There is a picture of a square - BCAD, and then there is a triangle EAD.
B. E C
A. D
FIGURE NOT DRAWN TO SCALE
In the figure above, the perimeter of the equilateral triangle AED IS 6. What is the area of the rectangle ABCD?
A) 6
B) 4
C) 4 radical 3
D) 3 radical 3
E) 2 radical 3
5 answers
wondering where E is.
clearly the base of ∆EAD is AD.
So, each of its sides has length 2.
Thus AD=2, and the area of ABCD is 2^2 = 4
Not sure what the EC is, but if you want the length, then since the altitude of ∆EAD is √3,
1^2 + (2+√3)^2 = EC^2
EC = √2+√6
So, each of its sides has length 2.
Thus AD=2, and the area of ABCD is 2^2 = 4
Not sure what the EC is, but if you want the length, then since the altitude of ∆EAD is √3,
1^2 + (2+√3)^2 = EC^2
EC = √2+√6
unless, of course, E is inside ABCD. In that case,
1^2 + (2-√3)^2 = EC^2
EC = √6-√2
1^2 + (2-√3)^2 = EC^2
EC = √6-√2
Why did you do
1^2 + ( 2+ radical3) ^2
Is that the Pythagorean theorem?
If it is why did you do
( 2+ radical3) ^2
Wouldn't it be
(Radical 3)^2 + EC = 2^2
1^2 + ( 2+ radical3) ^2
Is that the Pythagorean theorem?
If it is why did you do
( 2+ radical3) ^2
Wouldn't it be
(Radical 3)^2 + EC = 2^2
Maybe. I figured EC was the hypotenuse of
∆ECF where F is the midpoint of BC -- the side parallel to AD.
If I got it wrong, then you may be right.
Though, I doubt it, since that makes EC=1.
∆ECF where F is the midpoint of BC -- the side parallel to AD.
If I got it wrong, then you may be right.
Though, I doubt it, since that makes EC=1.
1 answer