Similar triangles, Trapezoids and Triangles that Share the Same Vertices

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This is an interesting question that requires understanding of dimensional changes. (They are everywhere.)

Question: If D and E are midpoints of AC and AB respectively and the area of ΔBFC = 20, what is the 
a. area of Δ DFB? 
b. area of Δ EFC?
c. area of Δ DFE? 
d. area of Δ ADE? 
e. area of trapezoid DECB? 
f. area of Δ ABC?








Solution:

DE is half the length of BC (D and E are midpoints so DE : BC =  AD  : AB = 1 : 2

Δ DFE and ΔCFB are similar and their area ratio is 1² : 2²  = 1 : 4  (If you are not sure about this part, read this link on similar triangles.)

so the area of Δ DFE = (1/4) of ΔBFC = 20 = 5 square units. 

The area of Δ DFB = the area of Δ EFC = 5 x 2 = 10 square units because Δ DFE and Δ DFB,   
Δ DFE and ΔEFC share the same vertexs D and E respectively, so the heights are the same. 
Thus the area ratio is still 1 to 2. 

Δ ADE and ΔDEF share the same base and their height ratio is 3 to 1, so the area of
Δ ADE is 5 x 3 = 15 square units.

[DE break the height into two equal length and the height ratio of Δ DFE and ΔCFB is 1 to 2 (due to similar triangles) so the height ratio of Δ ADE and ΔDEF is 3 to 1.]

The area of trapezoid DEBC is 45 square units.

The area of Δ ABC is 60 square units. 


Extra problems to practice (answer below): 

The ratio of   AD and AB is 2 to 3,  DE//BC and the area of Δ BFC is 126, what is the area of

a. Δ DFE ? 

b. Δ DFB ?

c.  Δ EFC ? 

d.  Δ ADE? 

e. How many multiples is it of Δ ABC to ΔBFC?










Answer key: 
a. 56 square units
b. 84 square units
c. 84 square units
d. 280 square units
e. 5 times multiples. 







 

Find the area of the petal, or the football shape.

Find the area of the football shape, or the petal shape.
The below Mathcounts mini presents two methods.

Circle and area revisited from Mathcounts mini

The first question is exactly the same as this one.
Besides the two methods on the videos, you can also use the following methods.

Solution III:
You can also look at this as a Venn Diagram question.
One quarter circle is A and the other is B, and both are congruent. (center at opposite corner vertexes)

The overlapping part is C.

A + B - C = 6^2 so C = A + B - 36 or 18pi - 36





                                                                                               

Solution IV:

If you use the area of the rectangle,
which is 6 x 12 minus, the area of the half circle with a radius 6, you get the two white spots that are un-shaded.

Use the area of the square minus that will again give you the answer.
\(6^{2}-\left( 6*12-\dfrac {6^{2}\pi } {2}\right)\)
= 18pi - 36




Similar triangles and triangles that share the same vertexes/or/and trapezoid

Another link from my blog

Similar triangles, dimensional change questions are all over the place so make sure you really
understand them.

Take care and happy problem solving !!

Trianges That Share the Same Vertex/Similar Triangles

Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here.

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?

Solution:
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 CD,  AB = 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.]