Maths-
General
Easy

Question

A Sphere has volume of 1837.35 cubic centimetres, what is the radius of the sphere ?

Hint:

Volume of sphere is equals 4 over 3 pi r cubed

The correct answer is: 7.597cm


    Explanation:
    • We have given a sphere has volume of 1837.35 cubic centimetres.
    • We have to find the radius of the sphere.
    Step 1 of 1:
    The sphere has volume of
    So,

    4 over 3 pi r cubed equals 1837.35 cm cubed

    r cubed equals 438.458

    r = 7.59cm
    Therefore the radius of the sphere is r = 7.597cm

    Related Questions to study

    General
    Maths-

    In the figure given below, find the area of:
    (i) the shaded portion and (ii) the unshaded portion.

    Hint:-
    i. Shaded portion = Outer rectangle - 4 Inner rectangles
    ii. Unshaded portion = Sum of all 4 inner rectangles.
    Step-by-step solution:-
    In the adjacent figure, Let rectangles ABCD be the outer rectangle & rectangles AHQP, rectangles IBED, rectangles GFCJ & rectangles SRKD be the 4 inner rectangles.
    For the outer rectangle-
    length = AB = DC = 25 cm & breadth = AD = BC = 15 cm
    For the inner rectangles-
    We observe from the given diagram that the verticle and horizontal bars of the shaded region is in the exact middle of the outer rectangle rectangles ABCD.
    ∴ Distance between points A & H = Distance between points I & B
    ∴ AH = IB ......................................................................................... (Equation i)
    Now, AB = AH + HI + IB
    ∴ AB = AH + HI + AH ..................................................................... (From Equation i)
    ∴ 25 = 2 AH + 5 ........................................................................... (From given information)
    ∴ 25 - 5 = 2 AH
    ∴ 20 = 2 AH
    ∴ AH = 10 ................................................................................... (Dividing both sides by 2)
    ∴ AH = IB = 10 cm .............................................................................. (Equation ii)
    Also, Distance between points D & K = Distance between points J & C
    ∴ DK = JC ......................................................................................... (Equation iii)
    Now, DC = DK + KJ + JC
    ∴ DC = DK + KJ + DK ..................................................................... (From Equation iii)
    ∴ 25 = 2 DK + 5 ........................................................................... (From given information)
    ∴ 25 - 5 = 2 DK
    ∴ 20 = 2 DK
    ∴ AH = 10 ................................................................................... (Dividing both sides by 2)
    ∴ DK = JC = 10 cm .............................................................................. (Equation iv)
    Also, Distance between points A & P = Distance between points S & D
    ∴ AP = SD ......................................................................................... (Equation v)
    Now, AD = AP + PS + SD
    ∴ AD = AP + PS + AP ..................................................................... (From Equation v)
    ∴ 15 = 2 AP + 5 ........................................................................... (From given information)
    ∴ 15 - 5 = 2 AP
    ∴ 10 = 2 AP
    ∴ AP = 5 ................................................................................... (Dividing both sides by 2)
    ∴ AP = SD = 5 cm .............................................................................. (Equation vi)
    Also, Distance between points B & E = Distance between points F & C
    ∴ BE = FC ......................................................................................... (Equation vii)
    Now, BC = BE + EF + FC
    ∴ BC = BE + EF + BE ..................................................................... (From Equation vii)
    ∴ 15 = 2 BE + 5 ........................................................................... (From given information)
    ∴ 15 - 5 = 2 BE
    ∴ 10 = 2 BE
    ∴ BE = 5 ................................................................................... (Dividing both sides by 2)
    ∴ BE = FC = 5 cm .............................................................................. (Equation viii)
    ow,
    i. Area of outer rectangle = Area of rectangles ABCD
    ∴ Area of outer rectangle = (AB × BC) …......... (Area of a rectangle = length × breadth)
    ∴ Area of outer rectangle = 25 × 15
    ∴ Area of outer rectangle = 375 cm2 ........................................................................... (Equation ix)
    Sum of areas of 4 inner rectangles = Area of  AHQP + Area of rectangles IBED + Area of rectangles GFCJ + Area of rectangles SRKD
    ∴ Sum of areas of 4 inner rectangles = (AH × AP) + (IB × BE) + (JC × FC) + (DK × SD) …......... (Area of a rectangle = length × breadth)
    ∴ Sum of areas of 4 inner rectangles = (10 × 5) + (10 × 5) + (10 × 5) + (10 × 5) ..................... (From given information & Equations ii, iv, vi & viii)
    ∴ Sum of areas of 4 inner rectangles = 50 + 50 + 50 + 50
    ∴ Sum of areas of 4 inner rectangles = 200 cm2 ............................................................. (Equation x)
    Area of Shaded region = Area of outer rectangle - sum of areas of 4 inner rectangles
    ∴ Area of Shaded region = 375 - 200 .................................................... (From Equations ix & x)
    ∴ Area of Shaded region = 175 cm2.
    ii. Area of Unshaded region = sum of Areas of 4 inner rectangles
    ∴ Area of Unshaded region = 200 cm2 .................................................................................... (From Equation x)
    Final Answer:-
    ∴ In the given figure, Areas of the shaded region and unshaded region are 175 cm2 & 200 cm2, respectively.

    In the figure given below, find the area of:
    (i) the shaded portion and (ii) the unshaded portion.

