Maths-
General
Easy

Question

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

Hint:

Volume of hemisphere equals 2 over 3 pi r cubed

The correct answer is: 683.84cm3


    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

    Related Questions to study

    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
    General
    Maths-

    A circle has a diameter of 120 metres. Your average walking speed is 4 kilometres per hour. How many minutes  will it take you to walk around the boundary of the circle 3 times? Round your answer to the nearest integer.

    ANS :- Time taken to walk  3 times around the circle is 11 mins (approx.)
    Explanation :-
    Given , the diameter of the circle is 120 m .The no. of round need to cover = 3
    Given , the speed = 4 km / hr
    circumference of circle pi d equals 120 pi m
    The total distance = no. of rounds × circumference of circle
    The total distance 2 cross times 120 pi equals 240 pi m equals 0.24 pi km
    We get text  time  end text equals fraction numerator text  total distance  end text over denominator text  speed  end text end fraction not stretchy rightwards double arrow text  time  end text equals fraction numerator 0.24 pi km over denominator 4 km divided by hr end fraction
    hrs = 0.1885 hrs
    As we know 1 hour = 60 mins
    As  time = 0.1885 × 60 mins = 11.3097 mins (11 is the nearest integer)

    A circle has a diameter of 120 metres. Your average walking speed is 4 kilometres per hour. How many minutes  will it take you to walk around the boundary of the circle 3 times? Round your answer to the nearest integer.

    Maths-General
    ANS :- Time taken to walk  3 times around the circle is 11 mins (approx.)
    Explanation :-
    Given , the diameter of the circle is 120 m .The no. of round need to cover = 3
    Given , the speed = 4 km / hr
    circumference of circle pi d equals 120 pi m
    The total distance = no. of rounds × circumference of circle
    The total distance 2 cross times 120 pi equals 240 pi m equals 0.24 pi km
    We get text  time  end text equals fraction numerator text  total distance  end text over denominator text  speed  end text end fraction not stretchy rightwards double arrow text  time  end text equals fraction numerator 0.24 pi km over denominator 4 km divided by hr end fraction
    hrs = 0.1885 hrs
    As we know 1 hour = 60 mins
    As  time = 0.1885 × 60 mins = 11.3097 mins (11 is the nearest integer)
    General
    Maths-

    Find the total area in the given figure.

    Step-by-step solution:-
    In the adjacent figure, Let rectangles ABCD & □ EFGC be the rectangle and square, respectively.
    ∴ from the given diagram, length = 50, breadth = 20, side = 30
    We have to find the area of entire figure which includes the rectangle & square mentioned above.
    ∴ Area of the given figure = Area of rectangles ABCD + Area of rectangles EFGC
    ∴ Area of the given figure = (length × breadth) + (side)2
    ∴ Area of the given figure = (50×20) + (30)2
    ∴ Area of the given figure = 1,000 + 900
    ∴ Area of the given figure = 1,900 square units.
    Final Answer:-
    ∴ Total area in the given figure is 190 square units.

    Find the total area in the given figure.

    Maths-General
    Step-by-step solution:-
    In the adjacent figure, Let rectangles ABCD & □ EFGC be the rectangle and square, respectively.
    ∴ from the given diagram, length = 50, breadth = 20, side = 30
    We have to find the area of entire figure which includes the rectangle & square mentioned above.
    ∴ Area of the given figure = Area of rectangles ABCD + Area of rectangles EFGC
    ∴ Area of the given figure = (length × breadth) + (side)2
    ∴ Area of the given figure = (50×20) + (30)2
    ∴ Area of the given figure = 1,000 + 900
    ∴ Area of the given figure = 1,900 square units.
    Final Answer:-
    ∴ Total area in the given figure is 190 square units.
    General
    Maths-

    A rectangular parking lot is 90 yards long and 35 yards wide. It costs about $.45 to pave each square foot of the parking lot with asphalt. About how much will it cost to pave the parking lot?

