DIFFICULT GEOMETRY

[payam]

In the figure, AB = AE = 8, BC = CD = 13, and DE = 2. What is the area of region
ABCDE ?
A. 76
B. 84
C. 92
D. 100
E. 108

[Harish Dorai]

Please Note: It is better to have a paper and pencil and draw the figure while reading through the explanation. Also requesting the members of the forum or the instructors to find a better approach to find the area of the triangle BCD in the figure.

My approach was as follows:

Area of the figure = Area of Trapezoid BAED + Area of Triangle BCD

Area of BAED = 1/2 x AE x (AB + DE) - This is the formula for Trapezoid.

= 1/2 x 8 x (8 + 2)
= 1/2 x 8 x 10
= 40

Finding the area of Triangle BCD is a little tricky. BC = CD, means BCD is an isosceles triangle. We can find the area if you know the side BD. To find that draw a line from D to line segment AB and it should be perpendicular to AE and let us assume it intersects AB at F. Now it will form a right triangle BFD. Now BF = AB - AF which is AB - DE (AF will be equal to DE).
So BF = 6

So BFD has BF = 6 and FD = 8 which is equal to AE. So the side BD = 10 (It is a 6,8,10 right triangle).

Now we know 3 sides of triangle. There is a formula to find the area of the triangle, if you know the length of 3 sides.

Let us assume the length of 3 sides as "A", "B" and "C". Then area of triangle can be calculated as:

Area = Square Root of S x (S-A) x (S-B) x (S-C), where S = (A + B + C)/2.

So let us assume for triangle BCD, A = 13, B = 13 and C = 10. So the value of S = 18.

Now applying the formula Area = Square Root of 18 x (18 - 13) x (18 - 13) x (18 - 10)

= Square Root of 18 x 5 x 5 x 8

= 60

So the total area of the entire figure is 40 + 60 = 100

Answer is (D)

[Harish Dorai]

Okay. Here is an alternate way to find the area of the triangle, with the formula that every one knows, Area = 1/2 x Base x Height.

We can find the height of the triangle as follows. The base of the triangle is BD = 10 (Please read the previous reply for the calculation). Since we know the triangle is isosceles(BC = CD), if you draw a line from point C to the side BD and perpendicular to side BD, it will bisect the side BD (This property is true for an isosceles triangle and Equilateral triangle). Let us assume this point of bisection be G. So the segment CG will now be the height of the triangle. Now the segment BG or DG will be equal to 5 (Since the point G bisects BD and BD = 10). We know BC or CD = 13. So we can apply Pythagorus theorm on right triangle BGC or DGC to find the length of CG and it will be 12.

So area of the triangle = 1/2 x Base x Height = 1/2 x BD x CG = 1/2 x 10 x 12 = 60

So area of the figure = 40 + 60 = 100

[moshe.v.]

ok, here is my way:
the area of the rectangle is 64 + the area of the triangle 60 =124 minus the overlapping triangle (24) = 100 (d) [/b]

[rfernandez]

Great solutions, everyone.

Harish, that first solution featured a great theorem, called Heron's Formula (which gives the area of a triangle from its side lengths). Few people know it, but I've found it helpful (but not necessary!) from time to time on some GMAT problems.