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yellowho: Master | Next Rank: 500 Posts; Posts: 233; Joined: Wed Aug 22, 2007 3:51 pmFollowed by: 2 members; Location: New York; Thanked: 7 times

by yellowho » Mon Feb 14, 2011 12:21 am

Stuck on this one:

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Source: Beat The GMAT — Problem Solving | Mon Feb 14, 2011 12:21 am

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon Feb 14, 2011 4:34 am

Area Square ABCD = s^2 = 625.
AD^2 = 625.
AD = 25.

Area of rhombus AHGD = bh = 500.
25(EH) = 500.
EH = 500/25 = 20.

Triangle AHE is a 3:4:5 triangle in which each side is multiplied by 5.
Thus, AE = 5*3 = 15.
Area of triangle AHE = 1/2*b*h = 1/2*15*20 = 150.

Since ED = 25-15 = 10, rectangle EHFD = bh = 10*20 = 200.

Shaded region = Square ABCD - Triangle AHE - Rectangle EHFD = 625-150-200 = 275.

Last edited by GMATGuruNY on Mon Feb 14, 2011 8:49 am, edited 2 times in total.

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yellowho: Master | Next Rank: 500 Posts; Posts: 233; Joined: Wed Aug 22, 2007 3:51 pmFollowed by: 2 members; Location: New York; Thanked: 7 times

by yellowho » Mon Feb 14, 2011 5:41 am

How do you know its a 3:4:5 triangle? Can you assume AHGD is a pivoted version of the square?

[quote="GMATGuruNY"][url=https://postimage.org/image/1g7hgtu04/][img]https://s2.postimage.org/1g7hgtu04/square_rhombus.jpg[/img][/url]

Area Square ABCD = s^2 = 625.
AD^2 = 625.
AD = 25.

Area of rhombus CEFD = bh = 500.
25(EH) = 500.
EH = 500/25 = 20.

Triangle AEH is a 3:4:5 triangle in which each side is multiplied by 5.
Thus, AE = 5*3 = 15.
Area of triangle AEH = 1/2*b*h = 1/2*15*20 = 150.

Since ED = 25-15 = 10, rectangle EHFD = bh = 10*20 = 200.

Shaded region = Square ABCD - Triangle AEH - Rectangle EHFD = 625-150-200 = 275.[/quote]

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon Feb 14, 2011 5:47 am

AEH is a right triangle with a leg of 20 and a hypotenuse of 25.
20:25 = 4:5.
The only right triangle with this proportion is a 3:4:5 triangle.
5*(3:4:5) = 15:20:25.

yellowho wrote:How do you know its a 3:4:5 triangle? Can you assume AHGD is a pivoted version of the square?

GMATGuruNY wrote:

Area Square ABCD = s^2 = 625.
AD^2 = 625.
AD = 25.

Area of rhombus CEFD = bh = 500.
25(EH) = 500.
EH = 500/25 = 20.

Triangle AEH is a 3:4:5 triangle in which each side is multiplied by 5.
Thus, AE = 5*3 = 15.
Area of triangle AEH = 1/2*b*h = 1/2*15*20 = 150.

Since ED = 25-15 = 10, rectangle EHFD = bh = 10*20 = 200.

Shaded region = Square ABCD - Triangle AEH - Rectangle EHFD = 625-150-200 = 275.

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yellowho: Master | Next Rank: 500 Posts; Posts: 233; Joined: Wed Aug 22, 2007 3:51 pmFollowed by: 2 members; Location: New York; Thanked: 7 times

by yellowho » Mon Feb 14, 2011 5:57 am

More specifically, how do you know 15 and 25 (hypotenuse and leg)?

[quote="GMATGuruNY"]AEH is a right triangle with a leg of 20 and a hypotenuse of 25.
20:25 = 4:5.
The only right triangle with this proportion is a 3:4:5 triangle.
5*(3:4:5) = 15:20:25.

[quote="yellowho"]How do you know its a 3:4:5 triangle? Can you assume AHGD is a pivoted version of the square?