    ANS :- $12757.5 is the cost of paving the parking lot  .
    Explanation :-
    Given, the length and breadth rectangular parking lot are 90 yards and 35 yards
    We know 1 yard = 3 foot , convert l and b to foot .
    Then we get area of parking lot = 28350 sq. foot
    Given , the cost per sq. foot is $ 0.45
    Total cost of paving parking lot = area of parking lot in sq. foot ×cost of paving parking lot per foot
    Total cost of paving parking lot = 28350 sq. foot × $0.45/ sq. foot = $12757.5
    The cost of paving the rectangular parking lot is  $12757.5

    A rectangular parking lot is 90 yards long and 35 yards wide. It costs about $.45 to pave each square foot of the parking lot with asphalt. About how much will it cost to pave the parking lot?

    Maths-General
    ANS :- $12757.5 is the cost of paving the parking lot  .
    Explanation :-
    Given, the length and breadth rectangular parking lot are 90 yards and 35 yards
    We know 1 yard = 3 foot , convert l and b to foot .
    Then we get area of parking lot = 28350 sq. foot
    Given , the cost per sq. foot is $ 0.45
    Total cost of paving parking lot = area of parking lot in sq. foot ×cost of paving parking lot per foot
    Total cost of paving parking lot = 28350 sq. foot × $0.45/ sq. foot = $12757.5
    The cost of paving the rectangular parking lot is  $12757.5
    parallel
    General
    Maths-

    A person is standing exactly at the centre of a circle. The distance of the person from the boundary of the circle is 3 ft. Find the area of the circle.

    ANS :- 28.274 sq. feet is the area of the circle.
    Explanation :-
    Given, the radius of circle is 3 ft
    Then we get area of circle = pi r squared
    equals pi left parenthesis 3 right parenthesis squared
    equals 9 pi f t squared
    = 28.274 sq. ft
    ∴The area of the circle is 28.274 sq. feet.

    A person is standing exactly at the centre of a circle. The distance of the person from the boundary of the circle is 3 ft. Find the area of the circle.

    Maths-General
    ANS :- 28.274 sq. feet is the area of the circle.
    Explanation :-
    Given, the radius of circle is 3 ft
    Then we get area of circle = pi r squared
    equals pi left parenthesis 3 right parenthesis squared
    equals 9 pi f t squared
    = 28.274 sq. ft
    ∴The area of the circle is 28.274 sq. feet.
    General
    Maths-

    The length and breadth of a rectangular plot are in the ratio 3: 1 and its perimeter is 128 m. Find the area of the plot ?

    Hint:-
    Perimeter of a rectangle = 2 (length + breadth)
    Area of a rectangle = length × breadth
    Step-by-step solution:-
     For the given rectangle-
     Length : Breadth = 3 : 1
     Let x be the common factor.
    ∴  Length = 3x …............................................ (Equation i)
    & Breadth = x …............................................ (Equation ii)
    Perimeter of a rectangle = 2 (length + breadth)
    ∴  Perimeter of a rectangle = 2 (3x + x) ............ (From Equations i & ii)
    ∴  Perimeter of a rectangle = 2 × 4x
    ∴  Perimeter of a rectangle = 8x
    ∴  128 = 8x ..................... (From given information)
    ∴  128 / 8 = x
    ∴  16 = x
    ∴  Length = 3x = 3 × 16 = 48 m ............................ (Equation iii)
    &  Breadth = x = 16 m ....................................... (Equation iv)
    Area of the given rectangle = length × breadth
    ∴  Area of the given rectangle = 48 ×16
    ∴  Area of the given rectangle = 768 m2
    Final Answer:-
    ∴ Area of the give rectangle is 768 m2.

    The length and breadth of a rectangular plot are in the ratio 3: 1 and its perimeter is 128 m. Find the area of the plot ?

