An equilateral triangle ABC is inscribed in square ADEF, forming three

It should be 'C'.
See the attached solution.

let's denote AB = BC = AC = 1. B belongs to DE, C belongs to EF
angle BAD = 15 degrees, EBC = 45 degrees
let's draw DH: H belongs to AB and DH is perpendicular to AB -> S(ABD) = 1/2*(AB)*(DH) = 1/2*1*(sin 15*cos15) = 1/2*(1/2)*sin(30) = 1/8 //DH = AB*sin(BAD)*cos(BDH) = AB*sin(BAD)*cos(BAD) = 1/2*sin(2*BAD)

let's draw EN: N belongs to BC and EN is perpendicular to BC -> S(BEC) = 1/2*(BC)*(EN) = 1/2*1*(sin45*cos45) = 1/2*1/sqrt(2)*1/sqrt(2)= 1/4

S(BEC)/S(ADB) = 2 -> C

Let's denote AB = BC = AC = 1. B belongs to DE, C belongs to EF
angle BAD = 15 degrees, EBC = 45 degrees
Assume the length of the side of the square as y and BE = x.
Now, A(BEC)/A(ADB) =(AD* DB)/ (BE*CE) ignoring 1/2 in denominator as well as numerator
as BE = CE = x and AD = y and DB = (y-x)
if you replace values and on solving, A(BEC)/A(ADB) = x^2 / (y^2 -xy)

Now, as per pythagorus, BC = AB = 2 * sqrt x
and AB^2 = AD^2 + DB^2
put values for AB = 2 * sqrt x and AD = y and DB = y-x
solve, you'll get x^2 / (y^2 -xy) = 2

Hence, c

Let Y be length of the square
ABC is a triangle such that B intersects DE and C intersects FE (This info is from problem description)

Since it is an quilateral triangle , each interior angle of triangle ABC is 60 degrees.
Therefore angle BAD = angle CAF = 15.
Tiangle AFC is right angled. So angle ACF = 75.

Now solve angles for triangles ADB abd BCE.

Tiangle BCE turns out to be an isocles right angled triangle.


Area of BCE = BE * CE/2

let BE =x


Area of BCE = x* x/2


Area of ADB = AD* DB/2

but DB+BE=y

->DB = y-x

Area of ADB = y* (y-x)/2


Since ADB is a right angled triangle.

y^2 +(y-x)^2 = 2 * x^2

->y^2 -xy = x^2/2

Area of BCE/Area of ADB = [x* x/2]/[ y* (y-x)/2]
= x* x/y* (y-x) =2

Fig

Thanks, knew i could count on you, by the way, whats the program you use to draw these diagrams? i've been using adobe photoshop and it looks too crude

adobe photoshop. I made template with grid.

By the way, second approach:

let x - a side of the equilateral triangle ABC
y - a side of the squre

1. The diagonal of squre = \(\sqrt{2}y = \frac{\sqrt{3}}{2}x+\frac{1}{2}x=\frac{x}{2}*(\sqrt{3}+1)\)
2. The area of squre = \(y^2=\frac{x^2}{8}*(\sqrt{3}+1)^2=\frac{x^2}{8}*(4+2\sqrt{3})=\frac{x^2}{4}*(2+sqrt{3})\)
3. The area of triangle ABC = \(\frac12*\frac{\sqrt{3}}{2}x*x=\frac{\sqrt{3}}{4}x^2\)
4. The area of triangle BEC = \(\frac12*\frac{x}{2}*x=\frac{x^2}{4}\)
5. The area of triangle ADB = 1/2 * (The area of squre - The area of triangle ABC - The area of triangle BEC) = \(\frac12*(\frac{x^2}{4}*(2+sqrt{3})-\frac{\sqrt{3}}{4}x^2 -\frac{x^2}{4})=\frac{x^2}{8}\)
6. ratio = \(\frac{\frac{x^2}{4}}{\frac{x^2}{8}}=2\)

we have the areas of both triangles, why do we need to maniipulate the second triangle with the pythg theorem?

Since |AB|=|BC|, x2 + 2xy + y2 +x2 = 2y2, so 2x2 +2xy =2x(x+y)= y2
The ratio of the area of triangle BEC to that of triangle ADB is y2/x(x+y) =2

AB and BC obviously are 2 sides of the triangle. x and y are the 2 parts of the side of the square

I used the pythg theorem to get the Area of ADB in terms of x.

It seems as if you are saying the diagonal is equal to a 30:60:90 triangle. For some reason this problem is really killing me! Can someone explain?

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 \((s^2/4)/(s^2/8) = 2\)

Oh wow - the answer is so obvious to me now. Thank you for the clarification!

Karishma this is an awesome explanation for this tough problem, thanks a lot

Just one small doubt, how do you know that ADB and AFC right triangles are equal? and both different from BCE, I have a bit of trouble making those sort of observations

Many thanks!
Cheers!
J

Regular polygons inscribed in other regular polygons or circles make symmetrical figures. Try drawing them out. The sides/angles you think are equal will usually be equal.
Angle DAF is 90 and BAC is 60 so angles DAB and FAC will be 15 degrees each. Also AD = AF and AB = AC. So triangles ADB and AFC are congruent. On the other hand, in triangle BEC, side BE = CE so it is a 45-45-90 triangle.