Heron’s Formula ( Part – 6 )

1. A floral design on a floor is made up of 16 tiles which are triangular , the sides of triangle being 18 cm , 56 cm and 70 cm . Find the cost of polishing the tiles at the rate of 50 p per cm^2 ?

Sol. The sides of triangles are a = 18 cm b = 56 cm c = 70 cm

The semi perimeter of triangle S = ( a + b + c ) / 2

The semi perimeter of triangle S = ( 18 + 56 + 70 ) / 2

The semi perimeter of triangle S = 144 / 2

The semi perimeter of triangle S = 72 cm

Using Heron’s Formula , area of triangle = √ s ( s – a ) ( s – b ) ( s – c )

area of triangle = √ 72 ( 72 – 18 ) ( 72 – 56 ) ( 72 – 70 )

area of triangle = √ 72 * 54 * 16 * 2

area of triangle = √ 2 * 2 * 2 * 3 * 3 * 2 * 3 * 3 * 3 * 2 * 2 * 2 * 2 * 2

area of triangle = 2 * 2 * 2 * 2 * 3 * 3 √ 6

area of triangle = 144 √ 6

area of triangle = 352.73 cm^2

Area of 16 tiles = 16 * 352.73

Area of 16 tiles = 5643.62 cm^2

The cost of polishing the tiles = 0.50 * 5643.62

The cost of polishing the tiles = Rs 2821.81 approx Rs 2822

maths ncert solutions

2. A floral design on a floor is made up of 10 tiles which are triangular , the sides of triangle being 18 cm , 56 cm and 70 cm . Find the cost of polishing the tiles at the rate of 50 p per cm^2 ?

Sol. The sides of triangles are a = 18 cm b = 56 cm c = 70 cm

The semi perimeter of triangle S = ( a + b + c ) / 2

The semi perimeter of triangle S = ( 18 + 56 + 70 ) / 2

The semi perimeter of triangle S = 144 / 2

The semi perimeter of triangle S = 72 cm

Using Heron’s Formula , area of triangle = √ s ( s – a ) ( s – b ) ( s – c )

area of triangle = √ 72 ( 72 – 18 ) ( 72 – 56 ) ( 72 – 70 )

area of triangle = √ 72 * 54 * 16 * 2

area of triangle = √ 2 * 2 * 2 * 3 * 3 * 2 * 3 * 3 * 3 * 2 * 2 * 2 * 2 * 2

area of triangle = 2 * 2 * 2 * 2 * 3 * 3 √ 6

area of triangle = 144 √ 6

area of triangle = 352.73 cm^2

Area of 16 tiles = 10 * 352.73

Area of 16 tiles = 3527.30 cm^2

The cost of polishing the tiles = 0.50 * 3527.30

The cost of polishing the tiles = Rs 1763.65 approx Rs 1764

3. A field in the shape of a trapezium whose parallel sides are 50 m and 20 m . The non parallel sides are 28 m and 26 m . Find the area of the field .

Sol. Draw CF || AD and CG || AB

In quadrilateral ADCF ,

CF || AD { By construction }

CD || AF { ABCD is a trapezium }

Therefore , BF = AB – AF = 50 – 20 = 30 cm

The sides of triangle are a = 26 cm b = 28 cm c = 30 cm

The semi perimeter of triangle S = ( a + b + c ) / 2

The semi perimeter of triangle S = ( 26 + 28 + 30 ) / 2

The semi perimeter of triangle S = 84 / 2

The semi perimeter of triangle S = 42 cm

Using Heron’s Formula , area of triangle = √ s ( s – a ) ( s – b ) ( s – c )

area of triangle = √ 42 ( 42 – 26 ) ( 42 – 28 ) ( 42 – 30 )

area of triangle = √ 42 * 16 * 14 * 12

area of triangle = √ 2 * 3 * 7 * 2 * 2 * 2 * 2 * 2 * 7 * 2 * 2 * 3

area of triangle = 2 * 2 * 2 * 2 * 3 * 7

area of triangle = 336 cm^2

Area of triangle BCF = ½ * BF * CG

½ * 30 * CG = 336

CG = ( 336 * 2 ) / 30

CG = 22.4 cm

So Area of trapezium = ½ * ( AB + CD ) * CG

Area of trapezium = ½ * ( 50 + 20 ) * 22.4

Area of trapezium = ½ * 70 * 22.4

Area of trapezium = 35 * 22.4

Area of trapezium = 784 cm^2

Milan Tomic

Hi. I’m Designer of Blog Magic. I’m CEO/Founder of ThemeXpose. I’m Creative Art Director, Web Designer, UI/UX Designer, Interaction Designer, Industrial Designer, Web Developer, Business Enthusiast, StartUp Enthusiast, Speaker, Writer and Photographer. Inspired to make things looks better.