedron with any given number of faces ?
Solution:
Yes, it is possible if the number of faces is 4 or more.

Question 4.
Is a square prism same as a cube ?
Solution:
Yes, a square prism is a cube.

Question 5.
Can a polyhedron have 10 faces, 20 edges and 15 vertices ?
Solution:
No, it is not possible as By Euler’s formula
F + V = E + 2
⇒ 10 + 15 = 20 + 2
⇒ 25 = 22
Which is not possible

Question 6.
Verify Euler’s formula for each of the following polyhedrons :

Solution:
(i) In this polyhedron,
Number of faces (F) = 7
Number of edges (E) = 15
Number of vertices (V) = 10
According to Euler’s formula,
F + V = E + 2
⇒ 7 + 10 = 15 + 2
⇒ 17 = 17
Which is true.
(ii) In this polyhedron,
Number of faces (F) = 9
Number of edges (E) = 16
Number of vertices (V) = 9
According to Euler’s formula,
F + V = E + 2
⇒ 9 + 9 = 16 + 2
⇒ 18 = 18
Which is true.
(iii) In this polyhedron,
Number of faces (F) = 9
Number of edges (E) =18
Number of vertices (V) = 11
According to Euler’s formula,
F + V = E + 2
⇒ 9 + 11 = 18 + 2
⇒ 20 = 20
Which is true.
(iv) In this polyhedron,
Number of faces (F) = 5
Number of edges (E) = 8
Number of vertices (V) = 5
According to Euler’s formula,
F + V = E + 2
⇒ 5 + 5 = 8 + 2
⇒ 10 = 10
Which is true.
(v) In the given polyhedron,
Number of faces (F) = 9
Number of edges (E) = 16
Number of vertices (V) = 9
According to Euler’s formula,
F + V = E + 2
⇒ 9 + 9 = 16 + 2
⇒ 18 = 18
Which is true.

Question 7.
Using Euler’s formula, find the unknown:

Solution:
We know that Euler’s formula is
F + V = E + 2
(i) F + 6 = 12 + 2
⇒ F + 6 = 14
⇒ F = 14 – 6 = 8
Faces = 8
(ii) F + V = E + 2
⇒ 5 + V = 9 + 2
⇒ 5 + V = 11
⇒ V = 11 – 5 = 6
Vertices = 6
(iii) F + V = E + 2
⇒ 20 + 12 = E + 2
⇒ 32 = E + 2
⇒ E = 32 – 2 = 30
Edges = 30

Hope given RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.1 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 18 Practical Geometry Ex 18.5

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.5

Other Exercises

Question 1.
Construct a quadrilateral ABCD given that AB = 4 cm, BC = 3 cm, ∠A = 75°, ∠B = 80° and ∠C = 120°.
Solution:
Steps of construction :
(i) Draw a line segment AB = 4 cm.

(ii) At A draw a ray AX making an angle of 75°.
(iii) At B draw another ray BY making an angle of 80° and cut off BC = 3 cm.
(iv) At C, draw another ray CZ making an angle of 120° which intersects AX at D.
Then ABCD is the required quadrilateral.

Question 2.
Construct a quadrilateral ABCD where AB = 5.5 cm, BC = 3.7 cm, ∠A = 60°, ∠B = 105° and ∠D = 90°.
Solution:
∠A = 60°, ∠B = 105° and ∠D = 90°
But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral)
⇒ 60° + 105° + ∠C + 90° = 360°
⇒ 255° + ∠C = 360°
⇒ ∠C = 360° – 255° = 105°
Steps of construction :
(i) Draw a line segment AB = 5.5 cm.
(ii) At A, draw a ray AX making an angle of

(iii) At B, draw another ray BY making an angle of 105° and cut off BC = 3.7 cm.
(iv) At C, draw a ray CZ making an angle of 105° which intersects AX at D.
Then ABCD is the required quadrilateral.

Question 3.
Construct a quadrilateral PQRS where PQ = 3.5 cm, QR = 6.5 cm, ∠P = ∠R = 105° and ∠S = 75°.
Solution:
∠P = 105°, ∠R = 105° and ∠S = 75°
But ∠P + ∠Q + ∠R + ∠S = 360° (Sum of angles of a quadrilateral)
⇒ 105° + ∠Q + 105° + 75° = 360°
⇒ 285° + ∠Q = 360°
⇒ ∠Q = 360° – 285° = 75°
Steps of construction :
(i) Draw a line segment PQ = 3.5 cm.

