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An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]
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13 Apr 2007, 10:25
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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. \(\sqrt 3\) C. 2 D. 5/2 E. \(\sqrt 5\)
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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]
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13 Apr 2007, 11:54
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It should be 'C'.
See the attached solution.



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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]
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14 Apr 2007, 02:16
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My solution is very similar to yours  I got 2 as well. Maybe a bit shorter. I am attaching my solution as a jpeg file. Tell me what you think.
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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]
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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. \(sqrt(3)\) C. 2 D. 5/2 E. \(sqrt(5)\)



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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]
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marcodonzelli 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?
4/3 sqrt3 2 5/2 sqrt5 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



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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]
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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 = (yx) 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 = yx solve, you'll get x^2 / (y^2 xy) = 2
Hence, c



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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]
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04 Mar 2008, 23:06
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chica wrote: is there any easier approach without trigonometry involved ?? I am confused with drawing here.. do not get it 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 = yx Area of ADB = y* (yx)/2 Since ADB is a right angled triangle. y^2 +(yx)^2 = 2 * x^2 >y^2 xy = x^2/2 Area of BCE/Area of ADB = [x* x/2]/[ y* (yx)/2] = x* x/y* (yx) =2



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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]
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07 Mar 2008, 10:46
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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]
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07 Mar 2008, 11:01
walker wrote: 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
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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]
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07 Mar 2008, 11:51
bmwhype2 wrote: 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.
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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]
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07 Mar 2008, 12:14
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 = \(\frac{\frac{x^2}{4}}{\frac{x^2}{8}}=2\)
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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]
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07 Mar 2008, 12:32
goalsnr wrote: chica wrote: is there any easier approach without trigonometry involved ?? I am confused with drawing here.. do not get it 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/2Area of ADB = AD* DB/2 but DB+BE=y >DB = yx Area of ADB = y* (yx)/2Since ADB is a right angled triangle.
y^2 +(yx)^2 = 2 * x^2
>y^2 xy = x^2/2Area of BCE/Area of ADB = [x* x/2]/[ y* (yx)/2] = x* x/y* (yx) =2 we have the areas of both triangles, why do we need to maniipulate the second triangle with the pythg theorem?
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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]
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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



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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]
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09 Mar 2008, 15:15
bmwhype2 wrote: goalsnr wrote: chica wrote: is there any easier approach without trigonometry involved ?? I am confused with drawing here.. do not get it 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/2Area of ADB = AD* DB/2 but DB+BE=y >DB = yx Area of ADB = y* (yx)/2Since ADB is a right angled triangle.
y^2 +(yx)^2 = 2 * x^2
>y^2 xy = x^2/2Area of BCE/Area of ADB = [x* x/2]/[ y* (yx)/2] = x* x/y* (yx) =2 we have the areas of both triangles, why do we need to maniipulate the second triangle with the pythg theorem? I used the pythg theorem to get the Area of ADB in terms of x.



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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]
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16 Dec 2013, 17:18
walker wrote: 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 = \(\sqrt{2}y = \frac{\sqrt{3}}{2}x+\frac{1}{2}x=\frac{x}{2}*(\sqrt{3}+1)\)
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?



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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]
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16 Dec 2013, 20:13
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



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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]
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16 Dec 2013, 23:07
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 454590 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 \((s^2/4)/(s^2/8) = 2\)
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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]
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17 Dec 2013, 06:14
Oh wow  the answer is so obvious to me now. Thank you for the clarification! VeritasPrepKarishma wrote: 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 454590 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 \((s^2/4)/(s^2/8) = 2\)



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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]
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VeritasPrepKarishma wrote: 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 454590 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 \((s^2/4)/(s^2/8) = 2\) 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



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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]
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15 Jan 2014, 21:16
jlgdr wrote: 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 454590 triangle.
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