Maths-
General
Easy

Question

If the base of a triangle becomes three times its height. What is the new area of a triangle?

Hint:

Use formula Area of the triangle = 1 half cross times b cross times h

The correct answer is: 3/2 h2


    Let Base = b and Height = h
    It is given that base = 3 cross times height
    ⇒ b = 3h
    Area of the triangle = 1 half cross times b cross times h
    1 half cross times 3 straight h cross times h
    =3 over 2h2

    Related Questions to study

    General
    Maths-

    The side of an equilateral triangle is 16 units. What will be the double of its area ?

    It is given that side of the triangle = 16 units
    Area of an equilateral triangle = fraction numerator square root of 3 over denominator 4 end fraction a2 = fraction numerator square root of 3 over denominator 4 end fraction 162
    = 64square root of 3 cm2
    Double of the area = 2 cross times 64square root of 3
    = 128square root of 3 = 221.7 cm2

    The side of an equilateral triangle is 16 units. What will be the double of its area ?

    Maths-General
    It is given that side of the triangle = 16 units
    Area of an equilateral triangle = fraction numerator square root of 3 over denominator 4 end fraction a2 = fraction numerator square root of 3 over denominator 4 end fraction 162
    = 64square root of 3 cm2
    Double of the area = 2 cross times 64square root of 3
    = 128square root of 3 = 221.7 cm2
    General
    Maths-

    In triangle ABC, AB = 8cm. If the altitudes corresponding to AB and BC are 4 cm and 5 cm respectively. Find the measure of BC.

    Base, AB = 8 cm and Height, H = CD = 4 cm
    Area of the triangle with altitude corresponding to AB
    1 half cross times b cross times h
    1 half cross times 8 cross times 4 = 16 cm2
    With Base = BC , Height, h = AE = 5 cm
    Area of the triangle with altitude corresponding to BC is
    1 half cross times B C cross times A E = 16 cm2
    1 half cross times B C cross times 5 = 16
    BC = 6.4 cm

    In triangle ABC, AB = 8cm. If the altitudes corresponding to AB and BC are 4 cm and 5 cm respectively. Find the measure of BC.

    Maths-General
    Base, AB = 8 cm and Height, H = CD = 4 cm
    Area of the triangle with altitude corresponding to AB
    1 half cross times b cross times h
    1 half cross times 8 cross times 4 = 16 cm2
    With Base = BC , Height, h = AE = 5 cm
    Area of the triangle with altitude corresponding to BC is
    1 half cross times B C cross times A E = 16 cm2
    1 half cross times B C cross times 5 = 16
    BC = 6.4 cm
    General
    Maths-

    The base and corresponding altitude of a parallelogram are 18 cm and 6 cm respectively. Find its area

    It is given that Base, b = 18 cm and Height, h = 6 cm
    Area of the parallelogram = b cross times h
    = 18 cross times 6 = 108 cm2

    The base and corresponding altitude of a parallelogram are 18 cm and 6 cm respectively. Find its area

    Maths-General
    It is given that Base, b = 18 cm and Height, h = 6 cm
    Area of the parallelogram = b cross times h
    = 18 cross times 6 = 108 cm2
    parallel
    General
    Maths-

    The ratio of the bases of two triangles is a : b. If the ratio of their corresponding altitudes is c : d, find the ratio of their areas (in the same order).

    It is given that ratio of bases of two triangle is a : b
    ⇒ Bases of the triangles = ax , bx
    Similarly, it is given that ratio of altitudes is c : d
    ⇒ Altitudes of the triangle = cy , dy
    Area of first triangle = 1 half cross times b cross times h
    1 half cross times a x cross times c y equals fraction numerator a c x y over denominator 2 end fraction
    Area of second triangle = 1 half cross times b x cross times d y equals fraction numerator b d x y over denominator 2 end fraction
    Ratio of the areas of two triangle
    fraction numerator text  area of first triangle  end text over denominator text  area of second triangle  end text end fraction equals fraction numerator 2 a c x y over denominator 2 b d x y end fraction
    fraction numerator a c over denominator b d end fraction
    Hence, ratio of area of two triangles is ac : bd

    The ratio of the bases of two triangles is a : b. If the ratio of their corresponding altitudes is c : d, find the ratio of their areas (in the same order).