(ii) At P, draw a ray PX making an angle of 105°.
(iii) At Q, draw another ray QY, making an angle of 75° and cut off QR = 6.5 cm.
(iv) At R, draw a ray RZ making an angle of 105° which intersects PX at S.
Then PQRS is the required quadrilateral.

Question 4.
Construct a quadrilateral ABCD when BC = 5.5 cm, CD = 4.1 cm, ∠A = 70°, ∠B = 110° and ∠D = 85°.
Solution:
∠A = 70°, ∠B = 110°, ∠D = 85°
But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral)
⇒ 70° + 110° + ∠C + 85° = 360°
⇒ 265° + ∠C = 360°
⇒ ∠C = 360° – 265° = 95°
Steps of construction:
(i) Draw a line segment BC = 5.5 cm.
(ii) At B, draw a ray BX making an angle of 110°.
(iii) At C, draw another ray CY making an angle of 95° and cut off CD = 4.1 cm.
(iv) At D, draw a ray DZ making an angle of 85° which intersects BX at A.

Then ABCD is the required quadrilateral.

Question 5.
Construct a quadrilateral ABCD, where ∠A = 65°, ∠B = 105°, ∠C = 75°, BC = 5.7 cm and CD = 6.8 cm.
Solution:
∠A = 65°, ∠B = 105°, ∠C = 75°
But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral)
⇒ 65° + 105° + 75° + ∠D = 360°
⇒ 245° + ∠D = 360°
⇒ ∠D = 360° – 245° = 115°
Steps of construction:
(i) Draw a line segment BC = 5.7 cm.
(ii) At B, draw a ray BX making an angle of

(iii) At C draw a another ray CY making an angle of 75° and cut off CD = 6.8 cm.
(iv) At D, draw a ray DZ making an angle of 115° which intersects BX at A.
Then ABCD is the required quadrilateral.

Question 6.
Construct a quadrilateral PQRS in which PQ = 4 cm, QR = 5 cm, ∠P = 50°, ∠Q = 110° and ∠R = 70°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 4 cm.
(ii) At P, draw a ray PX making an angle of 50°.
(iii) At Q, draw another ray QY making an angle of 110° and cut off QR = 5 cm.
(iv) At R, draw a ray RZ making an angle of 70° which intersects PX at S.

Then PQRS is the required quadrilateral.

Hope given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.5 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 18.4

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.4

Other Exercises

Question 1.
Construct a quadrilateral ABCD, in which AB = 6 cm, BC = 4 cm, CD = 4 cm, ZB = 95° and ∠C = 90°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4 cm.

(ii) At B, draw a ray BX making an angle of 95° and cut off BA = 6 cm.
(iii) At C, draw a ray CY making an angle of 90° and cut off CD = 4 cm.
Then ABCD is the required quadrilateral.

Question 2.
Construct a quadrilateral ABCD, where AB = 4.2 cm, BC = 3.6 cm, CD = 4.8 cm, ∠B = 30° and ∠C = 150°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 30° and cut of BA = 4.2 cm.
(iii) At C, draw another ray CY making an angle of 150° and cut off CD = 4.8 cm.

Then ABCD is the required quadrilateral.

Question 3.
Construct a quadrilateral PQRS, in which PQ = 3.5 cm, QR = 2.5 cm, RS = 4.1 cm, ∠Q = 75° and ∠R = 120°.
Solution:
Steps of construction :
(i) Draw a line segment QR = 2.5 cm.

(ii) At Q, draw a ray QX making an angle of 75° and cut off QP = 3.5 cm.
(iii) At R, draw another ray RY making an angle of 120° and cut off RS = 4.1 cm.
(iv) Join PS.
Then PQRS is the required quadrilateral.

Question 4.
Construct a quadrilateral ABCD given BC = 6.6 cm, CD = 4.4 cm, AD = 5.6 cm and ∠D = 100° and ∠C = 95°.
Solution:
Steps of construction :
(i) Draw a line segment CD = 4.4 cm.

