SOLUTION: Equilateral triangle △ABC has side length 2, M is the midpoint of
segment AC, and C is the midpoint of BD. What is the area of
△CDM?
(please check the link below as it conta
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-> SOLUTION: Equilateral triangle △ABC has side length 2, M is the midpoint of
segment AC, and C is the midpoint of BD. What is the area of
△CDM?
(please check the link below as it conta
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Question 1151273: Equilateral triangle △ABC has side length 2, M is the midpoint of
segment AC, and C is the midpoint of BD. What is the area of
△CDM?
(please check the link below as it contains the image for the question)
https://i.imgur.com/g3InsY0.png Found 2 solutions by ikleyn, jim_thompson5910:Answer by ikleyn(53751) (Show Source):
HINT 1. The area of the equilateral triangle is square units, where "a" is its side length.
So, in this case, the area of the equilateral triangle ABC is = .
HINT 2. The area of the triangle CDM is half the area of the triangle ABC.
Triangle ABC has base of BC = 2
Let h be the height of triangle ABC. It does not matter what h is for this thought experiment.
The area of triangle ABC is therefore, A = (1/2)*base*height = (1/2)*2*h = h.
Area of triangle ABC = h.
The base of triangle CDM is also 2, because C is the midpoint of BD, so BC = CD = 2.
The height of point M is half that of point A's height. Imagine that point A is (0, h). Through use of the midpoint formula, you'll find that the y coordinate of point M will be y = h/2.
So the height of triangle CDM is h/2.
area of triangle CDM = (1/2)*base*height
area of triangle CDM = (1/2)*2*(h/2)
area of triangle CDM = h/2
area of triangle CDM = (area of triangle ABC)/2
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