    Maths-General
    It is given that ratio of bases of two triangle is a : b
    ⇒ Bases of the triangles = ax , bx
    Similarly, it is given that ratio of altitudes is c : d
    ⇒ Altitudes of the triangle = cy , dy
    Area of first triangle = 1 half cross times b cross times h
    1 half cross times a x cross times c y equals fraction numerator a c x y over denominator 2 end fraction
    Area of second triangle = 1 half cross times b x cross times d y equals fraction numerator b d x y over denominator 2 end fraction
    Ratio of the areas of two triangle
    fraction numerator text  area of first triangle  end text over denominator text  area of second triangle  end text end fraction equals fraction numerator 2 a c x y over denominator 2 b d x y end fraction
    fraction numerator a c over denominator b d end fraction
    Hence, ratio of area of two triangles is ac : bd
    General
    Maths-

    The sides of triangle are 11 cm, 15 cm and 16 cm. What is the measure of altitude to the largest side?

     It is given that sides are in ratio 3 : 4 : 5
    ⇒ Length of the side a = 3x , b = 4x and c = 5 x
    Perimeter of the triangle = 144 m

    3x + 4x + 5x = 144

    12x = 144 ⇒  x = 12
    Now Using Pythagoras theorem,

    (5x)2 = (3x)2 + (4x)2

    25x2 = 9x2 + 16x2

    25x2 = 25x2  i.e. Pythagoras holds true
    ⇒ Given triangle is a right angled triangle
    Base, b = 3x =3(12) = 36 m and Height, h = 4x =4(12) = 48 m

    ⇒ Area of the triangle = 1 half cross times b cross times h

    1 half cross times 36 cross times 48 equals 864 straight m squared

    The sides of triangle are 11 cm, 15 cm and 16 cm. What is the measure of altitude to the largest side?

    Maths-General
     It is given that sides are in ratio 3 : 4 : 5
    ⇒ Length of the side a = 3x , b = 4x and c = 5 x
    Perimeter of the triangle = 144 m

    3x + 4x + 5x = 144

    12x = 144 ⇒  x = 12
    Now Using Pythagoras theorem,

    (5x)2 = (3x)2 + (4x)2

    25x2 = 9x2 + 16x2

    25x2 = 25x2  i.e. Pythagoras holds true
    ⇒ Given triangle is a right angled triangle
    Base, b = 3x =3(12) = 36 m and Height, h = 4x =4(12) = 48 m

    ⇒ Area of the triangle = 1 half cross times b cross times h

    1 half cross times 36 cross times 48 equals 864 straight m squared

    General
    Maths-

    The sides of triangle are 11 cm, 15 cm and 16 cm. What is the measure of altitude to the largest side?

    It is given that a = 11 cm , b = 15 cm and c = 16 cm.
    Using Heron’s formula
    Area of triangle = square root of s left parenthesis s minus a right parenthesis left parenthesis s minus b right parenthesis left parenthesis s minus c right parenthesis end root where s = fraction numerator a plus b plus c over denominator 2 end fraction
    s = fraction numerator 15 plus 16 plus 11 over denominator 2 end fraction = 21

                                                                  Area of triangle= square root of 21 left parenthesis 21 minus 11 right parenthesis left parenthesis 21 minus 15 right parenthesis left parenthesis 21 minus 16 right parenthesis end root
    square root of 21 left parenthesis 10 right parenthesis left parenthesis 6 right parenthesis left parenthesis 5 right parenthesis end root equals 30 square root of 7 cm squared
    Since we have to find altitude to the largest side, base of the triangle = 16 cm

    Also, area of triangle = 1 half cross times b cross times h

    ⇒ 1 half cross times 16 cross times h equals 30 square root of 7

    ⇒ straight h equals 15 over 4 square root of 7 equals 9.92 cm         (square root of 7 = 2.64)

    The sides of triangle are 11 cm, 15 cm and 16 cm. What is the measure of altitude to the largest side?

