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?

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

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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.

2007 State Harder Algebra Questions

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#16 2007 Mathcounts State Sprint: A data set for a class of 25 sixth graders has their ages listed as
the integer values of either 10 or 11 years. The median age in the data set is 0.36 years greater than the mean. How many 10-year-olds are in the class?

#23: 2007 Mathcounts State Sprint:  A box contains some green marbles and exactly four red marbles.
The probability of selecting a red marble is x%. If the number of green marbles is doubled, the probability of selecting one of the four red marbles from the box is (x - 15)%. How many green marbles are in the box before the number of green marbles is doubled?












#16: Solution I: 
 Since there are 25 students, the median has to be either 10 or 11 (Why?)
Since the mean has to be 10 (if all the children are all 10 year old) or higher, the median has to be 11.
Let there be x 10 year old students, and there are (25 - x) 11 year old students.

both sides * ----

 

 

Solution II: 

Let there be x 10 year old and y 11 year old. 

According to the given: 

10x + 11 y = 266
For x to be integer, y has to either 6 or 16. Further checking, only when y = 16, x = 9 works. 

#23: Let there be g numbers of Green marbles.
According to the given, we can set up two equations:

                                          equation 1

           equation 2

Equation 1 minus equation 2 and you have: 0 = xg + 60 -2xg + 30g
Move all the variables to the other side:
xg - 30g = g (x - 30) = 60
If g = 4, x = 45 (doesn't work)
If g = 6, x = 40 (it works.)

Mental Math Practices on Ratio, Percents and Proportion

Mental Math Practices: All questions can be solved in your head reasonably fast (less than 5 minutes) if you understand the concepts well.

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1. What is 12.5% of 128?

2. After 20% discount, you paid $144? What is the original price?

3. To make 20% profit, you charge your customer 144 dollars. What is the wholesale price?

4. You lost 20% in the stock market. What percentage increase will you break even?

5. A : B = 2 : 5 and A is 75 more than B, what is B?

6. A : B = 2 : 3 and A + B = 14, what is A and what is B?

7. A : B = 3 : 5 and B : C = 2 to 9, what is A : B?

8. Two triangles are similar and the length of their base ratio is 1: 2. The smaller triangle has a base that equals 5 and an area that is 110. What is the area of the larger triangle?

9. One side of the square increases 10% and the other side decreases 20 percent. What is the percentage change? Increase or decrease? By how much?

10. There are 60 students in a gym and 30% are boys. After a few more boys enter the gym, now 40% are boys. How many boys enter the room?

Beginning Algebra: II

Learn the basics here.

Here we are going to review and work on some other harder problems.

Question #1:   Tickets for the homecoming dance cost $20 for a single ticket or $35 for a couple. Ticket sales totaled $2280, and 128 people attended. How many tickets of each type were sold?

Solution: 
Let there be x tickets sold for singles and y tickets sold for couples. According to the given:
20 x + 35 y = 2280 --- equation 1
x  + 2 y = 128 --- equation 2

equation 2 times 20 - equation 1 and you have 20 x + 40 y - (20 x + 35 y) = 20 x 128 - 2280 = 280
5y = 280 so y = 56 and plug in to equation 2 and you get x = 16

There will be 16 tickets sold for singles and 56 tickets sold for couples.
Always check your answer to see if that is right.

Question #2: There are 20 questions on an algebra test, you got 5 points for each question you answer correctly and - 2 points for each questions you answer wrong. 
If a student gets 79 for his score and he answers all the questions, how many questions does he answer right? 

Solution I: 
If the student gets all the questions right, he'll get 5 x 20 = 100 points; however, he only gets 79 points. 
For each questions he gets wrong, he not only doesn't have the 5 points but on top of that, he gets 2 points deducted, thus it's minus 7 total. 

(100 - 79) / 7 = 3 so the student gets 3 wrong and 17 questions right. 
Check if that's correct: 17 x 5 - 2 x 3 = 79

Solution II:  
Let the student get x questions right, and that means he gets (20 - x) questions wrong.
5x - 2 (20 -x) = 79;    5x - 40 + 2x = 79;   7 x = 119 so x = 17

Question #3: My piggy bank contains only nickels, dimes, and quarters, and contains 20 coins worth a total of $3.30. If the total value of the quarters is five times the total value of all the other coins, how many dimes are in the piggy bank?

Solution:  
According to the given: 
N + D + Q = 20  --- equation 1
The total value of the quarters is 5 times the total value of all the other coins, which mean that if the total value of N + D = x, the total value of Q is 5 x
x + 5x = 330 (Make sure to use the same unit) so x = 55 (cents)
5x = 5 x 55 = 275 and 275/ 25 = 11 so there are 11 quarters

Plug in equation 1 and you have N + D = 20 - 11 = 9
Since 5N + 10D = x = 55, D and N could be
D        N
5         1
4         3
3         5
2         7... etc   2 + 7 = 9 so there are 2 dimes (7 nickels and 11 quarters)


Other applicable problems: (answer key below)
#1: Bernard has 41 coins consisting of nickels, dimes, and quarters, and they are worth a total of $4.00. If the number of dimes plus the number of quarters is one more than te number of nickels, then how many of each does he have?

#2:  Kim has $1.65 in nickels, dimes and quarters. She has 10 coins all together and the number of quarters is equal to the number of nickels and dimes combined. Haw many of each coin does she have?

#3:  A collection of 24 nickels, dimes, and quarters is worth $3.20. There are seven more nickels than dimes. How many of each are there in the collection?

#4: Continue with #3, if there are seven more dimes than nickels, how many of each are there in the collection? 











Answer key: 
#1: 6 quarters, 20 nickels and 15 dimes  
#2: 2 nickels, 3 dimes and 5 quarters 
#3: 9 quarters, 4 dimes and 11 nickels
#4: 7 quarters, 12 dimes and 5 nickels