Similar Triangles: Team question : Beginning level

9. In the figure below, quadrilateral CDEG is a square with CD = 3, and quadrilateral BEFH is a rectangle. If EB = 5, how many units is BH? Express your answer as a mixed number

Triangle BED is a 3-4-5 right triangle and is similar to triangle GEF.

BE : ED = GE : EF = 5 : 3 = 3 : FE

EF = 9/5 = BH  The answer

Dimensional Change

There are lots of questions on dimensional change and this is a very common one.

Make sure you understand the relationship among linear, 2-D (area) and 3-D (volume) ratio.

There are many similar triangles featured in the image on the left.
Each of the two legs of the largest triangles is split into 4 equal side lengths.





                                                                                            


Question : What is the area ratio of the sum of the two white trapezoids to the largest triangle? 
\(\dfrac {\left( 3+7\right) } {16}=\dfrac {10} {16}=\dfrac{5}{8}\)  

Question: If the area of the largest triangles are 400 square units, what is the area of the blue-colored trapezoid?
\(\dfrac {5} {16}\times 400\) =125 square units 

Again, each of the two legs are split into three equal segments. 

The volume ration of the cone on the top to the middle frustum to the 
bottom frustum is 1 : 7 : 19. 
 
Make sure you understand why.










 

Geometric mean of a right triangle

Geometric mean of a right triangle :

Video tutorial 

Notes

Practice I 

Practice II : focus on the height (or altitude)

Practice III : focus on the legs 

2013 Mathcounts Natinals Sprint # 28

2013 Mathcounts National Sprint #28 : In right triangle ABC, shown here,  line AC = 5 units and line BC = 12 units. Points D and E lie on  line AB and line BC respectively, so that line CD is perpendicular to line AB and E is the midpoint of line BC. Segments AE and CD intercept at point F. What is the ratio of AF to FE ? Express your answer as a common fraction.

                                                     Solution I : Using similar triangles

                                                        Solution II : Use Mass Point Geometry

Dimentional Change Questions III: Similar Shapes

There are numerous similar triangle questions on Mathcounts.

Here are the basics:

If two triangles are similar, their corresponding angles are congruent and their corresponding sides will have the same ratio or proportion.

Δ ABC and ΔDEF are similar. \(\frac{AB}{DE}\) = \(\frac{AC}{DF}\) = \(\frac{BC}{EF}\)= their height ratio = their perimeter ratio.







Once you know the linear ratio, you can just square the linear ratio to get the area ratio and cube the linear ratio to get the volume ratio. 

Practice Similarity of Triangles here.  Read the notes as well as work on the practice problems.  There is instant feedback online. 

Other practice sheets on Similar Triangles                                                        

Many students have trouble solving this problem when the two similar triangles are superimposed. 

Just make sure you are comparing smaller triangular base with larger triangular base and smaller triangular side with corresponding larger triangular side, etc... In this case:

\(\frac{BC}{DE}\)= \(\frac{AB}{AD}\) = \(\frac{AC}{AE}\)




Questions to ponder (Solutions below)

#1: Find the area ratio of Δ ABC to trapezoid BCDE to DEGF to FGIH. You can easily get those ratios using similar triangle properties. All the points are equally spaced and line \(\overline{BC}\)// \(\overline{DE}\) // \(\overline{FG}\) // \(\overline{HI}\). 

#2: Find the volume of the cone ABC to Frustum BCDE to DEGF to FGIH. Again, you can use the similar cone, dimensional change property to easily get those ratios.Same conditions as the previous question.

 

Answer key: 

#1:

 #2:

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

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

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:

2015 Mathcounts State Prep : Inscribed Cricle Radius and Similar Triangles

Question : \(\Delta\) ABC is an equilateral triangle. Circle "O" is the inscribed circle and it's radius is 15. 

What is the length of the radius of the smaller circle p which is tangent to circle "O" and the two sides?








Here is the link to the basics of inscribed circle radius as well as circumscribed circle radius of an equilateral triangle.

Solution I :
The length of the radius of an inscribed circle of an equilateral triangle is \(\dfrac {1} {3}\) of the height so you know AO is \(\dfrac {2} {3}\) of the height or 30 (the height is 15 + 30 = 45 unit long)

\(\Delta\) AEP is similar to \(\Delta\) AFO \(\rightarrow\) \(\dfrac {r} {15}=\dfrac {AP} {30}\)
so \(\overline {AP}\) = 2r.
\(\overline {AP}+\overline {PO}=30\)  \(\rightarrow\)2r + r + 15 = 30   \(\rightarrow\)  3r = 15 so r = 5 
or \(\dfrac {1} {3}\) of the larger radius 

Solution II:
\(\Delta\) APE is a 30-60-90 right triangle, so \(\overline {AP}\) = 2r
\(\overline {PO}\) = r + 15
\(\overline {AP}+\overline {PO}\)  \(\rightarrow\)  2r + r + 15 = 30  \(\rightarrow\)  3r  =  15 so  r  = 5 
or \(\dfrac {1} {3}\) of the larger radius 

This is an AMC-10 question.