    Maths-General
    Hint:-
    i. Shaded portion = Outer rectangle - 4 Inner rectangles
    ii. Unshaded portion = Sum of all 4 inner rectangles.
    Step-by-step solution:-
    In the adjacent figure, Let rectangles ABCD be the outer rectangle & rectangles AHQP, rectangles IBED, rectangles GFCJ & rectangles SRKD be the 4 inner rectangles.
    For the outer rectangle-
    length = AB = DC = 25 cm & breadth = AD = BC = 15 cm
    For the inner rectangles-
    We observe from the given diagram that the verticle and horizontal bars of the shaded region is in the exact middle of the outer rectangle rectangles ABCD.
    ∴ Distance between points A & H = Distance between points I & B
    ∴ AH = IB ......................................................................................... (Equation i)
    Now, AB = AH + HI + IB
    ∴ AB = AH + HI + AH ..................................................................... (From Equation i)
    ∴ 25 = 2 AH + 5 ........................................................................... (From given information)
    ∴ 25 - 5 = 2 AH
    ∴ 20 = 2 AH
    ∴ AH = 10 ................................................................................... (Dividing both sides by 2)
    ∴ AH = IB = 10 cm .............................................................................. (Equation ii)
    Also, Distance between points D & K = Distance between points J & C
    ∴ DK = JC ......................................................................................... (Equation iii)
    Now, DC = DK + KJ + JC
    ∴ DC = DK + KJ + DK ..................................................................... (From Equation iii)
    ∴ 25 = 2 DK + 5 ........................................................................... (From given information)
    ∴ 25 - 5 = 2 DK
    ∴ 20 = 2 DK
    ∴ AH = 10 ................................................................................... (Dividing both sides by 2)
    ∴ DK = JC = 10 cm .............................................................................. (Equation iv)
    Also, Distance between points A & P = Distance between points S & D
    ∴ AP = SD ......................................................................................... (Equation v)
    Now, AD = AP + PS + SD
    ∴ AD = AP + PS + AP ..................................................................... (From Equation v)
    ∴ 15 = 2 AP + 5 ........................................................................... (From given information)
    ∴ 15 - 5 = 2 AP
    ∴ 10 = 2 AP
    ∴ AP = 5 ................................................................................... (Dividing both sides by 2)
    ∴ AP = SD = 5 cm .............................................................................. (Equation vi)
    Also, Distance between points B & E = Distance between points F & C
    ∴ BE = FC ......................................................................................... (Equation vii)
    Now, BC = BE + EF + FC
    ∴ BC = BE + EF + BE ..................................................................... (From Equation vii)
    ∴ 15 = 2 BE + 5 ........................................................................... (From given information)
    ∴ 15 - 5 = 2 BE
    ∴ 10 = 2 BE
    ∴ BE = 5 ................................................................................... (Dividing both sides by 2)
    ∴ BE = FC = 5 cm .............................................................................. (Equation viii)
    ow,
    i. Area of outer rectangle = Area of rectangles ABCD
    ∴ Area of outer rectangle = (AB × BC) …......... (Area of a rectangle = length × breadth)
    ∴ Area of outer rectangle = 25 × 15
    ∴ Area of outer rectangle = 375 cm2 ........................................................................... (Equation ix)
    Sum of areas of 4 inner rectangles = Area of  AHQP + Area of rectangles IBED + Area of rectangles GFCJ + Area of rectangles SRKD
    ∴ Sum of areas of 4 inner rectangles = (AH × AP) + (IB × BE) + (JC × FC) + (DK × SD) …......... (Area of a rectangle = length × breadth)
    ∴ Sum of areas of 4 inner rectangles = (10 × 5) + (10 × 5) + (10 × 5) + (10 × 5) ..................... (From given information & Equations ii, iv, vi & viii)
    ∴ Sum of areas of 4 inner rectangles = 50 + 50 + 50 + 50
    ∴ Sum of areas of 4 inner rectangles = 200 cm2 ............................................................. (Equation x)
    Area of Shaded region = Area of outer rectangle - sum of areas of 4 inner rectangles
    ∴ Area of Shaded region = 375 - 200 .................................................... (From Equations ix & x)
    ∴ Area of Shaded region = 175 cm2.
    ii. Area of Unshaded region = sum of Areas of 4 inner rectangles
    ∴ Area of Unshaded region = 200 cm2 .................................................................................... (From Equation x)
    Final Answer:-
    ∴ In the given figure, Areas of the shaded region and unshaded region are 175 cm2 & 200 cm2, respectively.
    General
    Maths-

    The length and breadth of a rectangular ground are 46 m and 38 m respectively. Find the cost of levelling the ground at Rs. 18 per square metre.

    Hint:-
    Levelling a ground refers to the process of making the surface horizontal or flat.
    Area of a rectangle = length * breadth
    Step-by-step solution:-
    From the given information-
    length = 46m & breadth = 38m
    Since leveling would be done on the entire surface of the ground, we need to find the area of the given rectangular ground.
    ∴ Area of the given ground = length * breadth
    ∴ Area of the given ground = 46 * 38
    ∴ Area of the given ground = 1,748 m2 .................................................................................. (Equation i)
    ∴ cost of levelling the whole ground (@ Rs. 18 per sq mt) = 1,748 m2 * Rs. 18/ sq mt .............. (From Equation i)
    ∴ cost of levelling the whole ground (@ Rs. 18 per sq mt) = Rs. 31,464
    Final Answer:-
    ∴ Cost of levelling the given rectangular ground at Rs. 18 per sq mt is Rs. 31,464

    The length and breadth of a rectangular ground are 46 m and 38 m respectively. Find the cost of levelling the ground at Rs. 18 per square metre.

    Maths-General
    Hint:-
    Levelling a ground refers to the process of making the surface horizontal or flat.
    Area of a rectangle = length * breadth
    Step-by-step solution:-
    From the given information-
    length = 46m & breadth = 38m
    Since leveling would be done on the entire surface of the ground, we need to find the area of the given rectangular ground.
    ∴ Area of the given ground = length * breadth
    ∴ Area of the given ground = 46 * 38
    ∴ Area of the given ground = 1,748 m2 .................................................................................. (Equation i)
    ∴ cost of levelling the whole ground (@ Rs. 18 per sq mt) = 1,748 m2 * Rs. 18/ sq mt .............. (From Equation i)
    ∴ cost of levelling the whole ground (@ Rs. 18 per sq mt) = Rs. 31,464
    Final Answer:-
    ∴ Cost of levelling the given rectangular ground at Rs. 18 per sq mt is Rs. 31,464
    General
    Maths-

    Metallic spheres of radii 6 cm , 8 cm and 10 cm respectively are melted to form a single solid sphere . Find the radius of the resulting solid sphere ?

    Explanation:
    • We have given Metallic spheres of radii 6 cm , 8 cm and 10 cm respectively are melted to form a single solid sphere
    • We have to find radius of the resulting solid sphere.
    Step 1 of 1:
    We have given metallic sphere with radius 6cm, 8cm, 10cm
    Volume of each sphere will be

    V subscript 1 equals 4 over 3 pi r cubed

    equals 4 over 3 pi cross times 6 cubed
    equals 216 cross times 4 over 3 pi
    And

    V subscript 2 equals 4 over 3 pi r cubed

    equals 4 over 3 pi cross times 8 cubed
    equals 512 cross times 4 over 3 pi
    And

    V subscript 3 equals 4 over 3 pi r cubed

    equals 4 over 3 pi cross times 10 cubed
    equals 1000 cross times 4 over 3 pi
    Now, volume of big sphere will be equal to V subscript 1 plus V subscript 2 plus V subscript 3
    So,

    V equals V subscript 1 plus V subscript 2 plus V subscript 3
    4 over 3 pi r cubed equals 216 cross times 4 over 3 pi plus 512 cross times 4 over 3 pi plus 1000 cross times 4 over 3 pi

    4 over 3 pi r cubed equals 1728 cross times 4 over 3 pi

    r cubed equals 1728

    r = 12
    Therefore, The radius of big sphere is r = 12cm.

