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12345Views/nets

Solutions to the exercises

Exercise 1
a

When in doubt, let it be checked.

b

See the figure.

c

Plane P B Q H is a rhombus 6 2 + 3 2 = 45 cm.
Diagonal P Q = ( 72 ) cm.
Using goniometry you find P H Q = P B Q 78 ° . The other two angles are 102 ° .

Exercise 2
a

See the figure.

b

See the figure.

Exercise 3
a

The base of the pyramid consists of five isosceles triangles with a top angle of 72 ° and a base of 4 cm.
The two legs of these triangles have a length of A S = 2 ( sin ( 36 ) ) 3,4 cm. That is the radius of the circle containing the five vertices of the pentagon A B C D E .
You can calculate the height T S of the pyramid using Pythagoras' theorem in Δ A S T . You find: T S = 4 2 - A S 2 2,1 cm.
This information enables you to draw the views.

b

4 4 4

Exercise 4

The top is a cone with height 1 and the radius of the base circle is 1,5 m.
The bottom is a truncated cone and the radius of its base circle is 2 m en 1,5 m and its height 3 m. This truncated cone originates from a cone with a height of 12 m.

Exercise 5

The (stiff) cone skirt has the shape of a truncated cone.
The base circle has circumference 3 4 2 π 119 = 178,5 π . So the radius of the base circle is 89,25 cm.
The top circle has circumference 3 4 2 π 19 = 28,5 π . So the radius of the top circle is 14,25 cm.
With these you can draw the views. You could even calculate the height of the truncated cone ( 66,1 cm), but this is not necessary. The side view is shown here.

Exercise 6

The top plane en the bottom plane are regular octagons. these consist of eight isosceles triangles with top angle 45 o and base 5 cm. The two legs of these triangels heve a length of 2,5 ( sin ( 22,5 ) ) 6,5 cm. This is the radius of the circle containing the vertices of these octagons.
the side consists of 16 equilateral triangles with edges of 5 cm.

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