(ii) At C, draw a ray CX making an angle of 95° and cut off CB = 6.6 cm
(iii) At D, draw another ray DY making an angle of 100° and cut off DA = 5.6 cm.
(iv) Join AB.
Then ABCD is the required quadrilateral.

Question 5.
Construct a quadrilateral ABCD in which AD = 3.5 cm, AB = 4.4 cm, BC = 4.7 cm, ∠A = 125° and ∠B = 120°.
Solution:
Steps of construction :
(i) Draw a line segment AB 4.4 cm.

(ii) At A, draw a ray AX making an angle of 125° and cut off AD = 3.5 cm.
(iii) At B, draw another ray BY making an angle of 120° and cut off BC = 4.7 cm.
(iv) Join CD.
Then ABCD is the required quadrilateral.

Question 6.
Construct a quadrilateral PQRS in which ∠Q = 45°, ∠R = 90°, QR = 5 cm, PQ = 9 cm and RS = 7 cm.
Solution:
Steps of construction :
This quadrilateral is not possible to construct as shown in the figure.

Question 7.
Construct a quadrilateral ABCD in which AB = BC = 3 cm, AD = 5 cm, ∠A = 90° and ∠B = 105°.
Solution:
Steps of construction :
(i) Draw a line segment AB = 3 cm.

(ii) At A, draw a ray AX making an angle of 90° and cut off AD = 5 cm.
(iii) At B, draw another ray BY making an angle of 105° and cut off BC = 3 cm.
(iv) Join CD.
Then ABCD is the required quadrilateral.

Question 8.
Construct a quadrilateral BDEF where DE = 4.5 cm, EF = 3.5 cm, FB = 6.5 cm and ∠F = 50° and ∠E = 100°
Solution:
Steps of construction :
(i) Draw a line segment EF = 3.5 cm.
(ii) At E, draw a ray EX making an angle of 100° and cut off ED = 4.5 cm.
(iii) At F, draw another ray FY making an angle of 45° and cut off FB = 6.5 cm.
(iv) Join DB.

Then BDEF is the required quadrilateral.

Hope given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.4 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 18.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.3

Other Exercises

Question 1.
Construct a quadrilateral ABCD in which AB = 3.8 cm, BC = 3.4 cm, CD = 4.5 cm, AD = 5 cm and ∠B = 80°.
Solution:
Steps of construction :
(i) Draw a line segment AB = 3.8 cm.
(ii) At B, draw a ray BX making an angle of 80° and cut off BC = 3.4 cm.
(iii) With centre A and radius 5 cm and with centre C and radius 4.5 cm, draw arcs which intersect each other at D.
(iv) Join CD and AD.

ABCD is the required quadrilateral.

Question 2.
Construct a quadrilateral ABCD given that AB = 8 cm, BC = 8 cm, CD = 10 cm, AD = 10 cm and ∠A = 45°.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8 cm.

(ii) At A, draw a ray AX making an angle of 45° and cut off BC = 8 cm.
(iii) With centre A and C and radius 10 cm, draw arcs intersecting each other at D.
(iv) Join AD, CD.
Then ABCD is the required quadrilateral.

Question 3.
Construct a quadrilateral ABCD in which AB = 7.7 cm, BC = 6.8 cm, CD = 5.1 cm, AD = 3.6 cm and ∠C = 120°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6.8 cm.

(ii) At C, draw a ray CX making an angle of 120° and cut off CD = 5.1 cm.
(iii) With centre B and radius 7.7 cm and with centre D and radius 3.6 cm draw arcs which intersect each other at A.
(iv) Join AD and AB.
Then ABCD is the required quadrilateral.

Question 4.
Construct a quadrilateral ABCD in which AB = BC = 3 cm, AD = CD = 5 cm and ∠B = 120°
Solution:
Steps of construction :
(i) Draw a line segment AB = 3 cm.

(ii) At B, draw a ray BX making an angle of 120° and cut off BC = 3 cm.
(iii) With centres A and C, and radius 5 cm, draw arcs intersecting each other at D.
(iv) Join AD and CD.
Then ABCD is the required quadrilateral.

Question 5.
Construct a quadrilateral ABCD in which AB = 2.8 cm, BC = 3.1 cm, CD = 2.6 cm and DA = 3.3 cm and ∠A = 60°.
Solution:
Steps of construction :
(i) Draw a line segment AB = 2.8 cm.