    Maths-General
    Hint:-
    Perimeter of a rectangle = 2 (length + breadth)
    Area of a rectangle = length × breadth
    Step-by-step solution:-
     For the given rectangle-
     Length : Breadth = 3 : 1
     Let x be the common factor.
    ∴  Length = 3x …............................................ (Equation i)
    & Breadth = x …............................................ (Equation ii)
    Perimeter of a rectangle = 2 (length + breadth)
    ∴  Perimeter of a rectangle = 2 (3x + x) ............ (From Equations i & ii)
    ∴  Perimeter of a rectangle = 2 × 4x
    ∴  Perimeter of a rectangle = 8x
    ∴  128 = 8x ..................... (From given information)
    ∴  128 / 8 = x
    ∴  16 = x
    ∴  Length = 3x = 3 × 16 = 48 m ............................ (Equation iii)
    &  Breadth = x = 16 m ....................................... (Equation iv)
    Area of the given rectangle = length × breadth
    ∴  Area of the given rectangle = 48 ×16
    ∴  Area of the given rectangle = 768 m2
    Final Answer:-
    ∴ Area of the give rectangle is 768 m2.
    General
    Maths-

    Find the cost of fencing a square park with 16 m side if the cost of fencing is $16/m.

    ANS :- $ 1024 is the cost of fencing the square park .
    Explanation :-
    Given the side length of square park = 16 m
    Then perimeter (or) boundary length to be fenced = 4×16 = 64 m.
    Given the cost of fencing per metre = $16 / m
    Total cost of fencing = total perimeter in m × The cost of fencing per metre
    Total cost of fencing = 64 m × $16 / m = $ 1024
    ∴$ 1024 is the cost of fencing the square park .

    Find the cost of fencing a square park with 16 m side if the cost of fencing is $16/m.

    Maths-General
    ANS :- $ 1024 is the cost of fencing the square park .
    Explanation :-
    Given the side length of square park = 16 m
    Then perimeter (or) boundary length to be fenced = 4×16 = 64 m.
    Given the cost of fencing per metre = $16 / m
    Total cost of fencing = total perimeter in m × The cost of fencing per metre
    Total cost of fencing = 64 m × $16 / m = $ 1024
    ∴$ 1024 is the cost of fencing the square park .
    parallel
    General
    Maths-

    If the area of a triangle is 77 m2 and its height is 15 over 11m, find the length of its base

    ANS :- The base of the triangle is 112.93 m.
    Explanation :-
    Given, The height of the triangle is 15 over 11m (i.e h = 15 over 11m)
    Let the base length of the triangle be b
    Given, area of triangle = 77 sq.m
    Then we get area of triangle = ½ × b × h
    not stretchy rightwards double arrow 77 equals 1 divided by 2 cross times open parentheses b cross times 15 over 11 close parentheses
    77 cross times 11 cross times 2 equals 15 b not stretchy rightwards double arrow b equals 1694 over 15
    not stretchy rightwards double arrow b equals 112.93 straight m
    ∴Therefore, The base of the triangle is 112.93 m.

    If the area of a triangle is 77 m2 and its height is 15 over 11m, find the length of its base

    Maths-General
    ANS :- The base of the triangle is 112.93 m.
    Explanation :-
    Given, The height of the triangle is 15 over 11m (i.e h = 15 over 11m)
    Let the base length of the triangle be b
    Given, area of triangle = 77 sq.m
    Then we get area of triangle = ½ × b × h
    not stretchy rightwards double arrow 77 equals 1 divided by 2 cross times open parentheses b cross times 15 over 11 close parentheses
    77 cross times 11 cross times 2 equals 15 b not stretchy rightwards double arrow b equals 1694 over 15
    not stretchy rightwards double arrow b equals 112.93 straight m
    ∴Therefore, The base of the triangle is 112.93 m.
    General
    Maths-

    What is the area of a uniform path of width 2m along the outside of a square plot of land of length 20 m?

