Maths-
General
Easy

Question

Find the area of triangle whose base is 10 cm and height is 5 cm

Hint:

Use formula directly

The correct answer is: 25 cm2


    It is given that base, b = 10 cm and height, h = 5 cm
    Area of triangle = 1 half cross times b cross times h
    1 half cross times 10 cross times 5 equals 25 cm squared

    Related Questions to study

    General
    Maths-

    Find the area of a piece of cardboard which is in shape of an equilateral triangle of side 12 cm.

    It is given that side of the triangle =12 cm
    Area of equilateral triangle = fraction numerator square root of 3 over denominator 4 end fraction (side)2
    = fraction numerator square root of 3 over denominator 4 end fraction (12)2  = fraction numerator square root of 3 over denominator 4 end fraction (144)
    = 36square root of 3 cm2
    (Substituting the value of square root of 3 = 1.73, in the above equation)
    = 62.35 cm2

    Find the area of a piece of cardboard which is in shape of an equilateral triangle of side 12 cm.

    Maths-General
    It is given that side of the triangle =12 cm
    Area of equilateral triangle = fraction numerator square root of 3 over denominator 4 end fraction (side)2
    = fraction numerator square root of 3 over denominator 4 end fraction (12)2  = fraction numerator square root of 3 over denominator 4 end fraction (144)
    = 36square root of 3 cm2
    (Substituting the value of square root of 3 = 1.73, in the above equation)
    = 62.35 cm2
    General
    Maths-

    The difference between the sides of right angle triangle containing the right angle is 7 cm and its area is 60 cm2 . Calculate the perimeter of triangle.

    It is given that difference between perpendicular and base = 7
    i.e. b – h = 7  ⇒  b = 7 + h
    Now, Area of triangle = 60 cm2
    1 half cross times b cross times h = 60
    left parenthesis 7 plus h right parenthesis cross times h = 120
    h2 + 7h – 120 = 0
    h2 +15h – 8h – 120 = 0
    h(h + 15)-8(h + 15) = 0
    (h + 15)(h – 8) = 0
    h = 8 , - 15
    Since, perpendicular is always positive so h = 8
    Base, B = 7 + 8 = 15
    Using Pythagoras theorem ,
    H2 = B2 + P2
    H2 = 152 + 82 = 225 + 64 = 289
    H = square root of 289 = 17 cm
    Perimeter of triangle = Sum of all sides
    = 17 + 8 + 15 = 40 cm

    The difference between the sides of right angle triangle containing the right angle is 7 cm and its area is 60 cm2 . Calculate the perimeter of triangle.

    Maths-General
    It is given that difference between perpendicular and base = 7
    i.e. b – h = 7  ⇒  b = 7 + h
    Now, Area of triangle = 60 cm2
    1 half cross times b cross times h = 60
    left parenthesis 7 plus h right parenthesis cross times h = 120
    h2 + 7h – 120 = 0
    h2 +15h – 8h – 120 = 0
    h(h + 15)-8(h + 15) = 0
    (h + 15)(h – 8) = 0
    h = 8 , - 15
    Since, perpendicular is always positive so h = 8
    Base, B = 7 + 8 = 15
    Using Pythagoras theorem ,
    H2 = B2 + P2
    H2 = 152 + 82 = 225 + 64 = 289
    H = square root of 289 = 17 cm
    Perimeter of triangle = Sum of all sides
    = 17 + 8 + 15 = 40 cm
    General
    Maths-

    The Sum of parallel sides of a trapezoidal window is 800 inch and its area is 40000 sq inches. What is the height of the window?

    Ans :- 105 m.
    Explanation :-
    Given Sum of lengths of parallel sides = 800 inch
    Area of trapezium = 40,000 sq. inches
    Let height or depth of cross section  be h m
    text  Area of trapezium  end text equals 1 half left parenthesis text  height  end text right parenthesis left parenthesis text  sum of lengths of parallel sides)  end text
    40 comma 000 equals 1 half left parenthesis text  height  end text right parenthesis left parenthesis 800 right parenthesis not stretchy rightwards double arrow text  height  end text equals 80 comma 000 divided by 800 equals 100 text  inch.  end text
    Therefore , the height of the window is 100 inch .

