In a figure, AG is vertical to BC.
The points D and E are the middle points of AG and AC, respectively.
The point F is an intersection of BE and CD.
The length of DE is 2 cm.
The points D and E are the middle points of AG and AC, respectively.
The point F is an intersection of BE and CD.
The length of DE is 2 cm.
The area of triangle ABC is 36 cm2.
The area of the triangle CEF is 2 cm2.
Find the length of AG.
Answer
Since point E is the middle point of AC, the area of △BCE is a half of the area of △ABC, and it is 36 cm2 / 2 = 18 cm2.
Since the area of △CEF is 2 cm2, the area of △CBF is 18 - 2 = 16 cm2.
Since the area ratio of △CEF and △CBF is 2 : 16 = 1 : 8 and the height of both triangle is same, EF : BF which is a base of △CEF and △CBF is also 1 : 8.
△DEF and △BCF are homothetic and a homothetic ratio is EF : BF = 1 : 8.
As DE = 2cm, BC = 2 cm × 8/1 = 16 cm.
Therefore, AG = 36 cm2 × 2 / 16 cm = 4.5 cm.
4.5 cm
Solution
Since the area of △ABC is known, in order to find the length of AG which is the height of △ABC, the length of BC which is a base should just be known.
Since point E is the middle point of AC, the area of △BCE is a half of the area of △ABC, and it is 36 cm2 / 2 = 18 cm2.
Since the area of △CEF is 2 cm2, the area of △CBF is 18 - 2 = 16 cm2.
Since the area ratio of △CEF and △CBF is 2 : 16 = 1 : 8 and the height of both triangle is same, EF : BF which is a base of △CEF and △CBF is also 1 : 8.
△DEF and △BCF are homothetic and a homothetic ratio is EF : BF = 1 : 8.
DE : BC is also 1 : 8.
As DE = 2cm, BC = 2 cm × 8/1 = 16 cm.
Therefore, AG = 36 cm2 × 2 / 16 cm = 4.5 cm.