    Maths-General
    It is given that a = 11 cm , b = 15 cm and c = 16 cm.
    Using Heron’s formula
    Area of triangle = square root of s left parenthesis s minus a right parenthesis left parenthesis s minus b right parenthesis left parenthesis s minus c right parenthesis end root where s = fraction numerator a plus b plus c over denominator 2 end fraction
    s = fraction numerator 15 plus 16 plus 11 over denominator 2 end fraction = 21

                                                                  Area of triangle= square root of 21 left parenthesis 21 minus 11 right parenthesis left parenthesis 21 minus 15 right parenthesis left parenthesis 21 minus 16 right parenthesis end root
    square root of 21 left parenthesis 10 right parenthesis left parenthesis 6 right parenthesis left parenthesis 5 right parenthesis end root equals 30 square root of 7 cm squared
    Since we have to find altitude to the largest side, base of the triangle = 16 cm

    Also, area of triangle = 1 half cross times b cross times h

    ⇒ 1 half cross times 16 cross times h equals 30 square root of 7

    ⇒ straight h equals 15 over 4 square root of 7 equals 9.92 cm         (square root of 7 = 2.64)

    parallel
    General
    Maths-

    When a solid metal cube is completely submerged in a cylindrical vessel Containing milk with 30 cm diameter the level of milk rises by 1 41 over 99 cm. Find the length of the edge of the metal cube.

    Hint:
    We use principle of Archimedes to find the length of the cube.
    Explanations:
    Step 1 of 1:
    Let the length of the edge of the metal cube be a.
    Volume of cube = a3
    Given,  r = 30/2 cm and h = 1 41 over 99 cm
    By the principle of Archimedes,
    Volume of risen milk = volume of cube
    not stretchy rightwards double arrow pi open parentheses 30 over 2 close parentheses squared cross times 1 41 over 99 equals a cubed
    not stretchy rightwards double arrow a cubed equals 1000
    not stretchy rightwards double arrow a equals 10, discarding the negative value since distance cannot be negative.
    Final Answer:
    The length of the edge of the cube is 10cm.

    When a solid metal cube is completely submerged in a cylindrical vessel Containing milk with 30 cm diameter the level of milk rises by 1 41 over 99 cm. Find the length of the edge of the metal cube.

    Maths-General
    Hint:
    We use principle of Archimedes to find the length of the cube.
    Explanations:
    Step 1 of 1:
    Let the length of the edge of the metal cube be a.
    Volume of cube = a3
    Given,  r = 30/2 cm and h = 1 41 over 99 cm
    By the principle of Archimedes,
    Volume of risen milk = volume of cube
    not stretchy rightwards double arrow pi open parentheses 30 over 2 close parentheses squared cross times 1 41 over 99 equals a cubed
    not stretchy rightwards double arrow a cubed equals 1000
    not stretchy rightwards double arrow a equals 10, discarding the negative value since distance cannot be negative.
    Final Answer:
    The length of the edge of the cube is 10cm.
    General
    Maths-

    The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. Find the radius and volume of the cylinder.
     

    Hint:
    We plug in the values in formulae and solve the problem.
    Explanations:
    Step 1 of 2:
    Let the radius of vessel base be
    We have, 2 pi r equals 132 not stretchy rightwards double arrow r equals 132 cross times 7 over 22 cross times 1 half equals 21 cm
    Height  h = 25cm
    Step 2 of 2:
    Volume of the vessel = pi r squared h equals 22 over 7 cross times 21 squared cross times 25 equals 34650 cm cubed
    Final Answer:
    The radius is 21cm and volume of the cylinder is 34650cm3.

    The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. Find the radius and volume of the cylinder.
     

