SOLUTION: Equilateral triangle ABC with sides of 1cm has altitude AD. median AE of triangle ABD is drawn. What is the area, in cm^2, of triangle AEC?

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Question 1209435: Equilateral triangle ABC with sides of 1cm has altitude AD. median AE of triangle ABD is drawn. What is the area, in cm^2, of triangle AEC?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52800) About Me  (Show Source):
You can put this solution on YOUR website!
.
Equilateral triangle ABC with sides of 1cm has altitude AD. median AE
of triangle ABD is drawn. What is the area, in cm^2, of triangle AEC?
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The area of the triangle ABC is  a%5E2%2A%28sqrt%283%29%2F4%29 = sqrt%283%29%2F4 cm^2.


Every student, who studied the areas of triangles and the areas of equilateral
triangles, must know it, so I will not go in further explanations.


Next, when we draw altitude AD, we cut the area of triangle ABC in two equal half.


When we draw median AE in triangle ABD, we cut the area of triangle ABD in two equal parts.


Thus finally, the area of triangle AEC is  1 - 1%2F4 = 3%2F4  of the area of triangle ABC, i.e.  %283%2Asqrt%283%29%29%2F16  cm^2.


ANSWER.  The area of triangle AEC is  %283%2Asqrt%283%29%29%2F16 cm^2.

Solved.

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Actually, this problem is for / (or is from) mathematical mental competition,
where the answer should be given in the next instance to win.

I am absolutely serious: the true meaning of this question is that
it should be answered mentally quicker than you blink your eye.



Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Equilateral triangle ABC has side length s = 1 which leads to area %281%2F4%29s%5E2%2Asqrt%283%29+=+%281%2F4%29%2A1%5E2%2Asqrt%283%29+=+sqrt%283%29%2F4

Equilateral triangle ABC


Draw the altitude AD which goes from point A to side BC.
The altitude is perpendicular to BC.
For any equilateral triangle, an altitude is also a median.
D is the midpoint of BC.



The last point to introduce is E.
This is the midpoint of segment BD


Because segment EB is half of BD, and BD is half of BC, this makes EB have a length that is (1/2)*(1/2) = 1/4 of BC's length
EC is 3/4 the length of BC

Since triangles ABC and AEC have the same height we can say,
area of triangle AEC = (3/4)*(area of triangle ABC)

area of triangle AEC = %283%2F4%29%2A%28sqrt%283%29%2F4%29

area of triangle AEC = %283%2Asqrt%283%29%29%2F16 square cm exactly

area of triangle AEC = 0.324759526419 square cm approximately