    The Sum of parallel sides of a trapezoidal window is 800 inch and its area is 40000 sq inches. What is the height of the window?

    Maths-General
    Ans :- 105 m.
    Explanation :-
    Given Sum of lengths of parallel sides = 800 inch
    Area of trapezium = 40,000 sq. inches
    Let height or depth of cross section  be h m
    text  Area of trapezium  end text equals 1 half left parenthesis text  height  end text right parenthesis left parenthesis text  sum of lengths of parallel sides)  end text
    40 comma 000 equals 1 half left parenthesis text  height  end text right parenthesis left parenthesis 800 right parenthesis not stretchy rightwards double arrow text  height  end text equals 80 comma 000 divided by 800 equals 100 text  inch.  end text
    Therefore , the height of the window is 100 inch .
    parallel
    General
    Maths-

    George took a nonstop flight from Dallas to Los Angeles, a total flight distance of 1,233 miles. The plane flew at a speed of 460 miles per hour for the first 75 minutes of the flight and at a speed of 439 miles per hour for the remainder of the flight. To the nearest minute, for how many minutes did the plane fly at a speed of 439 miles per hour?

    The total distance between Los Angeles and Dallas is 1233 miles.
    So, total distance covered = 1233 miles.
    Initially,
    Speed of the plane = 460 miles per hour
    Time taken = 75 minutes
    We convert the time in hours as the speed is given in miles per hour.
    As, 60 minutes = 1 hour
    So, we get, 75 minutes =  1 over 60 cross times 75 = 5 over 4  = 1.25 hours
    Thus, time taken = 1.5 hours
    We know,
    text  Speed  end text equals fraction numerator text  Distance  end text over denominator text  Time  end text end fraction
    Rewriting the equation , we have
    text  Distance  end text equals text  Speed × time  end text
    Thus, the distance travelled initially is given by
    text  Distance  end text equals 460 cross times 1.25 equals 575
    Thus, the distance covered by the flight in the first 75 minutes while  the speed was 460 miles per hour = 575 miles
    Next,
    Remaining distance to travel = 1233 – 575 = 658 miles
    For the second part of the journey,
    Speed of the plane = 439 miles per hour
    Distance travelled = 658 miles
    Rewriting the above formula, we get
    text  Time  end text equals fraction numerator text  Distance  end text over denominator text  Speed  end text end fraction
    Using the above quantities, we get
    text  Time  end text equals 658 over 439 almost equal to 1.5 text  hours  end text
    Converting the time into minutes, we have,
    1.5 text  hours  end text equals 1.5 cross times 60 text  minutes  end text equals 90 text  minutes  end text
    Thus, the number of minutes for which the plane flies at speed 439 miles per hour = 90 minutes




    .

    George took a nonstop flight from Dallas to Los Angeles, a total flight distance of 1,233 miles. The plane flew at a speed of 460 miles per hour for the first 75 minutes of the flight and at a speed of 439 miles per hour for the remainder of the flight. To the nearest minute, for how many minutes did the plane fly at a speed of 439 miles per hour?