    Maths-General
    Hint:
    We plug in the values in formulae and solve the problem.
    Explanations:
    Step 1 of 2:
    Let the radius of vessel base be
    We have, 2 pi r equals 132 not stretchy rightwards double arrow r equals 132 cross times 7 over 22 cross times 1 half equals 21 cm
    Height  h = 25cm
    Step 2 of 2:
    Volume of the vessel = pi r squared h equals 22 over 7 cross times 21 squared cross times 25 equals 34650 cm cubed
    Final Answer:
    The radius is 21cm and volume of the cylinder is 34650cm3.
    General
    Maths-

    The volume of a metallic cylindrical pipe is 748 cubic.cm Its length is 14 cm., and its external radius is 9 cm. Find its thickness?

    Hint:
    We find the internal radius and subtract it from external radius to get the thickness.
    Explanations:
    Step 1 of 1:
    Let the internal radius of the pipe be r .
    Given, external radius R  = 9cm
    Length = height of the pipe h = 14cm
    We have volume of pipe = 748
    not stretchy rightwards double arrow pi h open parentheses R squared minus r squared close parentheses equals 748
    not stretchy rightwards double arrow pi cross times 14 left parenthesis 9 plus r right parenthesis left parenthesis 9 minus r right parenthesis equals 748
    not stretchy rightwards double arrow 81 minus r squared equals 748 cross times 7 over 22 cross times 1 over 14
    not stretchy rightwards double arrow r squared equals 81 minus 17 equals 64
    not stretchy rightwards double arrow r equals 8 cm
    Final Answer:
    The thickness of the pipe is 8 cm.

    The volume of a metallic cylindrical pipe is 748 cubic.cm Its length is 14 cm., and its external radius is 9 cm. Find its thickness?

    Maths-General
    Hint:
    We find the internal radius and subtract it from external radius to get the thickness.
    Explanations:
    Step 1 of 1:
    Let the internal radius of the pipe be r .
    Given, external radius R  = 9cm
    Length = height of the pipe h = 14cm
    We have volume of pipe = 748
    not stretchy rightwards double arrow pi h open parentheses R squared minus r squared close parentheses equals 748
    not stretchy rightwards double arrow pi cross times 14 left parenthesis 9 plus r right parenthesis left parenthesis 9 minus r right parenthesis equals 748
    not stretchy rightwards double arrow 81 minus r squared equals 748 cross times 7 over 22 cross times 1 over 14
    not stretchy rightwards double arrow r squared equals 81 minus 17 equals 64
    not stretchy rightwards double arrow r equals 8 cm
    Final Answer:
    The thickness of the pipe is 8 cm.
    parallel
    General
    Maths-

    The difference between the outer and inner curved surface areas of a 14 cm long cylinder is 88 sq.cm Find the outer and inner radii of the cylinder given that the volume of the metal is 176 cubic cm

    Hint:
    Forming the equations based on the given information, we will get two equations. Then we will find the radii by solving those two equations.
     Explanations:
    Step 1 of 3:
    Given, outer CSA – inner CSA = 88
    not stretchy rightwards double arrow 2 pi h left parenthesis R minus r right parenthesis equals 88, where  = length of cylinder,  R = outer radius, r = inner radius
    not stretchy rightwards double arrow R minus r equals 88 cross times 1 half cross times 7 over 22 cross times 1 over 14
    not stretchy rightwards double arrow R minus r equals 1…(i)
    Step 2 of 3:
    Also, given volume with  R – volume with  r = 176
    not stretchy rightwards double arrow pi h open parentheses R squared minus r squared close parentheses equals 176
    not stretchy rightwards double arrow pi cross times 14 cross times left parenthesis R plus r right parenthesis left parenthesis R minus r right parenthesis equals 176
    not stretchy rightwards double arrow R plus r equals 176 cross times 7 over 22 cross times 1 over 14 comma text  since  end text R minus r equals 1 comma text  from equation (i)  end text
    not stretchy rightwards double arrow R plus r equals 4…(ii)
    Step 3 of 3:
    Adding (i) and (ii), we get
    R plus r plus R minus r equals 4 plus 1
    not stretchy rightwards double arrow 2 R equals 5
    not stretchy rightwards double arrow R equals 5 over 2 equals 2.5 cm
    Putting R = 5/2 in equation (ii), we get
    5 over 2 plus r equals 4
    not stretchy rightwards double arrow r equals 4 minus 5 over 2 equals 3 over 2 equals 1.5 cm
    Final Answer:
    The outer and inner radii of the cylinder are 2.5 cm and 1.5 cm

