Showing posts with label triangles share the same vertex. Show all posts
Showing posts with label triangles share the same vertex. Show all posts

Tuesday, November 10, 2015

Mathcounts : Geometry -- Medians ; Squares in Isosceles Right Triangle, Similar Triangles, Triangles share the same vertex

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This one appears at 91 Mathcounts National Target #4. It's interesting.

Q: triangle ABC is equilateral and G is the midpoint. BI is the same length as BC and the question asks about the area of quadrilateral polygon GDBC.
We'll also find the area of triangle ADG and triagle DBI here. 

Solution I: The side length is 2 and it's an equilateral triangle so the area of triangle ABC is 
\(\dfrac {\sqrt {3}} {4}\times 2^{2}=\sqrt {3}\).
Using similar triangles GCF and ACE, you get \(\overline {GF}\)= \(\dfrac {\sqrt {3}} {2}\)
From there, you get area of  \(\Delta\)GIC = \(4*\dfrac {\sqrt {3}} {2}*\dfrac {1} {2}=\sqrt {3}\)
 \(\Delta\)FCH is a 30-60-90 degree right triangle so \(\overline {DH}=\sqrt {3}*\overline {BH}\)
Using similar triangles IDH and IGF, you get \(\dfrac {2+\overline {BH}} {3.5}=\dfrac {\sqrt {3}\overline {BH}} {\dfrac {\sqrt {3}} {2}}\) = 2\(\overline {BH}\);
2 +\(\overline {BH}\)= 7\(\overline {BH}\); \(\overline {BH}\)= \(\dfrac {1} {3}\)
\(\overline {DH}=\dfrac {\sqrt {3}} {3}\) = the height
 Area of \(\Delta\)DBI = 2 * \(\dfrac {\sqrt {3}} {3}\)*\(\dfrac {1} {2}\) = \(\dfrac {\sqrt {3}} {3}\)
Area of quadrilateral polygon GDBC =  \(\sqrt {3}-\dfrac {\sqrt {3}} {3}\)= \(\dfrac {2\sqrt {3}} {3}\)
The area of both  \(\Delta\)ABC and \(\Delta\)GIC is \(\sqrt {3}\) and both share the quadrilateral polygon GDBC (It's like a Venn diagram).
Thus, the area of both triangles are the same \(\dfrac {\sqrt {3}} {3}\).

Solution II:



There are some harder AMC-10 questions using the same technique.
Using triangles share the same vertex,  you get the two same area "a"s and "b"s because the length of the base is the same.
From the area we've found at solution I, we have two equations:
2a + b = \(\sqrt {3}\)
2b + a = \(\sqrt {3}\)
It's obvious a = b so you can find the area of the quadrilateral being the \(\dfrac {2} {3}\)of the equilateral triangle so the answer is \(\dfrac {2\sqrt {3}} {3}\).




Vinjai's solution III:

Draw the two extra lines and you can see that the three lines are medians, which break the largest triangle into 6 equal parts. [You can proof this using triangles share the same vertices : notes later.]
Quadrilateral polygon FECB is  \(\dfrac {2} {3}\)of the equilateral triangle so the answer is \(\dfrac {2\sqrt {3}} {3}\).







This is similar to 2005 Mathcounts National Target #4 questions. AMC might have similar ones.
Q: Both triangles are congruent and isosceles right triangles and each has a square embedded in it. If the area of the square on the left is 12 square unites, what is the area of the square on the right? 
Solution I: 


Solution II:

Tuesday, September 23, 2014

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. 







 

Sunday, November 25, 2012

Trianges That Share the Same Vertex/Similar Triangles

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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 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.]