    Hint:-
    Area of a square = side2
    Step-by-step solution:-
     As per given information-
     Side of the inner square plot = 20 m
     Area of the inner plot = Side2
    ∴  Area of the inner plot = 202
    ∴  Area of the inner plot = 400 m2 ............................................ (Equation i)
     Looking at the adjacent figure,
     For the outer square-
     Side = Side of the inner plot + 2 (width of the path)
    ∴  Side = 20 + 2 (2)
    ∴  Side = 20 + 4
    ∴ Side = 24 ............................................................................. (Equation ii)
    Area of the outer square = side2
    ∴  Area of the outer square = 242 ............................................. (From Equation ii)
    ∴  Area of the outer square = 576 m2 ....................................... (Equation iii)
    Area of the path = Area of the outer square - Area of the inner plot
    ∴  Area of the path = 576 - 400
    ∴  Area of the path = 176 m2

    What is the area of a uniform path of width 2m along the outside of a square plot of land of length 20 m?

    Maths-General
    Hint:-
    Area of a square = side2
    Step-by-step solution:-
     As per given information-
     Side of the inner square plot = 20 m
     Area of the inner plot = Side2
    ∴  Area of the inner plot = 202
    ∴  Area of the inner plot = 400 m2 ............................................ (Equation i)
     Looking at the adjacent figure,
     For the outer square-
     Side = Side of the inner plot + 2 (width of the path)
    ∴  Side = 20 + 2 (2)
    ∴  Side = 20 + 4
    ∴ Side = 24 ............................................................................. (Equation ii)
    Area of the outer square = side2
    ∴  Area of the outer square = 242 ............................................. (From Equation ii)
    ∴  Area of the outer square = 576 m2 ....................................... (Equation iii)
    Area of the path = Area of the outer square - Area of the inner plot
    ∴  Area of the path = 576 - 400
    ∴  Area of the path = 176 m2
    General
    Maths-

    A triangle has an area of 7.65 square feet. What is the area of the triangle in square inches?

    ANS :- The area of the triangle in square inches is 1101.6 sq. inches.
    Explanation :-
    We know 1 foot = 12 inches
    Squaring on both sides give, 1 text  sq.foot  end text equals 144 text  sq.inches  end text
    1 text  sq.foot  end text equals 144 text  sq.inches  end textis the conversion ratio
    Now multiply it with  7.65 on both sides
    Then we get 7.65 text  sq.foot  end text equals 7.65 cross times 144 text  sq.inches  end text equals 1101.6 text  sq.inches  end text
    ∴ The area of the triangle in square inches is 1101.6 sq. inches.

    A triangle has an area of 7.65 square feet. What is the area of the triangle in square inches?

    Maths-General
    ANS :- The area of the triangle in square inches is 1101.6 sq. inches.
    Explanation :-
    We know 1 foot = 12 inches
    Squaring on both sides give, 1 text  sq.foot  end text equals 144 text  sq.inches  end text
    1 text  sq.foot  end text equals 144 text  sq.inches  end textis the conversion ratio
    Now multiply it with  7.65 on both sides
    Then we get 7.65 text  sq.foot  end text equals 7.65 cross times 144 text  sq.inches  end text equals 1101.6 text  sq.inches  end text
    ∴ The area of the triangle in square inches is 1101.6 sq. inches.
    parallel
    General
    Maths-

    A rectangular floor of a room is 18 m long 12 m wide.Find the cost of covering the floor with carpet 6 m wide at Rs 40 per metre.