    Metallic spheres of radii 6 cm , 8 cm and 10 cm respectively are melted to form a single solid sphere . Find the radius of the resulting solid sphere ?

    Maths-General
    Explanation:
    • We have given Metallic spheres of radii 6 cm , 8 cm and 10 cm respectively are melted to form a single solid sphere
    • We have to find radius of the resulting solid sphere.
    Step 1 of 1:
    We have given metallic sphere with radius 6cm, 8cm, 10cm
    Volume of each sphere will be

    V subscript 1 equals 4 over 3 pi r cubed

    equals 4 over 3 pi cross times 6 cubed
    equals 216 cross times 4 over 3 pi
    And

    V subscript 2 equals 4 over 3 pi r cubed

    equals 4 over 3 pi cross times 8 cubed
    equals 512 cross times 4 over 3 pi
    And

    V subscript 3 equals 4 over 3 pi r cubed

    equals 4 over 3 pi cross times 10 cubed
    equals 1000 cross times 4 over 3 pi
    Now, volume of big sphere will be equal to V subscript 1 plus V subscript 2 plus V subscript 3
    So,

    V equals V subscript 1 plus V subscript 2 plus V subscript 3
    4 over 3 pi r cubed equals 216 cross times 4 over 3 pi plus 512 cross times 4 over 3 pi plus 1000 cross times 4 over 3 pi

    4 over 3 pi r cubed equals 1728 cross times 4 over 3 pi

    r cubed equals 1728

    r = 12
    Therefore, The radius of big sphere is r = 12cm.

    parallel
    General
    Maths-

    Find the area of a rectangular plot, one side of which measures 70 m and the diagonal is 74 m.

    Hint:-
    Area of a rectangle = length * breadth
    All angles in a rectangle are equal to 90°
    Diagonal of a rectangle divides it into 2 right angled triangles.
    Step-by-step solution:-
    In the adjacent diagram, we can see that the diagonal divides the given rectangle into 2 right angled triangles.
    Also, diagonal of the rectangle becomes the hypotenuse of the 2 triangles.
    Hypotenuse of the given triangles = diagonal of the given rectangle = 74m
    Now, 1 side of the rectangle = 1 side of the triangle = 70m
    Let the other side be x m
    By applying Pythagorean theorem, For a right angled triangle-
    Hypotenuse2 = sum of the squares of the remaining 2 sides
    ∴ (74)2 = x2 + (70)2
    ∴ 5,476 = x2 + 4,900
    ∴ 5,476 - 4,900 = x2
    ∴ x2 = 576
    ∴ x = 24 .............................. (Taking square root both the sides) ........................ (Equation i)
    Area of the given plot = length * breadth
    ∴ Area of the given plot = 70 * 24 ...................................................................................... (From given information & Equation i)
    ∴ Area of the given plot = 1,680 m2
    Final Answer:-
    ∴ Area of the given rectangular plot is 1,680 m2.

    Find the area of a rectangular plot, one side of which measures 70 m and the diagonal is 74 m.

    Maths-General
    Hint:-
    Area of a rectangle = length * breadth
    All angles in a rectangle are equal to 90°
    Diagonal of a rectangle divides it into 2 right angled triangles.
    Step-by-step solution:-
    In the adjacent diagram, we can see that the diagonal divides the given rectangle into 2 right angled triangles.
    Also, diagonal of the rectangle becomes the hypotenuse of the 2 triangles.
    Hypotenuse of the given triangles = diagonal of the given rectangle = 74m
    Now, 1 side of the rectangle = 1 side of the triangle = 70m
    Let the other side be x m
    By applying Pythagorean theorem, For a right angled triangle-
    Hypotenuse2 = sum of the squares of the remaining 2 sides
    ∴ (74)2 = x2 + (70)2
    ∴ 5,476 = x2 + 4,900
    ∴ 5,476 - 4,900 = x2
    ∴ x2 = 576
    ∴ x = 24 .............................. (Taking square root both the sides) ........................ (Equation i)
    Area of the given plot = length * breadth
    ∴ Area of the given plot = 70 * 24 ...................................................................................... (From given information & Equation i)
    ∴ Area of the given plot = 1,680 m2
    Final Answer:-
    ∴ Area of the given rectangular plot is 1,680 m2.
    General
    Maths-

    If the length of the rectangular sheet is 25 cm and its area is 500 cm2, find the breadth and hence find its perimeter.

    Hint:-
    i. Area of a rectangle = length * breadth
    ii. Perimeter of a rectangle = 2 (length + breadth)
    Step-by-step solution:-
    From the given information, length = 25 & Area = 500
    Let the breadth be x cm
    i. Area of the given rectanglular sheet = length * breadth
    ∴ 500 = 25 * x
    ∴ 500 / 25 = x
    ∴ x = 20 cm .................................................. (Equation i)
    ii. Perimeter of the rectanglular sheet = 2 (length + breadth)
    ∴ Perimeter of the rectanglular sheet = 2 (25 + 20) ............................................ (From Equation i)
    ∴ Perimeter of the rectanglular sheet = 2 (45)
    ∴Perimeter of the rectanglular sheet = 90 m
    Final Answer:-
    ∴ breadth of the given rectanglular sheet is 20 cm & perimeter is 90 cm.

    If the length of the rectangular sheet is 25 cm and its area is 500 cm2, find the breadth and hence find its perimeter.

    Maths-General
    Hint:-
    i. Area of a rectangle = length * breadth
    ii. Perimeter of a rectangle = 2 (length + breadth)
    Step-by-step solution:-
    From the given information, length = 25 & Area = 500
    Let the breadth be x cm
    i. Area of the given rectanglular sheet = length * breadth
    ∴ 500 = 25 * x
    ∴ 500 / 25 = x
    ∴ x = 20 cm .................................................. (Equation i)
    ii. Perimeter of the rectanglular sheet = 2 (length + breadth)
    ∴ Perimeter of the rectanglular sheet = 2 (25 + 20) ............................................ (From Equation i)
    ∴ Perimeter of the rectanglular sheet = 2 (45)
    ∴Perimeter of the rectanglular sheet = 90 m
    Final Answer:-
    ∴ breadth of the given rectanglular sheet is 20 cm & perimeter is 90 cm.
    General
    Maths-

    The length of a rectangular field is twice its breadth. If the perimeter of the field is 180 m. Find its area. Also find the cost of fencing it at Rs.12 per metre.

