Math Problem : KKK.6 Shadow of rectangular wall

As shown in Fig.1, there are three point A, B and P on the even ground.

There are ABCD of a 3m-high rectangular wall and PQ of a 9m-high pillar standing on the ground straight.

Fig.2 shows the picture looked at these from right above.

Wall ABCD is illuminated by the electric light in the position at the tip Q of the pillar.

Find the area of the shadow of the wall made into the ground.

Neither the size of an electric light nor the thickness of a wall shall be considered.

Answer

45 m²

Solution

Fig.3 is the figure which looked at Fig.1 from the side. 

Since △DAD' and △QPD' are homothetic in Fig.3 and as DA = 3m and QP = 9m, homothetic ratio is 3 : 9 = 1 : 3.

AP : D'P = (3 - 1) : 3 = 2 : 3.

In Fig.4, a shadow area is a shadow of wall ABCD.

Since AP : D'P = 2 : 3, △PAB and △PC'D' are homothetic.

A homothetic ratio is 2 : 3 and an area ratio is (2 × 2) : (3 × 3) = 4 : 9.

The area ratio of △PAB and a shadow area is 4 : (9 - 4) = 4 : 5.

As shown in Fig.5 as for △PAB, AF is a base and B(C) G is height.

△BEF and △PHF are homothetic and a homothetic ratio is BE : PH = 4 : 2 = 2 : 1.

FH = 6m × 1/(2+1) = 2m.

AF = 8m + 6m - 2m = 12 m.

As EG = HP = 2m, BG = 4m + 2m = 6m.

Thus, the area of △PAB is 12 × 6 / 2 = 36m².


Therefore, the area of the shadow to be found is 36 m² × 5/4= 45 m².