Area and volume > Volume
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Solutions to the exercises

Exercise 1

volume(semi cone) = 1 3 1 2 π 1 , 5 2 3 = 1 , 125 π
volume(quarter sphere) = 1 4 4 3 π 1 , 5 3 = 1 , 125 π
volume(diabolo) = 2 ( 1 3 π 1 , 5 2 2 , 25 - 1 3 π 0 , 5 2 0 , 75 ) = 3 , 25 π

Exercise 2

The volume under this hipped roof belongs to a solid consisting of a triangular prism (put onto its side) with two semi pyramids (that you can combine into one pyramid) on both sides. Its volume therefore is: 1 2 8 5 6 + 1 3 8 6 5 = 160 m3.

Exercise 3

The base is composed of five isosceles triangles with a top angle of 72°, a base length of 4 cm, two sides of 2 ( sin ( 36 o ) ) and a height of 2 ( tan ( 36 o ) ) . The area of the pentagon is therefore 5 1 2 4 2 ( tan ( 36 o ) ) 27 , 53 cm2.
The height of this regular five-sided pyramid is then ( 4 2 - ( 2 ( sin ( 36 o ) ) ) 2 ) 2 , 10 cm.
The volume of the figure is therefore 1 3 27 , 53 2 , 10 19 , 3 cm3.

Exercise 4

1 3 π 2 2 12 - 1 3 π 1 , 5 2 9 + 1 3 π 1 , 5 2 1 = 10 π

Exercise 5

(Mentally) saw through the pipe at the seam of the right-angled bend. Twist to straighten the pipe, as shown in the figure. The amount of steel (assuming there is no hole in the base plate) then becomes: 150 150 1 + π 25 2 650 - π 24 2 650 122560 mm3.
That is 122,56 cm3. The whole construction therefore weighs 956 gram.

Exercise 6

volume = 4 3 π 4 3 - 1 3 π ( 4 - ( 4 2 - 3 2 ) ) 2 ( 3 4 - ( 4 - ( 4 2 - 3 2 ) ) ) + π 3 2 3 337 , 8 m3

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