\(\Delta\) ABC is an isosceles triangle.

The radius of the smaller circle is 1 and the radius of the larger circle is 2,

A: what is the length of \(\overline {AP}\) ?

B. what is the area of \(\Delta\) ABC?







Solution for question A: 
\(\Delta\) AEP is similar to \(\Delta\) AFO \(\rightarrow\) \(\dfrac {1} {2}=\dfrac {AP} {AP +3}\)
 2\(\overline {AP}\) =  \(\overline {AP}\) + 3 \(\rightarrow\) AP = 3

Using Pythagorean theorem, you can get \(\overline {AE}\) = \(2\sqrt {2}\)
\(\Delta\) AEP is similar to \(\Delta\) ADC [This part is tricky. Make sure you see that !!]
 \(\rightarrow\) \(\dfrac {1} {\overline {DC}}=\dfrac {AE} {AD}\) = \(\dfrac {2\sqrt {2}} {8}\)
\(\overline {DC}\) = \(2\sqrt {2}\)  and \(\overline {BC}\) = 2 * \(2\sqrt {2}\) = \(4\sqrt {2}\)
The area of  \(\Delta\) ABC =  \(\dfrac{1}{2}\)*\(4\sqrt {2}\) * 8 =  \(16\sqrt {2}\)

Question: If you know the length of x and y, and the whole length of \(\overline {AB}\),

A: what is the ratio of a to b and 

B: what is the length of z.







Solution for question A:
\(\Delta\)ABC and \(\Delta\)AFE are similar so \(\dfrac {z} {x}=\dfrac {b} {a+b}\). -- equation 1
Cross multiply and you have z ( a + b ) = bx

\(\Delta\)BAD and \(\Delta\)BFE are similar so \(\dfrac {z} {y}=\dfrac {a} {a+b}\). -- equation 2
 Cross multiply and you have z ( a + b ) = ay

bx = ay so \(\dfrac {x} {y}=\dfrac {a} {b}\)  same ratio

Solution for question B: 
Continue with the previous two equations, if you add equation 1 and equation 2, you have:
\(\dfrac {z} {x}+\dfrac {z} {y}=\dfrac {b} {a+b}+\dfrac {a} {a+b}\)
\(\dfrac {zy+zx } {xy}=1\) \(\rightarrow\) z = \(\dfrac {xy} {x+y}\)

Applicable question: 

\(\overline {CD}=15\) and you know \(\overline {DB}:\overline {BC}=20:30=2:3\) 
 so  \(\overline {DB}=6\) and \(\overline {BC}=9\) 

\(\overline {AB}=\dfrac {20\times 30} {\left( 20+30\right) }\) = 12

Similar triangles, Trapezoids and Triangles that Share the Same Vertices

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

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. 







 

This Week's Work : Week 43 - for Inquisitive Young Mathletes

From Mathcounts Mini : Using Similarity to Solve Geometry Problems

Also try the warm-up, retry the problems in the video on your own and spend some time
pondering on those follow-up, harder problems.

#4, 5, 6 are state level problems. #7 is challenging.

From Mathcounts Mini : Tangent Segments and Similar Triangles

#8 is state/national level.

Have fun problem solving !! Mrs. Lin

Three Pole Problems : Similar triangles

Question: If you know the length of x and y, and the whole length of \(\overline {AB}\),

A: what is the ratio of a to b and 

B: what is the length of z.






 Solution for question A:
\(\Delta\)ABC and \(\Delta\)AFE are similar so \(\dfrac {z} {x}=\dfrac {b} {a+b}\). -- equation 1
Cross multiply and you have z ( a + b ) = bx

\(\Delta\)BAD and \(\Delta\)BFE are similar so \(\dfrac {z} {y}=\dfrac {a} {a+b}\). -- equation 2
 Cross multiply and you have z ( a + b ) = ay

bx = ay so \(\dfrac {x} {y}=\dfrac {a} {b}\)  same ratio

Solution for question B: 
Continue with the previous two equations, if you add equation 1 and equation 2, you have:
\(\dfrac {z} {x}+\dfrac {z} {y}=\dfrac {b} {a+b}+\dfrac {a} {a+b}\)
\(\dfrac {zy+zx } {xy}=1\) \(\rightarrow\) z = \(\dfrac {xy} {x+y}\)

Applicable question: 

\(\overline {CD}=15\) and you know \(\overline {DB}:\overline {BC}=20:30=2:3\) 
 so  \(\overline {DB}=6\) and \(\overline {BC}=9\) 

\(\overline {AB}=\dfrac {20\times 30} {\left( 20+30\right) }\) = 12

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 !!

