Geometry

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cartera: Senior | Next Rank: 100 Posts; Posts: 59; Joined: Tue Jun 10, 2008 6:49 pm; Thanked: 2 times

by cartera » Thu Jan 29, 2009 6:37 am

In the above rectangle ABCD where AD is the upper line of the rectangle and BC is the lower line of the rectangle , if the area is 84 and the distance between BF and FC 4:3, what is the size of the triangle?

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Source: Beat The GMAT — Problem Solving | Thu Jan 29, 2009 6:37 am

DanaJ: Site Admin; Posts: 2567; Joined: Thu Jan 01, 2009 10:05 amFollowed by: 550 members; Thanked: 712 times; GMAT Score:770

by DanaJ » Thu Jan 29, 2009 7:39 am

Let's clear all the points over there:
- A is the upper left point of the rectangle
- D is the upper right point of the rectangle
- C is the lower right point of the rectangle
Well, first of all we should notice that if we were to draw the height of triangle BEF stemming from E and dropping into BC, say we call it EG, then EG = AB = CD.
We also know that BC*AB = 84.
Now let's analyse what we know about BF:
1. BF/FC = 4/3, so FC = BF*3/4
2. BF + FC = BC, so by substituting the above, we get that BF*7/4 = BC.

Now, we've established that BC = BF*7/4 and that AB = EG. Since BC*AB = 84, this means that BF*7/4*EG = 84, which in turn leads us to BF*EG*1/2 = 24. Since EG is the height corresponding to the side BF of the BEF triangle, then this means that this will be the area of BEF.

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Thu Jan 29, 2009 12:35 pm

Dana's approach is, as always, great! Let's look at the question from a different angle and see how we can solve with less math.

Picture a rectangle. Let's draw a triangle with the entire side of the rectangle as it's base and put the vertex of the triangle on the opposite side of the rectangle.

If the dimensions of the rectangle are l and w, then the base of the triangle is l and the height is w. Therefore, the area of the triangle is (l*w)/2, or half the area of the rectangle. This rule is always true for triangles constructed within rectangles as described.

In the question posted above, the height of the triangle is still w, but the base isn't the entire length of the rectangle, it's only 4/7 of the length (since BF:FC is 4:3, we know that BF is 4 out of 7 parts of the length).

Therefore, instead of the triangle being 1/2 the area of the rectangle, it will be:

(4/7) * (half the area of the rectangle) = (4/7)*(84/2) = 168/7 = 24

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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