    Hint:-
    i. Fencing refers to blocking or covering the boundaries of a land with fences i.e. wooden blocks to prevent livestock from getting away.
    ii. Area of a rectangle = length × breadth
    iii. Perimeter of a rectangle = 2 (length + breadth)
    Step-by-step solution:-
    Let the breadth of the given rectangular field be x m.
    ∴ breadth = x ........................................................................ (Equation i)
    Now, from the given information-
    Length = 2 × breadth
    ∴ Length = 2 × x = 2x …............................................................. (Equation ii)
    Perimeter of the given field = 2 (length + breadth)
    ∴ Perimeter of the given field = 2 (2x + x) ................................. (From Equations i & ii)
    ∴ Perimeter of the given field = 2 × 3x
    ∴ Perimeter of the given field = 6x
    ∴ 180 = 6x .......................................... (From given information)
    ∴ 180 / 6 = x
    ∴ 30 = x
    Substituting x = 30 in Equations i & ii,
    breadth = x = 30 m ................................................................. (Equation iii)
    length = 2x = 2 × 30 = 60 m ..................................................... (Equation iv)
    ∴ Area of the given field = length × breadth
    ∴ Area of the given field = 60 * 30
    ∴ Area of the given field = 1,800 m2
    Since fences are put on the boundaries of the field,
    ∴ cost of fencing the whole field (@ Rs. 12 per mt) = Perimeter of the field × rate
    ∴ cost of fencing the whole field (@ Rs. 12 per mt) = 180 × Rs. 12/ mt ........................ (From given information)
    ∴ cost of fencing the whole field (@ Rs. 12 per mt) = Rs. 2,160
    Final Answer:-
    ∴ Area of the given field is 1,800 m2 & cost of fencing it @ Rs. 12 per mt is Rs. 2,160.

    The length of a rectangular field is twice its breadth. If the perimeter of the field is 180 m. Find its area. Also find the cost of fencing it at Rs.12 per metre.

    Maths-General
    Hint:-
    i. Fencing refers to blocking or covering the boundaries of a land with fences i.e. wooden blocks to prevent livestock from getting away.
    ii. Area of a rectangle = length × breadth
    iii. Perimeter of a rectangle = 2 (length + breadth)
    Step-by-step solution:-
    Let the breadth of the given rectangular field be x m.
    ∴ breadth = x ........................................................................ (Equation i)
    Now, from the given information-
    Length = 2 × breadth
    ∴ Length = 2 × x = 2x …............................................................. (Equation ii)
    Perimeter of the given field = 2 (length + breadth)
    ∴ Perimeter of the given field = 2 (2x + x) ................................. (From Equations i & ii)
    ∴ Perimeter of the given field = 2 × 3x
    ∴ Perimeter of the given field = 6x
    ∴ 180 = 6x .......................................... (From given information)
    ∴ 180 / 6 = x
    ∴ 30 = x
    Substituting x = 30 in Equations i & ii,
    breadth = x = 30 m ................................................................. (Equation iii)
    length = 2x = 2 × 30 = 60 m ..................................................... (Equation iv)
    ∴ Area of the given field = length × breadth
    ∴ Area of the given field = 60 * 30
    ∴ Area of the given field = 1,800 m2
    Since fences are put on the boundaries of the field,
    ∴ cost of fencing the whole field (@ Rs. 12 per mt) = Perimeter of the field × rate
    ∴ cost of fencing the whole field (@ Rs. 12 per mt) = 180 × Rs. 12/ mt ........................ (From given information)
    ∴ cost of fencing the whole field (@ Rs. 12 per mt) = Rs. 2,160
    Final Answer:-
    ∴ Area of the given field is 1,800 m2 & cost of fencing it @ Rs. 12 per mt is Rs. 2,160.
    parallel
    General
    Maths-

    What is the amount of air that can be held by a spherical ball of diameter 14 inches?

    Explanation:
    • We have given spherical ball of diameter 14 inches
    • We have to find the amount of air that can be held by a spherical ball of diameter 14 inches.
    Step 1 of 1:
    The diameter of the spherical ball is 14 inches
    The radius will be fraction numerator 14 text  inches  end text over denominator 2 end fraction equals 7 text  inches  end text
    So, The volume will be

    V equals 4 over 3 pi r cubed

    equals 4 over 3 pi left parenthesis 7 right parenthesis cubed
    equals 1436.755 inch cubed

    What is the amount of air that can be held by a spherical ball of diameter 14 inches?

    Maths-General
    Explanation:
    • We have given spherical ball of diameter 14 inches
    • We have to find the amount of air that can be held by a spherical ball of diameter 14 inches.
    Step 1 of 1:
    The diameter of the spherical ball is 14 inches
    The radius will be fraction numerator 14 text  inches  end text over denominator 2 end fraction equals 7 text  inches  end text
    So, The volume will be

    V equals 4 over 3 pi r cubed

    equals 4 over 3 pi left parenthesis 7 right parenthesis cubed
    equals 1436.755 inch cubed

    General
    Maths-

    A wire when bent in the form of a equilateral triangle encloses an area of 36 √3 sq.cm.Find the area enclosed by the same wire when bent to form a square and a rectangle whose length is 2 cm more than its width.

    Hint:-
    Area of an equilateral triangle = (√3 / 4) × side2
    Area of a square = Side2
    Area of a rectangle = length × breadth
    Step-by-step solution:-
    Area of the triangle = (√3 / 4) × side2
    ∴   36 √3 = (√3 / 4) × side2 …................................................ (From given information)
    ∴   36 = 1 / 4 × side2
    ∴   36 × 4 = side2
    ∴   144 = side2
    ∴   12 = side ................................................................................. (Taking square root both the sides)
    i.e. Side of the triangle = 12 cm
    Length of the wire used to make the triangle = Perimeter of this triangle = 3 × side = 3 × 12 = 36 cm ..................... (Equation i)
    We Know that the same wire is used to make the square and the triangle.
    ∴ Perimeter of the square = perimeter of the triangle
    ∴ Perimeter of the square = 36 ...................................................................................................................... (From Equation i)
    ∴   4 × side = 36 .................................................................................. (Perimeter of a square = 4 × side)
    ∴   side = 36/4 = 9 cm ..................................................................................................... (Equation ii)
    ∴   Area of the square = side2
    ∴   Area of the square = 92 ........................................................................................................................ (From Equation ii)
    ∴   Area of the square = 81 cm2
    For the rectangle,
    Let the width be x cm
    ∴  Width = x cm ............................................................................................................................................ (Equation iii)
    ∴  Length = width + 2 cm = x + 2 ..................................................................................................................... (Equation iv)
    We Know that the same wire is used to make the rectangle and the triangle.
    ∴  Perimeter of the rectangle = perimeter of the triangle
    ∴  Perimeter of the rectangle = 36 ...................................................................................................................... (From Equation i)
    ∴  2 (length + breadth) = 36 .................................................................................. (Perimeter of a rectangle= 2 × length + breadth)
    ∴  2 (x + x + 2) = 36 ..................................................................................................... (From Equations iii & iv)
    ∴  2 (2x + 2 ) = 36
    ∴  4x + 4 = 36
    ∴  4x = 36 - 4
    ∴  4x = 32
    ∴   x = 32/4 = 8 cm ................................................................................... (Equation v)
    Substituting Equation v in Equations iii & iv, we get-
    Width = x = 8
    Length = x + 2 = 8 + 2 = 10
    ∴ Area of the rectangle = length × breadth
    ∴ Area of the rectangle = 8 × 10
    ∴ Area of the rectangle = 80 cm2
    Final Answer:-
    ∴ Areas of the given Square and rectangle are 81 cm2 & 80 cm2, respectively.