(ii) At A draw a ray AX making an angle of 60° and cut off AD = 3.3 cm.
(iii) With centre B and radius 3.1 cm and with centre D and radius 2.6 cm, draw arc which intersect each other at C.
(iv) Join CB and CD.
Then ABCD is the required quadrilateral.

Question 6.
Construct a quadrilateral ABCD in which AB = BC = 6 cm, AD = DC = 4.5 cm and ∠B = 120°.
Solution:
Steps of construction:
The construction is not possible to draw as arcs of radius 4.5 cm from A and C, do not intersect at any point.

Hope given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.3 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 18 Practical Geometry Ex 18.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.2

Other Exercises

Question 1.
Construct a quadrilateral ABCD in which AB = 3.8 cm, BC = 3.0 cm, AD = 2.3 cm, AC = 4.5 cm and BD = 3.8 cm.
Solution:
Steps of construction :
(i) Draw a line segment AB = 3.8 cm.
(ii) With centre A and radius 2.3 cm and with centre B and radius 3.8 cm draw arcs intersecting each other at D.

(iii) Join AD and BD.
(iv) Again with centre A and radius 4.5 cm and with centre B and radius 3 cm, draw arcs intersecting each other at C.
(v) Join AC and BC and also CD.
Then ABCD is the required quadrilateral.

Question 2.
Construct a quadrilateral ABCD in which BC = 7.5 cm, AC = AD = 6 cm, CD = 5 cm and BD = 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment CD = 5 cm.

(ii) With centre C and D and radius 6 cm, draw line segments intersecting each other at A.
(iii) Join AC and AD.
(iv) Again with centre C and radius 7.5 cm and with centre D and radius 10 cm, draw arcs intersecting each other at B.
(v) Join CB, CA, DA, DB and AB.
Then ABCD is the required quadrilateral.

Question 3.
Construct a quadrilateral ABCD, when AB = 3 cm, CD = 3 cm, DA = 7.5 cm, AC = 8 cm and BD = 4 cm.
Solution:
Steps of construction :
This quadrilateral is not possible as
BD = 4 cm, AB = 3 cm and AD = 7.5 cm
The sum of any two sides of a triangle is greater than the third side.
But BD + AD = 4 + 3 = 7 cm

Question 4.
Construct a quadrilateral ABCD given AD = 3.5 cm, BC = 2.5 cm, CD = 4.1 cm, AC = 7.3 cm and BD = 3.2 cm.
Solution:
Steps of construction :
(i) Draw a line segment CD = 4.1 cm.
(ii) With centre C and radius 7.3 cm and with centre D and radius 3.5 cm, draw arcs intersecting each other at A.

(iii) Join AC and AD.
(iv) Again with centre C and radius 2.5 cm and with centre D and radius 3.2 cm, draw arcs intersecting each other at B.
(v) Join CB’, and DB’ and join AB’.
Then ABCD is the required quadrilateral.

Question 5.
Construct a quadrilateral ABCD given AD = 5 cm, AB = 5.5 cm, BC = 2.5 cm, AC = 7.1 cm and BD = 8 cm.
Solution:
Steps of construction:
(i) Draw a line segment AB = 5 cm.

(ii) With centre A and radius 7.1 cm and with centre B and radius 2.5 cm, draw arcs which intersect each other at C.
(iii) Join AC and BC.
(iv) Again with centre A and radius 5 cm and with centre B and radius 8 cm, draw arcs which intersect each other at D.
(v) Join AD and BD and CD.
Then ABCD is the required quadrilateral.

Question 6.
Construct a quadrilateral ABCD in which BC = 4 cm, CA = 5.6 cm, AD = 4.5 cm, CD = 5 cm and BD = 6.5 cm.
Solution:
Steps of construction:
(i) Draw a line segment CD = 5 cm.

(ii) With centre C and radius 5.6 cm and with centre D and radius 4.5 cm, draw arcs which intersect each other at A.
(iii) Join AC and AD.
(iv) Again with centre C and radius 4 cm and with centre D and radius 6.5 cm, draw arcs which intersect each other at B.
(v) Join BC and BD and AB.
Then ABCD is the required quadrilateral.

Hope given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.2 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 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.

(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.