    Maths-General
    The total distance between Los Angeles and Dallas is 1233 miles.
    So, total distance covered = 1233 miles.
    Initially,
    Speed of the plane = 460 miles per hour
    Time taken = 75 minutes
    We convert the time in hours as the speed is given in miles per hour.
    As, 60 minutes = 1 hour
    So, we get, 75 minutes =  1 over 60 cross times 75 = 5 over 4  = 1.25 hours
    Thus, time taken = 1.5 hours
    We know,
    text  Speed  end text equals fraction numerator text  Distance  end text over denominator text  Time  end text end fraction
    Rewriting the equation , we have
    text  Distance  end text equals text  Speed × time  end text
    Thus, the distance travelled initially is given by
    text  Distance  end text equals 460 cross times 1.25 equals 575
    Thus, the distance covered by the flight in the first 75 minutes while  the speed was 460 miles per hour = 575 miles
    Next,
    Remaining distance to travel = 1233 – 575 = 658 miles
    For the second part of the journey,
    Speed of the plane = 439 miles per hour
    Distance travelled = 658 miles
    Rewriting the above formula, we get
    text  Time  end text equals fraction numerator text  Distance  end text over denominator text  Speed  end text end fraction
    Using the above quantities, we get
    text  Time  end text equals 658 over 439 almost equal to 1.5 text  hours  end text
    Converting the time into minutes, we have,
    1.5 text  hours  end text equals 1.5 cross times 60 text  minutes  end text equals 90 text  minutes  end text
    Thus, the number of minutes for which the plane flies at speed 439 miles per hour = 90 minutes




    .
    General
    Maths-

    A Cylindrical container 21 m in radius and 72 m high is full of water. The water is emptied into a rectangular tank 63 m long and 24 m wide. Find the height of the water level in the tank?

    Hint:
    The volume of water remains same in both containers. So, we equate the volume of water in both the containers to get the height of water in rectangular tank.
    Explanations:
    Step 1 of 2:
    Let  h be the height of water in the tank.
    Volume of water in cylindrical container = pi left parenthesis 21 right parenthesis squared cross times 72 m3
    Volume of water in rectangular tank =63 cross times 24 cross times h m3
    Step 2 of 2:
    Given, volume water in container = volume of water in tank
    not stretchy rightwards double arrow pi left parenthesis 21 right parenthesis squared cross times 72 equals 63 cross times 24 cross times h
    not stretchy rightwards double arrow h equals 22 over 7 cross times 21 cross times 21 cross times 72 cross times 1 over 63 cross times 1 over 24
    not stretchy rightwards double arrow h equals66 m
    Final Answer:
    The height of the water level in the tank is 66 m.

    A Cylindrical container 21 m in radius and 72 m high is full of water. The water is emptied into a rectangular tank 63 m long and 24 m wide. Find the height of the water level in the tank?

    Maths-General
    Hint:
    The volume of water remains same in both containers. So, we equate the volume of water in both the containers to get the height of water in rectangular tank.
    Explanations:
    Step 1 of 2:
    Let  h be the height of water in the tank.
    Volume of water in cylindrical container = pi left parenthesis 21 right parenthesis squared cross times 72 m3
    Volume of water in rectangular tank =63 cross times 24 cross times h m3
    Step 2 of 2:
    Given, volume water in container = volume of water in tank
    not stretchy rightwards double arrow pi left parenthesis 21 right parenthesis squared cross times 72 equals 63 cross times 24 cross times h
    not stretchy rightwards double arrow h equals 22 over 7 cross times 21 cross times 21 cross times 72 cross times 1 over 63 cross times 1 over 24
    not stretchy rightwards double arrow h equals66 m
    Final Answer:
    The height of the water level in the tank is 66 m.
    General
    Maths-

    The base of an isosceles triangle is 24 cm and its area is 192 cm2 . Find its perimeter

    It is given that base, b = 24 cm and
    Area of an isosceles triangle = 192 cm2
    b over 2 square root of a squared minus b squared over 4 end root equals 192 not stretchy rightwards double arrow 24 over 2 square root of a squared minus 24 squared over 4 end root equals 192
    12 square root of a squared minus 144 end root equals 192 not stretchy rightwards double arrow square root of a squared minus 144 end root equals 16
    a squared minus 144 equals 16 squared equals 256 not stretchy rightwards double arrow a squared equals 256 plus 144
    a squared equals 400 not stretchy rightwards double arrow a equals 20
    So, length of two equal sides = 20 cm
    Now, perimeter of the triangle = sum of all sides
    = 20 + 20 + 24 = 64 cm

    The base of an isosceles triangle is 24 cm and its area is 192 cm2 . Find its perimeter