    A wire when bent in the form of a equilateral triangle encloses an area of 36 √3 sq.cm.Find the area enclosed by the same wire when bent to form a square and a rectangle whose length is 2 cm more than its width.

    Maths-General
    Hint:-
    Area of an equilateral triangle = (√3 / 4) × side2
    Area of a square = Side2
    Area of a rectangle = length × breadth
    Step-by-step solution:-
    Area of the triangle = (√3 / 4) × side2
    ∴   36 √3 = (√3 / 4) × side2 …................................................ (From given information)
    ∴   36 = 1 / 4 × side2
    ∴   36 × 4 = side2
    ∴   144 = side2
    ∴   12 = side ................................................................................. (Taking square root both the sides)
    i.e. Side of the triangle = 12 cm
    Length of the wire used to make the triangle = Perimeter of this triangle = 3 × side = 3 × 12 = 36 cm ..................... (Equation i)
    We Know that the same wire is used to make the square and the triangle.
    ∴ Perimeter of the square = perimeter of the triangle
    ∴ Perimeter of the square = 36 ...................................................................................................................... (From Equation i)
    ∴   4 × side = 36 .................................................................................. (Perimeter of a square = 4 × side)
    ∴   side = 36/4 = 9 cm ..................................................................................................... (Equation ii)
    ∴   Area of the square = side2
    ∴   Area of the square = 92 ........................................................................................................................ (From Equation ii)
    ∴   Area of the square = 81 cm2
    For the rectangle,
    Let the width be x cm
    ∴  Width = x cm ............................................................................................................................................ (Equation iii)
    ∴  Length = width + 2 cm = x + 2 ..................................................................................................................... (Equation iv)
    We Know that the same wire is used to make the rectangle and the triangle.
    ∴  Perimeter of the rectangle = perimeter of the triangle
    ∴  Perimeter of the rectangle = 36 ...................................................................................................................... (From Equation i)
    ∴  2 (length + breadth) = 36 .................................................................................. (Perimeter of a rectangle= 2 × length + breadth)
    ∴  2 (x + x + 2) = 36 ..................................................................................................... (From Equations iii & iv)
    ∴  2 (2x + 2 ) = 36
    ∴  4x + 4 = 36
    ∴  4x = 36 - 4
    ∴  4x = 32
    ∴   x = 32/4 = 8 cm ................................................................................... (Equation v)
    Substituting Equation v in Equations iii & iv, we get-
    Width = x = 8
    Length = x + 2 = 8 + 2 = 10
    ∴ Area of the rectangle = length × breadth
    ∴ Area of the rectangle = 8 × 10
    ∴ Area of the rectangle = 80 cm2
    Final Answer:-
    ∴ Areas of the given Square and rectangle are 81 cm2 & 80 cm2, respectively.
    General
    Maths-

    Volume of one big sphere is equal to the volume of 8 small spheres. Calculate the ratio of volume of big sphere to volume of small sphere.

    Explanation: 
    • We have given volume of one big sphere is equal to the volume of 8 small spheres
    • We have to find the ratio of volume of big sphere to volume of small sphere
    Step 1 of 1:
    We have given volume of one big sphere is equal to volume of 8small spheres.
    Let volume of big sphere be denoted as Vb and volume of small sphere be denoted as Vs
    So,
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell V subscript b equals 8 V subscript s end cell row cell V subscript b over V subscript s equals 8 over 1 end cell end table
    Therefore the ratio is 8:1

    Volume of one big sphere is equal to the volume of 8 small spheres. Calculate the ratio of volume of big sphere to volume of small sphere.

    Maths-General
    Explanation: 
    • We have given volume of one big sphere is equal to the volume of 8 small spheres
    • We have to find the ratio of volume of big sphere to volume of small sphere
    Step 1 of 1:
    We have given volume of one big sphere is equal to volume of 8small spheres.
    Let volume of big sphere be denoted as Vb and volume of small sphere be denoted as Vs
    So,
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell V subscript b equals 8 V subscript s end cell row cell V subscript b over V subscript s equals 8 over 1 end cell end table
    Therefore the ratio is 8:1
    parallel
    General
    Maths-

    In the figure, ABCD is a rectangle and PQRS is a square. Find the area of the shaded portion.

    step-by-step solution:-
    we can see from the diagram that the Area of shaded portion can be calculated by-
    Step- 1: Finding the area of rectangle ABCD (with length = 20 cm & breadth = 15 cm) and
    Step- 2: Subtracting the area of square PQRS (with side = 5 cm) from it.
    ∴ Area of shaded portion = Area of rectangles ABCD - Area of □ PQRS
    ∴ Area of shaded portion = length × breadth - (side)2
    ∴ Area of shaded portion = (20 × 15) - 52
    ∴ Area of shaded portion = 300 - 25
    ∴ Area of shaded portion = 275 cm2
    Final Answer:-
    ∴ Area of the shaded portion is 275 cm2

    In the figure, ABCD is a rectangle and PQRS is a square. Find the area of the shaded portion.

