Triangle problem

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farukqmul: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Fri Oct 14, 2011 12:46 pm; Location: Dhaka; GMAT Score:460

by farukqmul » Sat Jul 14, 2012 4:18 am

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Source: Beat The GMAT — Problem Solving | Sat Jul 14, 2012 4:18 am

theCEO: Master | Next Rank: 500 Posts; Posts: 363; Joined: Sun Oct 17, 2010 3:24 pmFollowed by: 3 members; Thanked: 115 times

by theCEO » Sat Jul 14, 2012 1:28 pm

The area of the shaded area is 1/4 of the area of triangle ADC.....

What are the choices?

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eagleeye: Legendary Member; Posts: 520; Joined: Sat Apr 28, 2012 9:12 pmFollowed by: 49 members; Thanked: 339 times; GMAT Score:770

by eagleeye » Sat Jul 14, 2012 2:15 pm

farukqmul wrote:

Let area of triangles DEC be x, ADE be y, and BDC be z.
Then we need to find shaded/unshaded = x/(y+z)

A couple of things that would make our life easier.
AC = 4EC , therefore AE:EC = 3:1
Also, AB=3AD, therefore AD:DB = 1:2
We also have area of triangle ADC = x+y.

Now triangle DEC and triangle AED share the same height (DQ)
For those of you who are salivating, I am NOT talking about Diary Queen )
Jokes apart, Since the height is the same , ratio of areas = ratio of bases
Then x/y = EC:AE = 1:3 => x/y = 1/3 => y=3x.

With the same idea, for triangles ADC and BDC, which share the same height CP, we have:
Area of triangle ADC/BDC = AD/BD
=> (x+y)/z = 1/2 => z = 2(x+y) = 2(x+3x) = 8x.

Hence, the required ratio = x/(y+z) = x/(3x+8x) = [spoiler]1/11 = 1:11[/spoiler]

Let me know if this helps

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Sun Jul 15, 2012 2:25 am

Note that, triangles AED and ECD has equal heights if we consider AE and EC as their bases, respectively. Hence, ratio of their areas are equal to the ratio of their bases.

Similarly, triangles ADC and BDC has equal heights if we consider AD and BD as their bases, respectively. Hence, ratio of their areas are equal to the ratio of their bases.

Let us assume that the area of the shaded region = area of the triangle ECD = x

AC = 4EC --> AE = 3EC --> Area of AED = 3*(Area of ECD) = 3x
AB = 3AD --> BD = 2AD --> Area of BDC = 2*(Area of ADC) = 2*(Area of AED + Area of ECD) = 2*(3x + x) = 2*4x = 8x

Hence, required ratio = x/(3x + 8x) = x/(11x) = 1/11

theCEO wrote:The area of the shaded area is 1/4 of the area of triangle ADC.....
What are the choices?

You can easily figure out that the figure is more or less drawn to scale. Now, does it look like the area of the shaded region is 1/4 of the unshaded area? Do not blindly follow mathematics.

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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 » Sun Jul 15, 2012 3:24 am

farukqmul wrote:

Use a COMMON TRIANGLE to satisfy the given ratios:

Triangle ABC is a 6-8-10 triangle that satisfies the two ratios given in the problem:
AC=8, EC=2, so AC = 4EC.
AB=6, AD=2, so AB = 3AD.

Area of triangle ABC = (1/2)(AC)(AB) = (1/2)(8*6) = 24.
Area of shaded triangle CDE = (1/2)(CE)(AD) = (1/2)(2*2) = 2.
Unshaded area = triangle ABC - shaded region = 24-2 = 22.
Shaded:unshaded = 2:22 = 1:11.

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farukqmul: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Fri Oct 14, 2011 12:46 pm; Location: Dhaka; GMAT Score:460

by farukqmul » Mon Jul 16, 2012 12:58 am

GMATGuruNY wrote:
farukqmul wrote:
Use a COMMON TRIANGLE to satisfy the given ratios:

Triangle ABC is a 6-8-10 triangle that satisfies the two ratios given in the problem:
AC=8, EC=2, so AC = 4EC.
AB=6, AD=2, so AB = 3AD.

Area of triangle ABC = (1/2)(AC)(AB) = (1/2)(8*6) = 24.
Area of shaded triangle CDE = (1/2)(CE)(AD) = (1/2)(2*2) = 2.
Unshaded area = triangle ABC - shaded region = 24-2 = 22.
Shaded:unshaded = 2:22 = 1:11.

The answer is correct but how can you say that it's a 6-8-10 triangle?Can you explain?

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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 Jul 16, 2012 7:42 am

farukqmul wrote:
GMATGuruNY wrote:
farukqmul wrote:
Use a COMMON TRIANGLE to satisfy the given ratios:

Triangle ABC is a 6-8-10 triangle that satisfies the two ratios given in the problem:
AC=8, EC=2, so AC = 4EC.
AB=6, AD=2, so AB = 3AD.

Area of triangle ABC = (1/2)(AC)(AB) = (1/2)(8*6) = 24.
Area of shaded triangle CDE = (1/2)(CE)(AD) = (1/2)(2*2) = 2.
Unshaded area = triangle ABC - shaded region = 24-2 = 22.
Shaded:unshaded = 2:22 = 1:11.
The answer is correct but how can you say that it's a 6-8-10 triangle?Can you explain?

When the answer choices are all fractions or ratios and the question stem does not specify any fixed values, ANY CASE that satisfies the conditions given in the problem will yield the SAME FRACTION OR RATIO.
The 6-8-10 triangle in my drawing above satisfies all of the conditions given in the problem.
Thus, the resulting ratio -- 1:11 -- must be the correct answer.

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imskpwr: Master | Next Rank: 500 Posts; Posts: 377; Joined: Wed Sep 14, 2011 10:45 amFollowed by: 1 member; Thanked: 10 times

by imskpwr » Tue Jul 17, 2012 2:51 am

farukqmul wrote:

What is Given?

relationship of two sides AB and AC with two other sides AD and AE respectively
This shows that ADE is kind of common triangle and we can find out area of triangle ADE in two ways:

1. considering AD as base.
2. considering AE as base.

AD = 1/3 of AB
AE = AC - EC = 3/4 of AC

area of ADE = 1/2 of AD x Height from E to AD.
= 1/2 x 1/3 x AB x Height from E to AD.
= 1/6 x AB x Height from E to AD........................(1)

area of ADE = 1/2 of AE x Height from D to AE.
= 1/2 x 3/4 x AC x Height from D to AE.
= 3/8 x AC x Height from D to AE........................(2)

Now Both these areas are equal b'cause they are for same triangle only.
Equating (1) and (2),

1/6 x AB x Height from E to AD = 3/8 x AC x Height from D to AE
1/3 x AB x Height from E to AD = 3/4 x AC x Height from D to AE......(3)

Now what is asked?
area of shaded region = 1/2 x EC x Height from D to AE
= 1/2 x 1/4 x AC x Height from D to AE...............(4)
area of complete Triangle = 1/2 x AB x Height from E to AD...........(5)

1-(4)/(5) will give the required ratio

On further simplification, ans will come out to be 1: 11

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