    Maths-General
    It is given that base, b = 24 cm and
    Area of an isosceles triangle = 192 cm2
    b over 2 square root of a squared minus b squared over 4 end root equals 192 not stretchy rightwards double arrow 24 over 2 square root of a squared minus 24 squared over 4 end root equals 192
    12 square root of a squared minus 144 end root equals 192 not stretchy rightwards double arrow square root of a squared minus 144 end root equals 16
    a squared minus 144 equals 16 squared equals 256 not stretchy rightwards double arrow a squared equals 256 plus 144
    a squared equals 400 not stretchy rightwards double arrow a equals 20
    So, length of two equal sides = 20 cm
    Now, perimeter of the triangle = sum of all sides
    = 20 + 20 + 24 = 64 cm
    parallel
    General
    Maths-

    Write an equation of the line with undefined slope and passing through (5,11).

    Hint:-
    1. X-intercept is the point on a line at which the given line intersects the x-axis.
    2. At this point y-coordinate = 0.
    3. A verticle line/ line parallel to the y-axis/ Y-axis has an undefined slope
    Step-by-step solution:-
    The given line has an undefined slope.
    ∴ When plotted on a graph, given line will be represented by a verticle line which is parallel to the y-axis.
    We know that any line that is parallel to the Y-axis can be represented as-
    x = a ....................................................................................... (Equation i)
    where a = x-intercept.
    Now, when a line is parallel to the y-axis, any point on the given line has its x-coordinate constant/ same.
    Since the given line passes through point (5,11) where the x-coordinate = 5
    The x-coordinate at any point on the line will be 5
    i.e. for y= 0 also, x = 5
    and x-intercept is the point at which y = 0
    ∴ x-intercept for the given line = a = 5 ................................ (Equation ii)
    ∴ Equation of the given line = x = 5 ..................................... (From Equation i & Equation ii)
    Final Answer:-
    ∴ x = 5 is the equation of the given line.

    Write an equation of the line with undefined slope and passing through (5,11).

    Maths-General
    Hint:-
    1. X-intercept is the point on a line at which the given line intersects the x-axis.
    2. At this point y-coordinate = 0.
    3. A verticle line/ line parallel to the y-axis/ Y-axis has an undefined slope
    Step-by-step solution:-
    The given line has an undefined slope.
    ∴ When plotted on a graph, given line will be represented by a verticle line which is parallel to the y-axis.
    We know that any line that is parallel to the Y-axis can be represented as-
    x = a ....................................................................................... (Equation i)
    where a = x-intercept.
    Now, when a line is parallel to the y-axis, any point on the given line has its x-coordinate constant/ same.
    Since the given line passes through point (5,11) where the x-coordinate = 5
    The x-coordinate at any point on the line will be 5
    i.e. for y= 0 also, x = 5
    and x-intercept is the point at which y = 0
    ∴ x-intercept for the given line = a = 5 ................................ (Equation ii)
    ∴ Equation of the given line = x = 5 ..................................... (From Equation i & Equation ii)
    Final Answer:-
    ∴ x = 5 is the equation of the given line.
    General
    Maths-

    Write an equation of the line with zero slope and passing through (3,7).

    Hint:-
    Equation of a line in slope point form-
    (y-y1) = m × (x - x1)
    Step-by-step solution:-
    As per given information-
    The given line passes through the point (3,7) and has a slope of 0.
    ∴ x1 = 3, y1 = 7 & m = 0
    Using Slope point form of a line, we can find the equation of the given line as-
    (y-y1) = m × (x-x1)
    ∴ y - 7 = 0 × (x - 3)
    ∴ y - 7 = 0
    ∴ y = 7

    Write an equation of the line with zero slope and passing through (3,7).

