The quadrangle ABCD in a figure is a square with 5 cm one-side.
All the length of AE, BF, CG, and DH is 2 cm.
Find the area of a shadow area.
All the length of AE, BF, CG, and DH is 2 cm.
Find the area of a shadow area.
Answer
This area is 2 × 5 = 10cm2.
Since HB and DF are parallel, △APH and △ASD are homothetic and a homothetic ratio is AH : AD = 3 : 5.
Thus, AP : PS = 3 : (5 - 3) = 3 : 2----<1>
Moreover,
According to <1> and <2>, if the continuous ratio is arranged as AP = 3 × 5 = 15, it will be set that AP : PS : SG = 15 : 10 : 9.
Therefore, the area ratio of a shadow area and parallelogram AECG is
(15 + 10 + 9) : 10 = 17 : 5.
The area of a shadow area is 10 cm2 × 5/17 = 50/17 cm2.
50/17 cm2
Solution
In the figure, since AE and GC are parallel and AG and EC are parallel, it turns out that a quadrangle AECG is a parallelogram.
This area is 2 × 5 = 10cm2.
Since HB and DF are parallel, △APH and △ASD are homothetic and a homothetic ratio is AH : AD = 3 : 5.
Thus, AP : PS = 3 : (5 - 3) = 3 : 2----<1>
Moreover,
since HP : DS = 3 : 5 and AP = DS, HP = SG, SG : AP = 3 : 5----<2>
According to <1> and <2>, if the continuous ratio is arranged as AP = 3 × 5 = 15, it will be set that AP : PS : SG = 15 : 10 : 9.
Therefore, the area ratio of a shadow area and parallelogram AECG is
(15 + 10 + 9) : 10 = 17 : 5.
The area of a shadow area is 10 cm2 × 5/17 = 50/17 cm2.
