KK.4 Shadow of quadratic prism

There is a big wall which stands vertically to the ground.

The bottom of the square pillar of Fig.2 is a trapezoid as shown in Fig.1.

The height is 30 cm.

It is placed as the side CD is parallel to the wall and the distance to the wall is 10 cm.

The angle to look up at the sun from the ground is 45 degrees.

The shadow area of Fig.2 expresses the shadow made on the ground of this quadratic prism.

Find the area of the whole shadow made to the wall by this quadratic prism.













Answer
175 cm2

Solution
It turns out that the shadow made to the wall as shown in Fig.3 becomes a pentagon. 



You can recognize easily if you draw the figure seen from just beside as Fig.4. 



The angle which looks up at the sun from the ground is 45 degrees. 

The straight line FP and GQ of Fig.3 express the line of light and the angle to the ground is 45 degrees. 

Since △FBP is a right-angled isosceles triangle and is FB=30 cm, BP = FB = 30 cm. 

Since BC = CO =10cm, OP = 10cm and X is 10 cm as well. 

Thinking the same way, since △GCQ is also a right-angled isosceles triangle, Y is 20 cm. 

Therefore, the shadow made to a wall becomes like Fig.5 and the area is 10 × 20 - 5 × 10 / 2 = 175cm2