    The difference between the outer and inner curved surface areas of a 14 cm long cylinder is 88 sq.cm Find the outer and inner radii of the cylinder given that the volume of the metal is 176 cubic cm

    Maths-General
    Hint:
    Forming the equations based on the given information, we will get two equations. Then we will find the radii by solving those two equations.
     Explanations:
    Step 1 of 3:
    Given, outer CSA – inner CSA = 88
    not stretchy rightwards double arrow 2 pi h left parenthesis R minus r right parenthesis equals 88, where  = length of cylinder,  R = outer radius, r = inner radius
    not stretchy rightwards double arrow R minus r equals 88 cross times 1 half cross times 7 over 22 cross times 1 over 14
    not stretchy rightwards double arrow R minus r equals 1…(i)
    Step 2 of 3:
    Also, given volume with  R – volume with  r = 176
    not stretchy rightwards double arrow pi h open parentheses R squared minus r squared close parentheses equals 176
    not stretchy rightwards double arrow pi cross times 14 cross times left parenthesis R plus r right parenthesis left parenthesis R minus r right parenthesis equals 176
    not stretchy rightwards double arrow R plus r equals 176 cross times 7 over 22 cross times 1 over 14 comma text  since  end text R minus r equals 1 comma text  from equation (i)  end text
    not stretchy rightwards double arrow R plus r equals 4…(ii)
    Step 3 of 3:
    Adding (i) and (ii), we get
    R plus r plus R minus r equals 4 plus 1
    not stretchy rightwards double arrow 2 R equals 5
    not stretchy rightwards double arrow R equals 5 over 2 equals 2.5 cm
    Putting R = 5/2 in equation (ii), we get
    5 over 2 plus r equals 4
    not stretchy rightwards double arrow r equals 4 minus 5 over 2 equals 3 over 2 equals 1.5 cm
    Final Answer:
    The outer and inner radii of the cylinder are 2.5 cm and 1.5 cm
    General
    Maths-

    The sides of a triangle are 16 cm, 30 cm and 34 cm. What is its area ?

    It is given that a = 16 cm , b = 30 cm and c = 34 cm
    Using Heron’s formula
    Area of triangle = square root of s left parenthesis s minus a right parenthesis left parenthesis s minus b right parenthesis left parenthesis s minus c right parenthesis end root where s = fraction numerator a plus b plus c over denominator 2 end fraction
    s = fraction numerator 30 plus 16 plus 34 over denominator 2 end fraction = 40
    Area of triangle
    square root of 40 left parenthesis 40 minus 30 right parenthesis left parenthesis 40 minus 34 right parenthesis left parenthesis 40 minus 16 right parenthesis end root
    square root of 40 left parenthesis 10 right parenthesis left parenthesis 6 right parenthesis left parenthesis 24 right parenthesis end root = 240 cm2

    The sides of a triangle are 16 cm, 30 cm and 34 cm. What is its area ?

    Maths-General
    It is given that a = 16 cm , b = 30 cm and c = 34 cm
    Using Heron’s formula
    Area of triangle = square root of s left parenthesis s minus a right parenthesis left parenthesis s minus b right parenthesis left parenthesis s minus c right parenthesis end root where s = fraction numerator a plus b plus c over denominator 2 end fraction
    s = fraction numerator 30 plus 16 plus 34 over denominator 2 end fraction = 40
    Area of triangle
    square root of 40 left parenthesis 40 minus 30 right parenthesis left parenthesis 40 minus 34 right parenthesis left parenthesis 40 minus 16 right parenthesis end root
    square root of 40 left parenthesis 10 right parenthesis left parenthesis 6 right parenthesis left parenthesis 24 right parenthesis end root = 240 cm2
    General
    Maths-

    The base of an isosceles right angled triangle is 30 cm. Find its area.