    Maths-General
    step-by-step solution:-
    we can see from the diagram that the Area of shaded portion can be calculated by-
    Step- 1: Finding the area of rectangle ABCD (with length = 20 cm & breadth = 15 cm) and
    Step- 2: Subtracting the area of square PQRS (with side = 5 cm) from it.
    ∴ Area of shaded portion = Area of rectangles ABCD - Area of □ PQRS
    ∴ Area of shaded portion = length × breadth - (side)2
    ∴ Area of shaded portion = (20 × 15) - 52
    ∴ Area of shaded portion = 300 - 25
    ∴ Area of shaded portion = 275 cm2
    Final Answer:-
    ∴ Area of the shaded portion is 275 cm2
    General
    Maths-

    A room is 10 m long and 6 m wide. How many tiles of size 20 cm by 10 cm are required to cover its floor?

    Hint:-
    Area of a rectangle = length × breadth
    Step-by-step solution:-
    Covering the floor of a room with tiles means putting tiles on the whole surface of the room.
    ∴ If we know the area of the whole room and the area of each tile, we can find the number of tiles required to cover the floor.
    Let the number of tiles required be n
    Now, from the given information,
    length of the room = l1 = 10m
    breadth of the room = b1 = 6m
    length of each tile = l2 = 20 cm = 0.2 m ............................................... (1m = 100 cm. ∴ 20 cm = 20/100 m = 0.2 m)
    breadth of each tile = b2 = 10 cm = 01. m ............................................ (1m = 100 cm. ∴ 10 cm = 10/100 m = 0.1 m)
    ∴ Area of the room = length × breadth
    ∴ Area of the room = 10 × 6
    ∴ Area of the room = 60 m2 ............................................................... (Equation i)
    Area of each tile = length × breadth
    ∴ Area of each tile = 0.2 × 0.1
    ∴ Area of each tile = 0.02 m2 ............................................................... (Equation ii)
    Area of the room = Area of each tile × total number of tiles required.
    ∴ 60 = 0.02 × total number of tiles required ................... (From Equations i & ii)
    ∴ 60 / 0.02 = total number of tiles required
    ∴ 3,000 = total number of tiles required
    Step-by-step solution:-
    Covering the floor of a room with tiles means putting tiles on the whole surface of the room.
    ∴ If we know the area of the whole room and the area of each tile, we can find the number of tiles required to cover the floor.
    Let the number of tiles required be n
    Now, from the given information,
    length of the room = l1 = 10m
    breadth of the room = b1 = 6m
    length of each tile = l2 = 20 cm = 0.2 m ............................................... (1m = 100 cm. ∴ 20 cm = 20/100 m = 0.2 m)
    breadth of each tile = b2 = 10 cm = 01. m ............................................ (1m = 100 cm. ∴ 10 cm = 10/100 m = 0.1 m)
    ∴ Area of the room = length × breadth
    ∴ Area of the room = 10 × 6
    ∴ Area of the room = 60 m2 ............................................................... (Equation i)
    Area of each tile = length × breadth
    ∴ Area of each tile = 0.2 × 0.1
    ∴ Area of each tile = 0.02 m2 ............................................................... (Equation ii)
    Area of the room = Area of each tile × total number of tiles required.
    ∴ 60 = 0.02 × total number of tiles required ................... (From Equations i & ii)
    ∴ 60 / 0.02 = total number of tiles required
    ∴ 3,000 = total number of tiles required
    Final Answer:-
    ∴ 3,000 tiles would be required to cover the floor of the given room.

    A room is 10 m long and 6 m wide. How many tiles of size 20 cm by 10 cm are required to cover its floor?

    Maths-General
    Hint:-
    Area of a rectangle = length × breadth
    Step-by-step solution:-
    Covering the floor of a room with tiles means putting tiles on the whole surface of the room.
    ∴ If we know the area of the whole room and the area of each tile, we can find the number of tiles required to cover the floor.
    Let the number of tiles required be n
    Now, from the given information,
    length of the room = l1 = 10m
    breadth of the room = b1 = 6m
    length of each tile = l2 = 20 cm = 0.2 m ............................................... (1m = 100 cm. ∴ 20 cm = 20/100 m = 0.2 m)
    breadth of each tile = b2 = 10 cm = 01. m ............................................ (1m = 100 cm. ∴ 10 cm = 10/100 m = 0.1 m)
    ∴ Area of the room = length × breadth
    ∴ Area of the room = 10 × 6
    ∴ Area of the room = 60 m2 ............................................................... (Equation i)
    Area of each tile = length × breadth
    ∴ Area of each tile = 0.2 × 0.1
    ∴ Area of each tile = 0.02 m2 ............................................................... (Equation ii)
    Area of the room = Area of each tile × total number of tiles required.
    ∴ 60 = 0.02 × total number of tiles required ................... (From Equations i & ii)
    ∴ 60 / 0.02 = total number of tiles required
    ∴ 3,000 = total number of tiles required
    Step-by-step solution:-
    Covering the floor of a room with tiles means putting tiles on the whole surface of the room.
    ∴ If we know the area of the whole room and the area of each tile, we can find the number of tiles required to cover the floor.
    Let the number of tiles required be n
    Now, from the given information,
    length of the room = l1 = 10m
    breadth of the room = b1 = 6m
    length of each tile = l2 = 20 cm = 0.2 m ............................................... (1m = 100 cm. ∴ 20 cm = 20/100 m = 0.2 m)
    breadth of each tile = b2 = 10 cm = 01. m ............................................ (1m = 100 cm. ∴ 10 cm = 10/100 m = 0.1 m)
    ∴ Area of the room = length × breadth
    ∴ Area of the room = 10 × 6
    ∴ Area of the room = 60 m2 ............................................................... (Equation i)
    Area of each tile = length × breadth
    ∴ Area of each tile = 0.2 × 0.1
    ∴ Area of each tile = 0.02 m2 ............................................................... (Equation ii)
    Area of the room = Area of each tile × total number of tiles required.
    ∴ 60 = 0.02 × total number of tiles required ................... (From Equations i & ii)
    ∴ 60 / 0.02 = total number of tiles required
    ∴ 3,000 = total number of tiles required
    Final Answer:-
    ∴ 3,000 tiles would be required to cover the floor of the given room.
    General
    Maths-

    A Solid is in the form of a right circular cone mounted on a hemisphere. The radius of hemisphere is 3.5 cm and height of the cone is 4 cm . The solid is placed in a cylindrical vessel , full of water , in such a way that the whole solid is submerged in water. If the radius of cylindrical vessel is 5 cm and its height is 10.5 cm . Find the volume of water left in the cylindrical vessel