    Maths-General
    Hint:-
    Equation of a line in slope point form-
    (y-y1) = m × (x - x1)
    Step-by-step solution:-
    As per given information-
    The given line passes through the point (3,7) and has a slope of 0.
    ∴ x1 = 3, y1 = 7 & m = 0
    Using Slope point form of a line, we can find the equation of the given line as-
    (y-y1) = m × (x-x1)
    ∴ y - 7 = 0 × (x - 3)
    ∴ y - 7 = 0
    ∴ y = 7
    General
    Maths-

    Write an equation of the line with the slope -1 and y-intercept 10

    Hint:-
    1. Equation of a line in slope intercept form-
    y = mx + b
    2. Y-intercept is that point on a line where the line intersects with the y-axis i.e. the point at which x-coordinate is 0.
    Step-by-step solution:-
    As per given information-
    The given line has a slope of -1 and value of y-intercept 10.
    ∴ m = -1 & b = 10
    Using Slope intercept form of a line, we can find the equation of the given line as-
    y = mx + b
    ∴ y = -1 * x + 10
    ∴ y = -x + 10
    Final Answer:-
    ∴ y = -x + 10 is the equation of the given line

    Write an equation of the line with the slope -1 and y-intercept 10

    Maths-General
    Hint:-
    1. Equation of a line in slope intercept form-
    y = mx + b
    2. Y-intercept is that point on a line where the line intersects with the y-axis i.e. the point at which x-coordinate is 0.
    Step-by-step solution:-
    As per given information-
    The given line has a slope of -1 and value of y-intercept 10.
    ∴ m = -1 & b = 10
    Using Slope intercept form of a line, we can find the equation of the given line as-
    y = mx + b
    ∴ y = -1 * x + 10
    ∴ y = -x + 10
    Final Answer:-
    ∴ y = -x + 10 is the equation of the given line
    parallel
    General
    Maths-

    A line passes through point (3, 1) and has slope = -2/3. Find its equation.

    Hint:-
    Equation of a line in slope point form-
    (y-y1) = m × (x - x1)
    Step-by-step solution:-
    As per given information-
    The given line passes through the point (3,1) and has a slope of -2/3.
    ∴ x1 = 3, y1 = 1 & m = - 2/3
    Using Slope point form of a line, we can find the equation of the given line as-
    (y-y1) = m × (x  -x1)
    ∴ y - 1 = - 2/3 × (x - 3)
    ∴ y - 1 = -2/3 x - (- 2/3) × 3
    ∴ y - 1 = - 2/3 x - (- 2)
    ∴ y - 1 = -2/3 x + 2
    ∴ y = - 2/3 x + 2 + 1
    ∴ y = - 2/3 x + 3
    Final Answer:-
    ∴ y = -2/3 x + 3 is the equation of the given line.
    Final Answer:-∴ y = -2/3 x + 3 is the equation of the given line.

    A line passes through point (3, 1) and has slope = -2/3. Find its equation.

    Maths-General
    Hint:-
    Equation of a line in slope point form-
    (y-y1) = m × (x - x1)
    Step-by-step solution:-
    As per given information-
    The given line passes through the point (3,1) and has a slope of -2/3.
    ∴ x1 = 3, y1 = 1 & m = - 2/3
    Using Slope point form of a line, we can find the equation of the given line as-
    (y-y1) = m × (x  -x1)
    ∴ y - 1 = - 2/3 × (x - 3)
    ∴ y - 1 = -2/3 x - (- 2/3) × 3
    ∴ y - 1 = - 2/3 x - (- 2)
    ∴ y - 1 = -2/3 x + 2
    ∴ y = - 2/3 x + 2 + 1
    ∴ y = - 2/3 x + 3
    Final Answer:-
    ∴ y = -2/3 x + 3 is the equation of the given line.
    Final Answer:-∴ y = -2/3 x + 3 is the equation of the given line.
    General
    Maths-

    A metal pipe has internal radius of 2.5 cm, the pipe is 2mm thick all around. Find the weight in kilogram of 2 meters of the pipe if 1 cubic cm of the metal weighs 7.7 gm.
     