    Hint:-
    Area of a rectangle = length × breadth
    Step-by-step solution:-
    As per the given information-
    Floor is to be fully covered with the carpet.
    ∴ Area of the carpet = Area of the floor …...................................................................... (Equation i)
    Now, For the floor-
    Length = 18m
    Breadth = 12m
    ∴  Area of the floor = length × breadth
    ∴  Area of the floor = 18 × 12
    ∴  Area of the floor = 216 m2 ....................................................................................... (Equation ii)
    From Equations i & ii-
     Area of the carpet = 216 m2
    ∴  length × breadth = 216
    ∴ length × 6 = 216 ..................................................................................................... (From given information)
    ∴  length = 216 / 6
    ∴  length = 36 m
    Cost of carpet @ Rs. 40 per mt = 36 m × Rs. 40 per mt
    ∴ Cost of carpet @ Rs. 40 per mt = Rs. 1,440
    Final Answer:-
    ∴ Cost of covering the floor with carpet is Rs. 1,440.

    A rectangular floor of a room is 18 m long 12 m wide.Find the cost of covering the floor with carpet 6 m wide at Rs 40 per metre.

    Maths-General
    Hint:-
    Area of a rectangle = length × breadth
    Step-by-step solution:-
    As per the given information-
    Floor is to be fully covered with the carpet.
    ∴ Area of the carpet = Area of the floor …...................................................................... (Equation i)
    Now, For the floor-
    Length = 18m
    Breadth = 12m
    ∴  Area of the floor = length × breadth
    ∴  Area of the floor = 18 × 12
    ∴  Area of the floor = 216 m2 ....................................................................................... (Equation ii)
    From Equations i & ii-
     Area of the carpet = 216 m2
    ∴  length × breadth = 216
    ∴ length × 6 = 216 ..................................................................................................... (From given information)
    ∴  length = 216 / 6
    ∴  length = 36 m
    Cost of carpet @ Rs. 40 per mt = 36 m × Rs. 40 per mt
    ∴ Cost of carpet @ Rs. 40 per mt = Rs. 1,440
    Final Answer:-
    ∴ Cost of covering the floor with carpet is Rs. 1,440.
    General
    Maths-

    The diagonal of a square is 4 √2 cm. Find the length of another diagonals if the diagonal of another square whose area is double that of the first square.

    Hint:-
    Area of a square = side2
    All angles in a square are equal to 90°
    Diagonal of a square divides it into 2 right angled triangles.
    Step-by-step solution:-
    In the adjacent diagram, we can see that the diagonal divides the given square into 2 right angled triangles.
    Also, diagonal of the square becomes the hypotenuse of the 2 triangles.
    Hypotenuse of the given triangles = diagonal of the given square = 4 √2 cm
    Let the sides of the given square be x cm.
    ∴ side of the triangles (other than hypotenuse) = x cm.
    By applying Pythagorean theorem, For a right angled triangle-
    Hypotenuse2 = sum of the squares of the remaining 2 sides
    ∴  (4 √2)2 = x2 + x2
    ∴  16 × 2 =  2 x 2
    ∴  32 = 2 x 2
     i.e. 2 x 2 = 32
    ∴  x2 = 32 / 2
    ∴  x2 = 16
    ∴  x = 4 .............................. (Taking square root both the sides) ........................ (Equation i)
    Area of the given square = side2
    ∴ Area of the given square = 42 ...................................................................................... (From Equation i)
    ∴ Area of the given square = 16 cm2 ................................................................................... (Equation ii)
    As per given information-
    Area of another square = 2 × Area of original square
    ∴ Area of another square = 2 × 16 ..................................................................................... (From Equation ii)
    ∴ Area of another square = 32
    ∴ Side2 = 32 ........................................................................................... (Area of a square = side2)
    ∴ Side2 = 32
    ∴ Side = 4 √2 ......................................................................................... (Equation iii)
    Apllying Pythagorean theorem, For a right angled triangle-
    Hypotenuse2 = sum of the squares of the remaining 2 sides
    ∴ Hypotenuse2 = (4 √2)2 + (4 √2)2 ...................................................................................... (From Equation iii)
    ∴ Hypotenuse2 = 2 (4 √2)2
    ∴ Hypotenuse2 = 2 (16 × 2)
    ∴ Hypotenuse2 = 2 (32)
    ∴ Hypotenuse2 = 64
    ∴ Hypotenuse = 8 ......................................................................................................... (Taking square root both the sides)
    Final Answer:-
    ∴ Length of the diagonal of another square is 8 cm.