    Explanation:
    • We have given a Solid is in the form of a right circular cone mounted on a hemisphere. The radius of hemisphere is 3.5 cm and height of the cone is 4 cm . The solid is placed in a cylindrical vessel , full of water , in such a way that the whole solid is submerged in water. If the radius of cylindrical vessel is 5 cm and its height is 10.5 cm
    • We have to find volume of water left in the cylindrical vessel.
    Step 1 of 1:
    We have given radius of the hemisphere is 3.5cm
    Now the solids is in the form of a right circular cone mounted on a hemisphere, then radius of the base  of the cone will be equal to radius of the hemisphere.
    Radius of the base of the cone is
    Height of the cone is
    So,
    Volume of the solid = volume of the cone + volume of hemisphere.
    So,

    V subscript text solid  end text end subscript equals 1 third pi r squared h plus 2 over 3 pi r cubed

    equals 1 third pi r squared left parenthesis h plus 2 r right parenthesis

    equals 1 third pi 3.5 squared left parenthesis 4 plus 7 right parenthesis

    = 141.109
    Now the radius of the base of the cylindrical vessel is 5cm
    Height of the cylindrical vessel is 10.5cm
    So, Volume of the water in the cylindrical vessel will be

    V equals pi r squared h

    equals 22 over 7 cross times 25 cross times 10.5

    = 825cm3
    Now, when the solid is completely submerged in the cylindrical vessel full of water,
    The
    Volume of the water left in the vessel = volume of the water in the vessel – volume of solid

    = (825 - 141.16)cm3

    = 683.84cm3

    A Solid is in the form of a right circular cone mounted on a hemisphere. The radius of hemisphere is 3.5 cm and height of the cone is 4 cm . The solid is placed in a cylindrical vessel , full of water , in such a way that the whole solid is submerged in water. If the radius of cylindrical vessel is 5 cm and its height is 10.5 cm . Find the volume of water left in the cylindrical vessel

    Maths-General
    Explanation:
    • We have given a Solid is in the form of a right circular cone mounted on a hemisphere. The radius of hemisphere is 3.5 cm and height of the cone is 4 cm . The solid is placed in a cylindrical vessel , full of water , in such a way that the whole solid is submerged in water. If the radius of cylindrical vessel is 5 cm and its height is 10.5 cm
    • We have to find volume of water left in the cylindrical vessel.
    Step 1 of 1:
    We have given radius of the hemisphere is 3.5cm
    Now the solids is in the form of a right circular cone mounted on a hemisphere, then radius of the base  of the cone will be equal to radius of the hemisphere.
    Radius of the base of the cone is
    Height of the cone is
    So,
    Volume of the solid = volume of the cone + volume of hemisphere.
    So,

    V subscript text solid  end text end subscript equals 1 third pi r squared h plus 2 over 3 pi r cubed

    equals 1 third pi r squared left parenthesis h plus 2 r right parenthesis

    equals 1 third pi 3.5 squared left parenthesis 4 plus 7 right parenthesis

    = 141.109
    Now the radius of the base of the cylindrical vessel is 5cm
    Height of the cylindrical vessel is 10.5cm
    So, Volume of the water in the cylindrical vessel will be

    V equals pi r squared h

    equals 22 over 7 cross times 25 cross times 10.5

    = 825cm3
    Now, when the solid is completely submerged in the cylindrical vessel full of water,
    The
    Volume of the water left in the vessel = volume of the water in the vessel – volume of solid

    = (825 - 141.16)cm3

    = 683.84cm3

    parallel
    General
    Maths-

    A path of uniform width 4m runs around the outside a rectangular field 24m by 18 m.Find the area of the path ?

    Hint:-
    Area of a rectangle = Length × breadth
    Step-by-step solution:-
    From the adjacent diagram,
    Let the outer rectangle represent the path around the field.
    ∴  Length of the outer rectangle = length of the field + 2 (width of the path)
    ∴  Length of the outer rectangle = 24 + 2 (4)
    ∴  Length of the outer rectangle = 24 + 8
    ∴ Length of the outer rectangle = 32 m ........................................................... (Equation i)
    Also, breadth of the outer rectangle = breadth of the field + 2 (width of the path)
    ∴ breadth of the outer rectangle = 18 + 2 (4)
    ∴  breadth of the outer rectangle = 18 + 8
    ∴  breadth of the outer rectangle = 26 m .................................................. (Equation ii)
    Area of the outer rectangle = length × breadth
    ∴ Area of the outer rectangle = 32 × 26 ......................................................... (From Equations i & ii)
    ∴ Area of the outer rectangle = 832 m2 ............................................................ (Equation iii)
    Area of the field = length × breadth
    ∴  Area of the room = 24 × 18 ......................................................................... (From given information)
    ∴  Area of the room = 432 m2 .......................................................................... (Equation iv)
    Now, Area of the path = Area of outer rectangle - Area of the inner field
    ∴  Area of the path = 832 - 432 ............................................................ (From Equations iii & iv)
    ∴  Area of the path = 400 m2
    Final Answer:-
    ∴ Area of the path is Rs. 400 m2.

    A path of uniform width 4m runs around the outside a rectangular field 24m by 18 m.Find the area of the path ?

    Maths-General
    Hint:-
    Area of a rectangle = Length × breadth
    Step-by-step solution:-
    From the adjacent diagram,
    Let the outer rectangle represent the path around the field.
    ∴  Length of the outer rectangle = length of the field + 2 (width of the path)
    ∴  Length of the outer rectangle = 24 + 2 (4)
    ∴  Length of the outer rectangle = 24 + 8
    ∴ Length of the outer rectangle = 32 m ........................................................... (Equation i)
    Also, breadth of the outer rectangle = breadth of the field + 2 (width of the path)
    ∴ breadth of the outer rectangle = 18 + 2 (4)
    ∴  breadth of the outer rectangle = 18 + 8
    ∴  breadth of the outer rectangle = 26 m .................................................. (Equation ii)
    Area of the outer rectangle = length × breadth
    ∴ Area of the outer rectangle = 32 × 26 ......................................................... (From Equations i & ii)
    ∴ Area of the outer rectangle = 832 m2 ............................................................ (Equation iii)
    Area of the field = length × breadth
    ∴  Area of the room = 24 × 18 ......................................................................... (From given information)
    ∴  Area of the room = 432 m2 .......................................................................... (Equation iv)
    Now, Area of the path = Area of outer rectangle - Area of the inner field
    ∴  Area of the path = 832 - 432 ............................................................ (From Equations iii & iv)
    ∴  Area of the path = 400 m2
    Final Answer:-
    ∴ Area of the path is Rs. 400 m2.
    General
    Maths-