    Hint:
    We find the volume of the pipe. Then we multiply it by the given parameter to find the weight and express it in the required unit.
    Explanations:
    Step 1 of :
    Given, inter radius be r = 2.5cm and outer radius be  R = (2.5+0.2)= 2.7cm (since 1cm=10mm)
    Length be h = 2m = 200 cm (since 1m = 100cm)
    Therefore, volume of metal in pipe = pi h open parentheses R squared minus r squared close parentheses
    equals pi h left parenthesis R plus r right parenthesis left parenthesis R minus r right parenthesis
    equals 22 over 7 cross times 200 cross times left parenthesis 2.7 plus 2.5 right parenthesis left parenthesis 2.7 minus 2.5 right parenthesis
    equals 22 over 7 cross times 200 cross times 5.2 cross times 0.2
    equals 653.71 cm3
    Step 2 of 2:
    Given, 1 cm3 of the metal weighs 7.7gm
    therefore653.71 cm3 of the metal in pipe will weigh 653.717.7 = 5033.57gm = 5.03kg
    Final Answer:
    The required weight of pipe is 5.03kg

    A metal pipe has internal radius of 2.5 cm, the pipe is 2mm thick all around. Find the weight in kilogram of 2 meters of the pipe if 1 cubic cm of the metal weighs 7.7 gm.
     

    Maths-General
    Hint:
    We find the volume of the pipe. Then we multiply it by the given parameter to find the weight and express it in the required unit.
    Explanations:
    Step 1 of :
    Given, inter radius be r = 2.5cm and outer radius be  R = (2.5+0.2)= 2.7cm (since 1cm=10mm)
    Length be h = 2m = 200 cm (since 1m = 100cm)
    Therefore, volume of metal in pipe = pi h open parentheses R squared minus r squared close parentheses
    equals pi h left parenthesis R plus r right parenthesis left parenthesis R minus r right parenthesis
    equals 22 over 7 cross times 200 cross times left parenthesis 2.7 plus 2.5 right parenthesis left parenthesis 2.7 minus 2.5 right parenthesis
    equals 22 over 7 cross times 200 cross times 5.2 cross times 0.2
    equals 653.71 cm3
    Step 2 of 2:
    Given, 1 cm3 of the metal weighs 7.7gm
    therefore653.71 cm3 of the metal in pipe will weigh 653.717.7 = 5033.57gm = 5.03kg
    Final Answer:
    The required weight of pipe is 5.03kg
    General
    Maths-

    Top of a metallic plate is in the form of a parallelogram. If one of its sides is 8 inches and its altitude is 3 inches. Find the area of the metallic plate

    Ans :- 150 sq. inches.
    Explanation :-
    Given , Base = 8 inches  ; height (or ) altitude =  3 inches
    Area of parallelogram = base × height = 8 × 3 = 24 sq. inches .
    Therefore, the area of the metallic plate is 24 sq. inches .

    Top of a metallic plate is in the form of a parallelogram. If one of its sides is 8 inches and its altitude is 3 inches. Find the area of the metallic plate

    Maths-General
    Ans :- 150 sq. inches.
    Explanation :-
    Given , Base = 8 inches  ; height (or ) altitude =  3 inches
    Area of parallelogram = base × height = 8 × 3 = 24 sq. inches .
    Therefore, the area of the metallic plate is 24 sq. inches .
    parallel
    General
    Maths-

    Find the equation of the given line.

    Hint:-
    1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis - a - vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1.
    2. Equation of a line in slope point form-
    (y-y1) = m × (x-x1)

    Step-by-step solution:-
    From the given diagram, we can see that-
    rise = increase in the y-coordinates of the given line = 4 units and
    run = increase in the x-coordinates of the given line = 2 units
    ∴ Slope of the given line = m = rise / run
    ∴ Slope of the given line = m = 4 / 2
    ∴ Slope of the given line = m = 2
    We observe from the given diagram that-
    The given line passes through origin (0,0)
    i.e. x1 = 0; y1 = 0 and m = 2
    Hence, we can use the slope point formula of a line to find the equation of the given line-
    (y-y1) = m × (x - x1)
    ∴ y - 0 = 2 × (x - 0)
    ∴ y = 2 × x
    ∴ y = 2x
    Final Answer:-
    ∴ y = 2x is the equation of the given line.

    Find the equation of the given line.

