Triangle problem

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

What are the choices?

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

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.

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?

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.

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