Application of Mathematics

In: Science

Submitted By mmgathee
Words 1486
Pages 6
Title: Application of Mathematics

Question one
Volume of figure one
Volume = area * length
Length = 7500M
Area = area of the outer big rectangle minus the area of the curve and the smaller inner rectangle * Area of the bigger rectangle = height * width
=76.8 M2 * Area under the curve
We use the equation y=aX2 and solve for a with x and y coordinates as (3.6, 4.2), as this are the coordinates on the peak of the curve. a=0.3240740741 Our equation is: y= 0.3240740741x2 * Area under the curve =07.20.3240740741X2
= 40.32M2
Area of the inner rectangle= height * width
= 7.2 * 2
=14.4 m2
Total area =76.8-14.4-40.32
VOLUME = 7500*22.08
Volume of figure two.
The volume is found by multiplying the total area by height .
Height = 7500 M
Area is obtained by subtracting the area of the semicircle and inner rectangle from outer rectangle
Area of outer rectangle= width * height
=152.4 M2
Area of the semicircle = R2
= * 3.652
=41.853868 M2
Area of the inner rectangle = height * width
= 5*7.3
=36.5 M2
Total area = 152.4- 41.85868-36.5
=74.046132 M2
Volume= 74.046132 M2 * 7500 M
=555,345.99 M3

Volume of figure three.
This is found by multiplying the height by the total area
Total area
Area of the outer rectangle minus the areas of the two 110 degrees circles .
Area of the rectangle = height * width
= 11 * 6
= 66 metres squared
Area of the circular parts =
110360 * 2 * 2.12* * 2= 16.9331844
Total area = 66- 16.9331844
=49.066815 M2
Volume = 49.066815* 7500
= 368,001.117 M3

B) * Figure one
The cost of constructing tunnel one will be the cost of concrete per cubic meter multiplied by the volume of of the tunnel.
= 120* 165,600
= £19,872,000 * Figure two
Cost = 555,345.99 *120
= £66,641,518.80 * Figure three
Cost =…...

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