    Maths-General
    Hint:-
    1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis - a - vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1.
    2. Equation of a line in slope point form-
    (y-y1) = m × (x-x1)

    Step-by-step solution:-
    From the given diagram, we can see that-
    rise = increase in the y-coordinates of the given line = 4 units and
    run = increase in the x-coordinates of the given line = 2 units
    ∴ Slope of the given line = m = rise / run
    ∴ Slope of the given line = m = 4 / 2
    ∴ Slope of the given line = m = 2
    We observe from the given diagram that-
    The given line passes through origin (0,0)
    i.e. x1 = 0; y1 = 0 and m = 2
    Hence, we can use the slope point formula of a line to find the equation of the given line-
    (y-y1) = m × (x - x1)
    ∴ y - 0 = 2 × (x - 0)
    ∴ y = 2 × x
    ∴ y = 2x
    Final Answer:-
    ∴ y = 2x is the equation of the given line.
    General
    Maths-

    The top view of a swimming pool is in the form of a parallelogram with a base and height of 60 feet and 45 feet respectively. Find the area of a top view of the swimming pool.

    Ans :- 150 sq. inches.
    Explanation :-
    Given , Base = 60 feet ; height = 45 feet
    Area of parallelogram = base × height = 60 × 45 = 2,700 sq. feet .
    Therefore, the area of the top view of the swimming pool is 2,700 sq. feet.

    The top view of a swimming pool is in the form of a parallelogram with a base and height of 60 feet and 45 feet respectively. Find the area of a top view of the swimming pool.

    Maths-General
    Ans :- 150 sq. inches.
    Explanation :-
    Given , Base = 60 feet ; height = 45 feet
    Area of parallelogram = base × height = 60 × 45 = 2,700 sq. feet .
    Therefore, the area of the top view of the swimming pool is 2,700 sq. feet.
    General
    Maths-

    Calculate the area of equilateral triangle whose height is 25 cm

    It is given that height, h = 25 cm
    Let AB = x
    We know that all angles are equal to 60°.
    Sin A = equals P over H equals fraction numerator A D over denominator A B end fraction
    Sin 60° = 25 over x
    fraction numerator square root of 3 over denominator 2 end fraction equals 25 over x
    straight x equals fraction numerator 50 over denominator square root of 3 end fraction equals 50 over 3 square root of 3 cm
    Since all sides are equal BC = 50 over 3 square root of 3
    We know that Area of triangle = 1 half cross times b cross times straight h
    equals 1 half cross times 50 over 3 square root of 3 cross times 25
    = 208.33square root of 3
    = 360.84 cm2                       ( ∴ square root of 3 = 1.73 )
    NOTE – We can use Pythagoras theorem too. Since perpendicular of an equilateral triangle bisects the side, implies that BD = x over 2 . So, in triangle ABD
    AB2 = AD2 + BD2
    Find x and then use Area of triangle = 1 half cross times b cross times straight h

    Calculate the area of equilateral triangle whose height is 25 cm

    Maths-General
    It is given that height, h = 25 cm
    Let AB = x
    We know that all angles are equal to 60°.
    Sin A = equals P over H equals fraction numerator A D over denominator A B end fraction
    Sin 60° = 25 over x
    fraction numerator square root of 3 over denominator 2 end fraction equals 25 over x
    straight x equals fraction numerator 50 over denominator square root of 3 end fraction equals 50 over 3 square root of 3 cm
    Since all sides are equal BC = 50 over 3 square root of 3
    We know that Area of triangle = 1 half cross times b cross times straight h
    equals 1 half cross times 50 over 3 square root of 3 cross times 25
    = 208.33square root of 3
    = 360.84 cm2                       ( ∴ square root of 3 = 1.73 )
    NOTE – We can use Pythagoras theorem too. Since perpendicular of an equilateral triangle bisects the side, implies that BD = x over 2 . So, in triangle ABD
    AB2 = AD2 + BD2
    Find x and then use Area of triangle = 1 half cross times b cross times straight h
    parallel

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