    It is given that Base of the triangle = 30 cm
    We know that in a right isosceles triangle, base and height are equal and hypotenuse is the third side.
    So, Area of the triangle = 1 half cross times b cross times h equals 1 1 half cross times 30 cross times 30
    = 450 cm2

    The base of an isosceles right angled triangle is 30 cm. Find its area.

    Maths-General
    It is given that Base of the triangle = 30 cm
    We know that in a right isosceles triangle, base and height are equal and hypotenuse is the third side.
    So, Area of the triangle = 1 half cross times b cross times h equals 1 1 half cross times 30 cross times 30
    = 450 cm2
    parallel
    General
    Maths-

    Driver A charges $25 fees and $0.10 per extra mile driven. Driver B charges no fees but charges $0.60 per mile driven. Find the Application distance in miles for which the two drivers will cost the same money. Also, find the cost.

    Hint:-
    1. Extra charges = number of miles driven × rate per mile
    2. Total cost = Constant fees + Extra charges.
    Step-by-step solution:-
    Let x be the number of miles driven and y be the total fees charged.
    We know that-
    Extra charges = Number of miles driven × Rate per mile
    ∴ Extra charges = x * rate per mile ..................................................... (Equation i)
    Now, for Driver A-
    Total fees charged = y = Constant fees + Extra charges
    ∴ Total fees charged = y = Constant fees + x × rate per mile ................... (From Equation i)
    ∴ Total fees charged = y = 25 + x * 0.10 .................................................... (From given information)
    ∴ Total fees charged = y = 25 + 0.10 x ....................................................... (Equation ii)
    For Driver B-
    Total fees charged = y = Constant fees + Extra charges
    ∴ Total fees charged = y = Constant fees + x × rate per mile ................... (From Equation i)
    ∴ Total fees charged = y = 0 + x × 0.60 .................................................... (From given information)
    ∴ Total fees charged = y = 0.60 x ....................................................... (Equation iii)
    Since we need to find the distance for which cost of both drivers is the same-
    Equating Equations ii & iii, we get-
    25 + 0.10 x = 0.60 x
    ∴ 25 = 0.60 x - 0.10 x
    ∴ 25 = 0.50 x
    ∴ 25 / 0.50 = x ...................................................................................... (Dividing both sides by 0.50)
    ∴ 50 = x
    For a distance of 50 miles, the cost charged by both the drivers will be the sam.
    Now, at 50 miles i.e. x = 50,
    y = 0.60 x ….............................................. (Using Equation iii)
    ∴ y = 0.60 × 50
    ∴ y = 30
    Note:-
    Alternatively, we can find the total cost by using Equation ii,
    y = 25 + 0.10 x
    ∴ y = 25 + 0.10 × 50
    ∴ y = 25 + 5
    ∴ y = 30.
    Hence, the answer remains the same.

    Driver A charges $25 fees and $0.10 per extra mile driven. Driver B charges no fees but charges $0.60 per mile driven. Find the Application distance in miles for which the two drivers will cost the same money. Also, find the cost.

    Maths-General
    Hint:-
    1. Extra charges = number of miles driven × rate per mile
    2. Total cost = Constant fees + Extra charges.
    Step-by-step solution:-
    Let x be the number of miles driven and y be the total fees charged.
    We know that-
    Extra charges = Number of miles driven × Rate per mile
    ∴ Extra charges = x * rate per mile ..................................................... (Equation i)
    Now, for Driver A-
    Total fees charged = y = Constant fees + Extra charges
    ∴ Total fees charged = y = Constant fees + x × rate per mile ................... (From Equation i)
    ∴ Total fees charged = y = 25 + x * 0.10 .................................................... (From given information)
    ∴ Total fees charged = y = 25 + 0.10 x ....................................................... (Equation ii)
    For Driver B-
    Total fees charged = y = Constant fees + Extra charges
    ∴ Total fees charged = y = Constant fees + x × rate per mile ................... (From Equation i)
    ∴ Total fees charged = y = 0 + x × 0.60 .................................................... (From given information)
    ∴ Total fees charged = y = 0.60 x ....................................................... (Equation iii)
    Since we need to find the distance for which cost of both drivers is the same-
    Equating Equations ii & iii, we get-
    25 + 0.10 x = 0.60 x
    ∴ 25 = 0.60 x - 0.10 x
    ∴ 25 = 0.50 x
    ∴ 25 / 0.50 = x ...................................................................................... (Dividing both sides by 0.50)
    ∴ 50 = x
    For a distance of 50 miles, the cost charged by both the drivers will be the sam.
    Now, at 50 miles i.e. x = 50,
    y = 0.60 x ….............................................. (Using Equation iii)
    ∴ y = 0.60 × 50
    ∴ y = 30
    Note:-
    Alternatively, we can find the total cost by using Equation ii,
    y = 25 + 0.10 x
    ∴ y = 25 + 0.10 × 50
    ∴ y = 25 + 5
    ∴ y = 30.
    Hence, the answer remains the same.
    General
    Maths-

