Oh wow - the answer is so obvious to me now. Thank you for the clarification!
WholeLottaLove wrote:
An equilateral triangle ABC is inscribed in square ADEF, forming three right triangles: ADB, ACF and BEC. What is the ratio of the area of triangle BEC to that of triangle ADB?
A. 4/3
B. √(3)
C. 2
D. 5/2
E. √(5)
This is the closest I could get to a real answer:
The sides of the square = (Y) while the sides of the equilateral triangle = (S) Isosceles triangle CBE = 45:45:90 and as such, their lengths are equal to side CB=√2 so EB = EC = S/√2. If the sides of the square = y then sides DB = FC = y-(s/√2).
We know:
EB = EC = S/√2
DB = FC = y-(s/√2)
Area CBE = 1/2 (b*h)
Area = 1/2*(s/√2)*(s/√2)
Area = s^2 / 4
Area ADB = 1/2 (b*h)
Area = 1/2 (y-[s/√2])*(y)
Area = 1/2 y^2 - sy/√2
Area = (y^2)/2 - sy/√2*2
Area = (y^2)/2 - sy√2 / (√2 * √2 * 2)
Area = (y^2)/2 - sy√2 / (2 * 2)
Area = (y^2)/2 - sy√2 / (4)
Area = (2y^2)/4 - sy√2 / (4)
Area = [(2y^2) - sy√2] / (4)
Area = [y(2y - s√2)] / (4)
Difference in area:
[s^2 / 4] - [y(2y - s√2)] / (4)
[s^2 - y(2y - s√2)] / (4)But, that's unlike ANY answer choice
You need to get y in terms of s to get the ratio.
y - side of square
s - side of equilateral triangle
Note that the diagonal of a square is \(\sqrt{2}*side = \sqrt{2}*y\)
Also note that the diagonal is composed of 2 parts, the altitude of the equilateral triangle + the altitude of triangle BEC
Altitude of equilateral triangle is given by \(\sqrt{3}/2 * Side = \sqrt{3}/2 * s\)
Altitude of 45-45-90 triangle is half of the hypotenuse (which is s here). So altitude of BEC = s/2
So \(\sqrt{2}*y = \sqrt{3}/2 * s + s/2\)
So \(y = s * (\sqrt{3} + 1)/(2\sqrt{2})\)
Now, area of square = \(y^2 = s^2(\sqrt{3} + 1)^2/8\)
Area of triangle ABC = \((\sqrt{3}/4) * s^2\)
Area of triangle BEC \(= s^2/4\)
Area of triangle ADB = Area of triangle AFC \(= 1/2 * (s^2(\sqrt{3} + 1)^2/8 - (\sqrt{3}/4) * s^2 - s^2/4) = s^2/8\)
Hence the required ratio is \((s^2/4)/(s^2/8) = 2\)