(E) Let's break this problem down into steps. 1) To find the area of the quadrilateral ABCE, we are going to subtract the area of triangle AEF from the area of rectangle ABCF. Let's figure out the length of all the sides of triangle AEF and rectangle ABCF.

2) Angle B = 90 degrees. Because BD and AB are 8 and 6, respectively, triangle ABD is a 3-4-5 triangle. AD = 10.

3) If ED = 5 and AD = 10, then AE = 10 – 5, or 5.

4) As alternate angles, angle AEF = angle CED, so triangle AEF must be similar to triangle CED. Because the two triangles are similar and have the same length hypotenuse, the two triangles must be identical.

5) Since AB = EF + EC = 6, and since EF = EC, 2EF = 6. Thus, EF and EC = 3.

6) AF = BC, BC = CD, and BC + CD = BD = 8, so AF, BC, and CD all equal 4.

7) The rectangle ABCF would then have an area = 6 × 4 = 24. The area of a triangle is (base × height)/2. Triangle AEF, therefore, has an area of (4 × 3)/2 = 6.

8) The area of ABCE is 24 – 6 = 18.

The correct answer is choice (E). ---------

I don't get how you guys get ED = 5. If AD=10 and AD = ED + AE that's right but we don't know whether the point E is the mid point of the line AD.

Post subject: Re: math, t.1, qt.6: geometry, quadrilateral, triangles

Posted: Sat Jul 07, 2012 4:05 am

Joined: Tue Apr 13, 2010 8:48 am Posts: 478

Answer should be C) 14, as the sides of rectangle are 6,6,4,4 that add up to an area of 20. Deducting the area of the triangle, 6, it gives the result 14.

Post subject: Re: math, t.1, qt.6: geometry, quadrilateral, triangles

Posted: Sat Jul 07, 2012 4:08 am

Joined: Thu Jul 05, 2012 11:55 am Posts: 64

Yes, usually there're lots of possible solutions to geometry problems. This one is not an exception. Any way is good if it's convenient to you and solves the question.

One of the possible solutions is to first calculate AD, as we do in the original explanation. Then we notice that triangles ECD and ABD are similar. Since ED : AD = 5 : 10 = 1 : 2, then we know that: DC : DB = 1 : 2 and EC : AB = 1 : 2. Therefore CD = 4, so BC = 8 – 4 = 4. EC = (1/2) × 6 = 3.

At the end we can either use the formula for trapezoid AECB: The area = (AB + EC) × BC / 2 = (6 + 3) × 4 / 2 = 18.

Or we can finish it as in the original solution: FE = 6 – 3 = 3. The area of AECB = the area of ABCF – the area of AFE The area of AECB = 6 × 4 – 4 × 3 / 2 = 24 – 6 = 18.

Once again, I'd like to notice that any solution that is convenient to you and solves the problem in a reasonable amount of time is good. The convenience of using one or another geometrical property is very individual.

Post subject: Re: math, t.1, qt.6: geometry, quadrilateral, triangles

Posted: Sat Jul 07, 2012 5:13 am

Joined: Thu Jul 05, 2012 11:55 am Posts: 64

questioner wrote:

The area of ABD = (1/2) × 8 × 6 = 24

This is correct.

questioner wrote:

The area of ECD = (1/2) × 3 (EC is half of AB) × 4 = 6

How do you know that "EC is half of AB"? In order to know this, you have to do step 2: Because BD and AB are 8 and 6, respectively, triangle ABD is a 3-4-5 triangle. AD = 10. Then you can state that the triangles EDC and ADB are similar, because they are both right and the angle D is the same. Using their similarity you can say that ED is half of AD, so EC is half of AB and CD is half of BD.

If you put this solution into steps, as the solution in the explanation, then you will have 7-8 steps. It is a good way to solve, as also the way in the explanation. Somebody likes similar triangles more, somebody likes equal triangles. It's just a matter of convenience.

Concept used --> The most common Pythagorean triangles are (3, 4, 5) and (6, 8, 10).

Since we know 6 and 8 are the lengths of the biggest triangle. We can say that the 3rd side is 10 in length. We are given that ED = 5; Therefore, AE = 5. Triangles AFE and DCE are right angled; Therefore they are (3, 4, 5) in length. Since BD = 8, therefore BC = 4. Area of the quadrilateral is (1/2) (3 + 6) × 4 = 18

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