    The perimeter of a rectangle is 28 cm and its length is 8 cm. Find its:
    (i) breadth (ii) area

    step-by-step solution:-

    i. We will use the formula for perimeter of a rectangle and substitute the value of perimeter given in the question.
    Let the breadth be x cm
    Perimeter of the given rectangle = 28
    ∴ 2 (length + breadth) = 28
    ∴ 2 × (8 + x) = 28
    ∴ 2 × 8 + 2 × x = 28 .............. (Opening the bracket and multiplying 2 with the whole term)
    ∴ 16 + 2x = 28
    ∴ 2x = 28 - 16
    ∴ 2x = 12
    ∴ x = 12 over 2
    ∴ x = breadth = 6 cm ....... (Equation i)
    ii. We will use the formula for area of rectangle and substitute the value of breadth from equation i and length from given information.
    Area of the given rectangle = length × breadth
    ∴ Area of the given rectangle = 8 × 6 ............ (From Equation i & given information)
    ∴ Area of the given rectangle = 48 cm2
    Final Answer:-
    ∴ For a rectangle with perimeter 28 cm and length 8 cm, its breadth is 6 cm and area is 48 cm2

    The perimeter of a rectangle is 28 cm and its length is 8 cm. Find its:
    (i) breadth (ii) area

    Maths-General
    step-by-step solution:-

    i. We will use the formula for perimeter of a rectangle and substitute the value of perimeter given in the question.
    Let the breadth be x cm
    Perimeter of the given rectangle = 28
    ∴ 2 (length + breadth) = 28
    ∴ 2 × (8 + x) = 28
    ∴ 2 × 8 + 2 × x = 28 .............. (Opening the bracket and multiplying 2 with the whole term)
    ∴ 16 + 2x = 28
    ∴ 2x = 28 - 16
    ∴ 2x = 12
    ∴ x = 12 over 2
    ∴ x = breadth = 6 cm ....... (Equation i)
    ii. We will use the formula for area of rectangle and substitute the value of breadth from equation i and length from given information.
    Area of the given rectangle = length × breadth
    ∴ Area of the given rectangle = 8 × 6 ............ (From Equation i & given information)
    ∴ Area of the given rectangle = 48 cm2
    Final Answer:-
    ∴ For a rectangle with perimeter 28 cm and length 8 cm, its breadth is 6 cm and area is 48 cm2

    General
    Maths-

    How many square tiles of the side 20 cm will be needed to pave a footpath which is 2 m wide and surrounds a rectangular plot 40 m long and 22 m wide.

    Hint:-
    Area of a rectangle = length × breadth
    Step-by-step solution:-
    Paving the floor of a footpath with tiles means putting tiles on the whole surface of the footpath.
    ∴ If we know the area of the footpath and the area of each tile, we can find the number of tiles required to cover the footpath.
    Let the number of tiles required be n
    Now, from the given information,
    length of the plot = l1 = 40 m
    breadth of the plot = b1 = 22 m
    Side of each tile = 20 cm = 0.2 m (1m = 100 cm)
    ∴ Area of the plot = length × breadth
    ∴ Area of the plot = 40 × 22
    ∴ Area of the plot = 880 m2 ............................................................... (Equation i)
    In the adjacent diagram-
    For the outer rectangle-
    Length = Length of plot + 2 (width of footpath)
    Length = 40 + 2 (2)
    Length = 40 + 4
    Length = 44 m .................................................................................. (Equation ii)
    Breadth = Breadth of plot + 2 (width of footpath)
    Breadth = 22 + 2 (2)
    Breadth = 22 + 4
    Breadth = 26 m .................................................................................. (Equation ii)
    Area of Outer rectangle = length × breadth
    ∴  Area of Outer rectangle = 44 × 26 ..................................................... (From Equations ii & iii)
    ∴  Area of Outer rectangle = 1,144 m2 .................................................. (Equation iv)
    Area of footpath = Area of outer rectangle - Area of inner plot
    ∴  Area of footpath = 1,144 - 880 ......................................................... (From Equations i & iv)
    ∴  Area of footpath = 264 m2 ............................................................... (Equation v)
    Area of each tile = Side2
    ∴  Area of each tile = (0.2)2
    ∴  Area of each tile = 0.04 m2 ............................................................... (Equation vi)
    Area of the footpath = Area of each tile × total number of tiles required.
    ∴  264 = 0.04 × total number of tiles required ................... (From Equations v & vi)
    ∴  264 / 0.04 = total number of tiles required
    ∴  6,600 = total number of tiles required
    Final Answer:-
    ∴ 6,600 tiles would be required to cover the footpath.

    How many square tiles of the side 20 cm will be needed to pave a footpath which is 2 m wide and surrounds a rectangular plot 40 m long and 22 m wide.

    Maths-General
    Hint:-
    Area of a rectangle = length × breadth
    Step-by-step solution:-
    Paving the floor of a footpath with tiles means putting tiles on the whole surface of the footpath.
    ∴ If we know the area of the footpath and the area of each tile, we can find the number of tiles required to cover the footpath.
    Let the number of tiles required be n
    Now, from the given information,
    length of the plot = l1 = 40 m
    breadth of the plot = b1 = 22 m
    Side of each tile = 20 cm = 0.2 m (1m = 100 cm)
    ∴ Area of the plot = length × breadth
    ∴ Area of the plot = 40 × 22
    ∴ Area of the plot = 880 m2 ............................................................... (Equation i)
    In the adjacent diagram-
    For the outer rectangle-
    Length = Length of plot + 2 (width of footpath)
    Length = 40 + 2 (2)
    Length = 40 + 4
    Length = 44 m .................................................................................. (Equation ii)
    Breadth = Breadth of plot + 2 (width of footpath)
    Breadth = 22 + 2 (2)
    Breadth = 22 + 4
    Breadth = 26 m .................................................................................. (Equation ii)
    Area of Outer rectangle = length × breadth
    ∴  Area of Outer rectangle = 44 × 26 ..................................................... (From Equations ii & iii)
    ∴  Area of Outer rectangle = 1,144 m2 .................................................. (Equation iv)
    Area of footpath = Area of outer rectangle - Area of inner plot
    ∴  Area of footpath = 1,144 - 880 ......................................................... (From Equations i & iv)
    ∴  Area of footpath = 264 m2 ............................................................... (Equation v)
    Area of each tile = Side2
    ∴  Area of each tile = (0.2)2
    ∴  Area of each tile = 0.04 m2 ............................................................... (Equation vi)
    Area of the footpath = Area of each tile × total number of tiles required.
    ∴  264 = 0.04 × total number of tiles required ................... (From Equations v & vi)
    ∴  264 / 0.04 = total number of tiles required
    ∴  6,600 = total number of tiles required
    Final Answer:-
    ∴ 6,600 tiles would be required to cover the footpath.
    parallel

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