    The diagonal of a square is 4 √2 cm. Find the length of another diagonals if the diagonal of another square whose area is double that of the first square.

    Maths-General
    Hint:-
    Area of a square = side2
    All angles in a square are equal to 90°
    Diagonal of a square divides it into 2 right angled triangles.
    Step-by-step solution:-
    In the adjacent diagram, we can see that the diagonal divides the given square into 2 right angled triangles.
    Also, diagonal of the square becomes the hypotenuse of the 2 triangles.
    Hypotenuse of the given triangles = diagonal of the given square = 4 √2 cm
    Let the sides of the given square be x cm.
    ∴ side of the triangles (other than hypotenuse) = x cm.
    By applying Pythagorean theorem, For a right angled triangle-
    Hypotenuse2 = sum of the squares of the remaining 2 sides
    ∴  (4 √2)2 = x2 + x2
    ∴  16 × 2 =  2 x 2
    ∴  32 = 2 x 2
     i.e. 2 x 2 = 32
    ∴  x2 = 32 / 2
    ∴  x2 = 16
    ∴  x = 4 .............................. (Taking square root both the sides) ........................ (Equation i)
    Area of the given square = side2
    ∴ Area of the given square = 42 ...................................................................................... (From Equation i)
    ∴ Area of the given square = 16 cm2 ................................................................................... (Equation ii)
    As per given information-
    Area of another square = 2 × Area of original square
    ∴ Area of another square = 2 × 16 ..................................................................................... (From Equation ii)
    ∴ Area of another square = 32
    ∴ Side2 = 32 ........................................................................................... (Area of a square = side2)
    ∴ Side2 = 32
    ∴ Side = 4 √2 ......................................................................................... (Equation iii)
    Apllying Pythagorean theorem, For a right angled triangle-
    Hypotenuse2 = sum of the squares of the remaining 2 sides
    ∴ Hypotenuse2 = (4 √2)2 + (4 √2)2 ...................................................................................... (From Equation iii)
    ∴ Hypotenuse2 = 2 (4 √2)2
    ∴ Hypotenuse2 = 2 (16 × 2)
    ∴ Hypotenuse2 = 2 (32)
    ∴ Hypotenuse2 = 64
    ∴ Hypotenuse = 8 ......................................................................................................... (Taking square root both the sides)
    Final Answer:-
    ∴ Length of the diagonal of another square is 8 cm.
    General
    Maths-

    1 square foot ≅ _____ square yard

    ANS :- Option 2
    Explanation :-
    We know 1 yard  = 3 foot
    Squaring on both sides give , 1 text  yard  end text squared equals 9 text  foot  end text squared
    1 text  yard  end text squared equals 9 text  foot  end text squared  is the conversion ratio
    Now multiply it with  on both sides
    Then we get 1 text  yard  end text squared equals 9 text  foot  end text squared
    ∴ Option 2 is correct .

    1 square foot ≅ _____ square yard

    Maths-General
    ANS :- Option 2
    Explanation :-
    We know 1 yard  = 3 foot
    Squaring on both sides give , 1 text  yard  end text squared equals 9 text  foot  end text squared
    1 text  yard  end text squared equals 9 text  foot  end text squared  is the conversion ratio
    Now multiply it with  on both sides
    Then we get 1 text  yard  end text squared equals 9 text  foot  end text squared
    ∴ Option 2 is correct .
    parallel

    card img

    With Turito Academy.

    card img

    With Turito Foundation.

    card img

    Get an Expert Advice From Turito.

    Turito Academy

    card img

    With Turito Academy.

    Test Prep

    card img

    With Turito Foundation.