[quote="GMATGuruNY"][url=https://postimage.org/image/1g7hgtu04/][img]https://s2.postimage.org/1g7hgtu04/square_rhombus.jpg[/img][/url]

Area Square ABCD = s^2 = 625.
AD^2 = 625.
AD = 25.

Area of rhombus CEFD = bh = 500.
25(EH) = 500.
EH = 500/25 = 20.

Triangle AEH is a 3:4:5 triangle in which each side is multiplied by 5.
Thus, AE = 5*3 = 15.
Area of triangle AEH = 1/2*b*h = 1/2*15*20 = 150.

Since ED = 25-15 = 10, rectangle EHFD = bh = 10*20 = 200.

Shaded region = Square ABCD - Triangle AEH - Rectangle EHFD = 625-150-200 = 275.[/quote][/quote][/quote]

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon Feb 14, 2011 8:48 am

yellowho wrote:More specifically, how do you know 15 and 25 (hypotenuse and leg)?

As noted above:
Area Square ABCD = s^2 = 625.
AD^2 = 625.
AD = 25.

AD is not only a side of square ABCD but also the base of rhombus AHGD.
Thus, in rhombus AHGD:
b=25, h = EH.
The formula for the area of a parallelogram (including a rhombus) is A = bh.
We're told that the area of rhombus AHGD is 500:
bh = 500
25*(EH) = 500.
EH = 500/25 = 20.

EH is not only the height of rhombus AHGD but also one of the legs of right triangle AHE.
Thus, in right triangle AHE:
EH = 20.
AH = 25 (because AD=25, and all the sides of a rhombus are equal).
Thus, in triangle AHE, leg:hypotenuse = 20:25 = 4:5.
Since the only right triangle with this proportion is a 3:4:5 triangle, we know:
AE:EH:AH = 3:4:5 = 15:20:25.
Thus, AE = 15.

Does this help?

Last edited by GMATGuruNY on Mon Feb 14, 2011 1:36 pm, edited 1 time in total.

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garuhape: Senior | Next Rank: 100 Posts; Posts: 74; Joined: Fri Dec 17, 2010 12:37 pm; Location: Munich; Thanked: 3 times

by garuhape » Mon Feb 14, 2011 12:50 pm

Or you use: a^2+b^2=c^3

you have 20^2+b^2=25^2 <=> 400+b^2=625 <=> b^2=225 <=> b=15

Hope this helped

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yellowho: Master | Next Rank: 500 Posts; Posts: 233; Joined: Wed Aug 22, 2007 3:51 pmFollowed by: 2 members; Location: New York; Thanked: 7 times

by yellowho » Mon Feb 14, 2011 7:29 pm

Yes thanks. The missing knowledge for me was that i didn't know the rhombus has all equal sides. Side question here, if two sides of a triangle follow one of the special triangles you automatically know the third? What if you don't know whether it is a right triangle.

[quote="GMATGuruNY"][quote="yellowho"]More specifically, how do you know 15 and 25 (hypotenuse and leg)?
[/quote]

[url=https://postimage.org/image/1g7hgtu04/][img]https://s2.postimage.org/1g7hgtu04/square_rhombus.jpg[/img][/url]

As noted above:
Area Square ABCD = s^2 = 625.
AD^2 = 625.
AD = 25.

AD is not only a side of square ABCD but also the base of rhombus AHGD.
Thus, in rhombus AHGD:
b=25, h = EH.
The formula for the area of a parallelogram (including a rhombus) is A = bh.
We're told that the area of rhombus AHGD is 500:
bh = 500
25*(EH) = 500.
EH = 500/25 = 20.

EH is not only the height of rhombus AHGD but also one of the legs of right triangle AHE.
Thus, in right triangle AHE:
EH = 20.
AH = 25 (because AD=25, and all the sides of a rhombus are equal).
Thus, in triangle AHE, leg:hypotenuse = 20:25 = 4:5.
Since the only right triangle with this proportion is a 3:4:5 triangle, we know:
AE:EH:AH = 3:4:5 = 15:20:25.
Thus, AE = 15.

Does this help?[/quote]

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