Maths-
General
Easy

Question

The base of a triangle is four times its height. What is the area of the triangle?

Hint:

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

The correct answer is: 2h2


    Let Base = b and Height = h
    It is given that base = 4 cross times height
    ⇒ b = 4h
    Area of the triangle = 1 half cross times b cross times h
    1 half cross times 4 straight h cross times h
    =  2h2

    Related Questions to study

    General
    Maths-

    If the perimeter of an equilateral triangle is 342 units. Then find its area.

    Let side of the triangle = a units
    It is given that Perimeter of the triangle = 342 units
    cross times a = 342
    a = 114 units
    Area of  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 1142  = 3249square root of 3 sq. units
    = 5627.43 sq. units

    If the perimeter of an equilateral triangle is 342 units. Then find its area.

    Maths-General
    Let side of the triangle = a units
    It is given that Perimeter of the triangle = 342 units
    cross times a = 342
    a = 114 units
    Area of  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 1142  = 3249square root of 3 sq. units
    = 5627.43 sq. units
    General
    Maths-

    A star is made up of 4 equilateral triangles and a square. If the sides of the triangles are 8 units, what is the surface area of the star ?

    It is given that side of equilateral triangle = 8 units
    Area of 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 82  = 16 square root of 3square units
    Since there are 4 triangles, so area of four equilateral triangles
    = 4 cross times 16 square root of 3= 64square root of 3 sq. units
    = 110.85 sq. units
    In the figure, length of the side of square = 8 units
    Now, Area of square = (side)2 = 82 = 64 sq. units
    Surface Area of the star
    = Area of 4 equilateral triangles + Area of the square
    =  110.85 + 64 = 174.85 sq. units

    A star is made up of 4 equilateral triangles and a square. If the sides of the triangles are 8 units, what is the surface area of the star ?

    Maths-General
    It is given that side of equilateral triangle = 8 units
    Area of 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 82  = 16 square root of 3square units
    Since there are 4 triangles, so area of four equilateral triangles
    = 4 cross times 16 square root of 3= 64square root of 3 sq. units
    = 110.85 sq. units
    In the figure, length of the side of square = 8 units
    Now, Area of square = (side)2 = 82 = 64 sq. units
    Surface Area of the star
    = Area of 4 equilateral triangles + Area of the square
    =  110.85 + 64 = 174.85 sq. units
    General
    Maths-

    If the side AC of a given triangle is 18 units and the height of the triangle is 10 units, what is the area of triangle ABC ?

    It is given that Base, b = AC = 18 units and Height, h = 10 units
    Area of the triangle = 1 half cross times b cross times h
    1 half cross times 18 cross times 10 = 90 sq. units

    If the side AC of a given triangle is 18 units and the height of the triangle is 10 units, what is the area of triangle ABC ?

    Maths-General
    It is given that Base, b = AC = 18 units and Height, h = 10 units
    Area of the triangle = 1 half cross times b cross times h
    1 half cross times 18 cross times 10 = 90 sq. units
    parallel
    General
    Maths-

    A triangle has an area of 90 𝑚2 and a base of 12 m, find the height of such a triangle?

    It is given that base of the triangle = 12 m
    area of the triangle = 90 m2
    1 half cross times b cross times h = 90
    1 half cross times 12 cross times h = 90
    h = 90 over 6 = 15 m

    A triangle has an area of 90 𝑚2 and a base of 12 m, find the height of such a triangle?

    Maths-General
    It is given that base of the triangle = 12 m
    area of the triangle = 90 m2
    1 half cross times b cross times h = 90
    1 half cross times 12 cross times h = 90
    h = 90 over 6 = 15 m
    General
    Maths-

    If the height to base ratio of a triangle ABC is 3:4 and the area is 864 square units. Determine the height and base of this triangle.

    It is given that height to base ratio is 3 : 4
    ⇒ Height, h = 3x and Base, b = 4x
    Area of triangle = 1 half cross times b cross times h
    864 = 1 half cross times 4 x cross times 3 x
    864  = 6x2
    144 = x2
    12 = x
    Height = 3x = 3(12) = 36 units and Base = 4x = 4(12) = 48 units

    If the height to base ratio of a triangle ABC is 3:4 and the area is 864 square units. Determine the height and base of this triangle.

    Maths-General
    It is given that height to base ratio is 3 : 4
    ⇒ Height, h = 3x and Base, b = 4x
    Area of triangle = 1 half cross times b cross times h
    864 = 1 half cross times 4 x cross times 3 x
    864  = 6x2
    144 = x2
    12 = x
    Height = 3x = 3(12) = 36 units and Base = 4x = 4(12) = 48 units
    General
    Maths-

    If the side of an equilateral triangular park is 20 units. What will be half of its area ?

    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 202 = 100 cm2
    Half of the areaequals 1 half cross times 100 square root of 3
    50 square root of 3 = 86.60 cm2

    If the side of an equilateral triangular park is 20 units. What will be half of its area ?

    Maths-General
    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 202 = 100 cm2
    Half of the areaequals 1 half cross times 100 square root of 3
    50 square root of 3 = 86.60 cm2
    parallel
    General
    Maths-

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

    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

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

    Maths-General
    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
    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
    parallel
    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
    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

    parallel
    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)

    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.
    parallel

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