There is a right-angled isosceles triangle ABC.
There is another right triangle DBE width is shorter than ABC by 3 cm and length is longer than ABC by 6 cm.
The figure shows two triangles overlap.
The point F is an intersection of AC and DE.
The ratio of the area of triangle CEF to ADF is 3 : 8.
Find the area of triangle ABC.
There is another right triangle DBE width is shorter than ABC by 3 cm and length is longer than ABC by 6 cm.
The figure shows two triangles overlap.
The point F is an intersection of AC and DE.
The ratio of the area of triangle CEF to ADF is 3 : 8.
Find the area of triangle ABC.
Answer
Line EG is drawn in parallel with AB from E and △GEC is created.
△GEC is also a right-angled isosceles triangle and GE = CE = 3cm.
Moreover, since GE and DB are parallel, △ADF and △GEF are homothetic and a homothetic ratio is DA : GE = 6cm : 3cm = 2 : 1.
Next, as shown in Fig. 2, FH is drawn in parallel with GE from F.
220.5 cm2
Solution
Line EG is drawn in parallel with AB from E and △GEC is created.
△GEC is also a right-angled isosceles triangle and GE = CE = 3cm.
Moreover, since GE and DB are parallel, △ADF and △GEF are homothetic and a homothetic ratio is DA : GE = 6cm : 3cm = 2 : 1.
An area ratio is 2 × 2 : 1 × 1 = 4 : 1.
The area ratio of △ADF and △CEF is 8 : 3 and the area ratio of △ADF and △GEF is 4 : 1 = 8 : 2.
When the area of △GEF is set to 2, the area of △GCE is 3 - 2 = 1.
Thus, the area ratio of △GEF and △GCE is 2 : 1.
FG : GC is also 2 : 1.
Next, as shown in Fig. 2, FH is drawn in parallel with GE from F.
Since FG : GC = 2 : 1, HE : EC is also 2 : 1, then HE = 3cm × 2 = 6cm.
Moreover, as for the homothetic ratio of △ADF and △GEF is 2 : 1, BH : HE which is a ratio of the height of two triangles is also 2 : 1.
Thus, it is understood that BH = HE × 2 = 6cm × 2 =12cm.
Therefore, as BC = AB = 12 + 6 + 3 = 21cm, the area of△ABC is 21 × 21 / 2 = 220.5cm2.
