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ABC is an equilateral triangle. DE and DF are perpendiculars drawn from D to the sides shown.
DE=4 cm and DF = 26 cm. Find the length, in cm, of the altitude AG.
https://ibb.co/4jLR8TF
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Connect points D and G by a straight line.
This line divides triangle ABC in two triangles ABD and ACD.
So, the area of the whole triangle ABC is the sum of the areas of triangles ABD and ACD
= + . (1)
Let "a" be the side length of the equilateral triangle ABC.
Then the area of triangle ABC is = ;
the area of triangle ABD is = = ;
the area of triangle ACD is = = .
So, we can write equation (1) in the form
= + . (2)
In this equation, reduce the common factor in both sides. You will get then
AG = 4 + 26 = 30.
At this point, the solution is complete, and you get this
ANSWER. AG = 30 cm.
Solved.
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This problem is a classic,
and the solution, which I provided in my post, is a classic version, too.
It provides a clear idea in clean form with the minimum calculations, and makes
the problem and the solution interesting, attractive, informative and educative.
This problem is one of the most brilliant among simple problems about triangles.
Such problems should be treated with reverence, and their classical solutions
should be passed on from generation to generation in that beautiful form as we
know them from our predecessors.
ABC is an equilateral triangle. DE and DF are perpendiculars drawn from D to the sides shown. DE=4 cm and DF = 26 cm.
Find the length, in cm, of the altitude AG.
https://ibb.co/4jLR8TF
As ΔABC is equilateral, ∡B = ∡C = 60o
From right ΔBED, hypotenuse BD =
From right ΔCFD, hypotenuse DC =
So, one side of equilateral triangle ABC, BC = BD + DC =
AC is also
As an altitude from ANY vertex of ANY equilateral triangle, BISECTS that vertex' opposite side, altitude AG
BISECTS BC, and therefore, BG = GC = =
Since AC = , and GC = , we use the PYTHAGOREAN THEOREM formula to get:
30 cm = AG