This Week's Work : Week 9 -- for Inquisitive Young Mathletes

Part 1:
See below for links:
They are all related to dimensional change and similar polygons.

Dimensional change questions I 

Dimensional change questions II 
 
Dimensional change questions III : Similar Triangles 

Par II:
Tangent Segments and Similar Triangles from Mathcounts Mini 

If you have more time, download the extra word problems to see if you can solve them at
reasonable speed and accuracy.

Online timed test and problem of the week will be sent out through e-mail.
Time: 40  minutes without a calculator.

The Monty Hall Problem explained

2013 Mathcounts State Prep: Similar Triangles and Height to the Hypotenuse

There are many concepts you can learn from this image, which cover numerous similar right triangles, ratio/proportion/dimensional change and the height to the hypotenuse.

Question:
#1: Δ ABC is a 3-4-5 right triangle. What is the height to the hypotenuse? 
Solution: 
Use the area of a triangle to get the height to the hypotenuse. 
Let the height to the hypotenuse be "h"
The area of Δ ABC is  \(\Large\frac{3*4}{2}\)= \(\Large\frac{5*h}{2}\)
Both sides times 2 and consolidate: h =  \(\Large\frac{3*4}{5}\) = \(\Large\frac{12}{5}\)

Practice: What is the height to the hypotenuse?

Question:
#2: How many similar triangles can you spot?
Solution: 
There are 4 and most students have difficulty comparing the largest one with the other smaller ones.
Δ ABC is similar to Δ ADE, Δ FBD, ΔGEC. Make sure you really understand this and can apply this to numerous similar triangle questions. 

Question: 
#3: What is the area of  □ DEGF if \(\overline{BF}\) = 9 and \(\overline{GC}\) = 4
Solution: 
Using the two similar triangles Δ FBD and  ΔGEC (I found using symbols to find the corresponding legs
 to be much easier than using the lines.), you have \(\frac{\Large{\overline{BF}}}{\Large{\overline{FD}}}\) = \(\frac{\Large{\overline{GE}}}{\Large{\overline{GC}}}\).
s (side length of the square) = \({\overline{GE}}\) =  \({\overline{FD}}\)
Plug in the given and you have 9 * 4 = s² so the area of □ DEGF is 36 square units. (each side then is square root of 36 or 6)

Question: 
#4: Δ ABC is a 9-12-15 right triangle. What is the side length of the square? 
Solution :
The height to the hypotenuse is\(\frac{\Large{9*12}}{\Large{15}}\) = \(\frac{\Large{36}}{\Large{5}}\)
Δ ABC is similar to Δ ADE. Using base and height similarities, you have \(\frac{\Large{\overline{BC}}}{\Large{\overline{DE}}}\) = \(\frac{\Large{15}}{\Large{S}}\) = \(\frac{\frac{\Large{36}}{\Large{5}}}{\frac{\Large{36}}{\Large{5}} - \Large{S}}\)
Cross multiply and you have 108 - 15*S = \(\frac{\Large{36}}{\Large{5}}\) *S
S =\(\frac{\Large{180}}{\Large{37}}\)

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.]

Similar Triangles I

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

There are numerous questions on similar triangles at competition math.
Practice how to spot them and use them to solve tougher chapter, median hard state questions.

For example, There are three similar triangles in this image.

Triangle ABC is similar to triangle BDC and triangle ADB.

Using consistent symbols will help you set up the right proportion comparisons much easier.

BD² = AD x DC
BA² = AD x AC
BC² = CD x AC  

All because of the similarities.
That is also where those ratios work.











#1: What is the height to the hypotenuse of a 3-4-5 right triangle? 
Solution: 
 The area of a triangle is base x height over 2.
 Area of a 3-4-5 right triangle  = (3 x 4) / 2 = (5 x height to the hypotenuse) / 2
 Both sides times 2 : (3 x 4) = 5 x height to the hypotenuse
 The height to the hypotenuse = 12/5 

The same goes with 6-8-10 right triangle, the height to the hypotenuse is 4.8.
It's very straightforward so make sure you know how and why that works.

Hope this helps.