There is a parallelogram ABCD as shown in a figure.
The points E and F are points of dividing the side BC into three equally.
The point G is a point of the middle of the side CD.
Find the area ratio of the area of a shadow area and the area of parallelogram ABCD.
The points E and F are points of dividing the side BC into three equally.
The point G is a point of the middle of the side CD.
Find the area ratio of the area of a shadow area and the area of parallelogram ABCD.
Answer
Solution
An auxiliary line HG is drawn as shown in Fig.1.
The area of △AJG of a shadow area is GI × D'J' / 2.
When it is set that BC = 3 and D'C ' = 2, the area of a parallelogram is 3 × 2 = 6.
HI = BF × 1/2 = 2 × 1/2 = 1.
GI =3 - 1 = 2.
△GIJ and △EFJ are homothetic and a homothetic ratio is 2:1.
When GI is set to the base of △GIJ, the height of △GIJ is G'J' = 1 × 2/(1+2) = 2/3.
Thus, △AJG = GI x D'J' / 2 = 2 × (1+ 2/3) / 2 = 5/3.
Therefore, △AJG : Parallelogram = 5/3 : 6 = 5 : 18.
5 : 18
An auxiliary line HG is drawn as shown in Fig.1.
The area of △AJG of a shadow area is GI × D'J' / 2.
When it is set that BC = 3 and D'C ' = 2, the area of a parallelogram is 3 × 2 = 6.
HI = BF × 1/2 = 2 × 1/2 = 1.
GI =3 - 1 = 2.
△GIJ and △EFJ are homothetic and a homothetic ratio is 2:1.
When GI is set to the base of △GIJ, the height of △GIJ is G'J' = 1 × 2/(1+2) = 2/3.
Thus, △AJG = GI x D'J' / 2 = 2 × (1+ 2/3) / 2 = 5/3.
Therefore, △AJG : Parallelogram = 5/3 : 6 = 5 : 18.
