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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?

Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

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01 Mar 2008, 23:17

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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?

Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

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04 Mar 2008, 06:42

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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

Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

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04 Mar 2008, 11:44

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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 = (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

Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

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04 Mar 2008, 22:06

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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 = 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

Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

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07 Mar 2008, 10:01

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
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

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 = \(\frac{\frac{x^2}{4}}{\frac{x^2}{8}}=2\)
_________________

Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

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07 Mar 2008, 11:32

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/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

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

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

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09 Mar 2008, 14:15

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/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

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.

Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

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16 Dec 2013, 19:13

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)

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)

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

Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

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17 Dec 2013, 05:14

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)

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

Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

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15 Jan 2014, 05:24

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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)

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

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

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.
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