Sunday, November 25, 2012

Trianges That Share the Same Vertex/Similar Triangles

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Besides similar triangles, "triangles that share the same vertex" also appears regularly on geometry questions. 

Look , for example, at the left image. 
It's easy to see AD being the height to base CE for Δ AEC.

However, it's much harder to see the same AD being the external height of  Δ ABC if BC is the base.

Question to ponder based on the above image:
#1: If  BC : CE = 2 to 5 and the area of Δ ABE is 98, what is the area of Δ ABC and Δ ACE?

Since both triangles share the same height AD, if you use BC and CE  as the bases, the area ratio stays constant as 2 to 5.  Thus the area of Δ ABC = (2/7) * 98 = 28 and the area of
Δ AEC = (5/7) *98 = 70.

Knowing the above concept would help you solve the ostensibly hard trapezoid question.

Question #2: If ABCD is a trapezoid where AB is parallel to CDAB = 12 units and CD = 18 units.
If the area of  Δ AEB = 60 square units, what is the area of ΔCED, Δ AED and ΔBEC?

Solution:  Δ AEB and ΔCED are similar (Why? Make sure you understand this?) Thus if AB to CD = 12 to 18 or 2 to 3 ratio (given), the ratio of the area of the two similar triangles is 22 to 32 = 4 to 9 ratio.
The area of  Δ AEB = 60 square units (given), so the area of ΔCED is 9 * (60/4) = 135 square units
Δ AEB and Δ AED share the same vertex. The height is the same with base BE and ED.
Thus the area ratio is still 2 to 3. The area of Δ AEB = 60 square units (given),
 the area of Δ AED = 3 * (60/2) = 90 square units.

Using the same reasoning you get the area of ΔBEC = 90 square units. [Keep in mind that Δ AED and ΔBEC don't have to be congruent, but they do have the same area.]