    The largest triangle is inscribed in a semi-circle of radius 4 cm. Find the area inside the semi-circle which is not occupied by the triangle.

    It is given that radius of the triangle = 4 cm
    Base of the triangle = 2r = 8 cm
    Height of the triangle = r = 4 cm
    Area of the triangle = equals 1 half cross times b cross times h equals 1 half cross times 8 cross times 4
    = 16 cm2
    Area of semi-circle = 1 half pi r squared equals 1 half =   (3.14) 42
    = 25.12 cm2
    Area not occupied by the triangle
    = Area of semi-circle – Area of triangle
    = 25.12 – 16 = 9.12 cm2

    The largest triangle is inscribed in a semi-circle of radius 4 cm. Find the area inside the semi-circle which is not occupied by the triangle.

    Maths-General
    It is given that radius of the triangle = 4 cm
    Base of the triangle = 2r = 8 cm
    Height of the triangle = r = 4 cm
    Area of the triangle = equals 1 half cross times b cross times h equals 1 half cross times 8 cross times 4
    = 16 cm2
    Area of semi-circle = 1 half pi r squared equals 1 half =   (3.14) 42
    = 25.12 cm2
    Area not occupied by the triangle
    = Area of semi-circle – Area of triangle
    = 25.12 – 16 = 9.12 cm2
    General
    Maths-

    A rectangular paper of 22 cm by 16 cm size can be exactly wrapped to cover the curved surface of a cylinder of 16 cm height. Find the volume of the cylinder?

    Hint:
    Observe that the paper is wrapped by its length. So, the circumference of the cylinder is equal to the length of the paper. We use this information to get the volume.
    Explanations:
    Step 1 of 1:
    Let the base radius of the cylinder be r .
    We have 2 pi r equals 22
    not stretchy rightwards double arrow r equals 22 cross times 7 over 22 cross times 1 half
    not stretchy rightwards double arrow r equals 3.5 cm
    Given, h = 16cm
    Volume of the cylinder
    equals pi r squared h
    equals 22 over 7 cross times 35 over 10 cross times 35 over 10 cross times 16
    equals 616 cm3
    Final Answer:
    The volume of the cylinder is 616cm3.

    A rectangular paper of 22 cm by 16 cm size can be exactly wrapped to cover the curved surface of a cylinder of 16 cm height. Find the volume of the cylinder?

    Maths-General
    Hint:
    Observe that the paper is wrapped by its length. So, the circumference of the cylinder is equal to the length of the paper. We use this information to get the volume.
    Explanations:
    Step 1 of 1:
    Let the base radius of the cylinder be r .
    We have 2 pi r equals 22
    not stretchy rightwards double arrow r equals 22 cross times 7 over 22 cross times 1 half
    not stretchy rightwards double arrow r equals 3.5 cm
    Given, h = 16cm
    Volume of the cylinder
    equals pi r squared h
    equals 22 over 7 cross times 35 over 10 cross times 35 over 10 cross times 16
    equals 616 cm3
    Final Answer:
    The volume of the cylinder is 616cm3.
    parallel

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