Please give me feedback if there is any error or what concepts to be included for future practice tests. Thanks a lot in advance.
Online practices on finding the area of a triangle or irregular polygons.
Click here for the timed online test. Type in your first name and only write down "number" answers.
Before trying out the online timed test, here are some reminders.
a. If the irregular polygon include the origin, then using that as a starting point for shoestring method might save some calculation time.
b. There are Pythagorean Triples hidden in lots of questions that ask for the area of a triangle given three side lengths. Thus, if you find the side length that matches the triples or multiples of the triples, it might be much easier to use that instead of heron's formula.
c. For quadrilaterals, if the diagonals are perpendicular to each other, such as rhombus, kite or others, it's much easier to use D1* D2 / 2 to get the area.
d. Sometimes breaking the polygon into different smaller parts is the easiest way to find the answer.
e. Deliberate practices are the key to steady progress.
Have fun problem solving!!
Tuesday, December 11, 2012
Monday, December 10, 2012
Line Intercept, Slope, Area/Geometry Questions
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Download this year's Mathcounts handbook here.
This question is similar to 2010 #3 Mathcouts Chapter Team problem.
Solution I: Using the given slope, you can find the equation for AB ,
which is: y = -2 x + b, plug in (4, 7) and you have 7 = -2 x 4 + b so b = 15
The y intercept for AB is (0, 15)
Now you have four points, (0, 0), (0, 15), (4, 7) and (15, 0).
Using shoestring method to find area of any polygon, you have the area of polygon
AEDO = 82.5 square units.
Solution II:
After finding the y intercept for AB.
You break the area into three parts.
The side of the shaded region is 4 x 7 = 28.
The area of triangle ACE = 4 * (15-7) / 2 = 16
The area of triangle
EBD = (external height)7 * (15-4) / 2 = 38.5
So the total area is 82.5 square units.
Applicable problems: Answer key below.
#1: The slope of AB is -3 and it intercepts with CD at point E, which is (3, 3). If the x intercepts of CD is (10,0), what is the area of AEDO?
#2: The slope of AB is 2 and it intercepts with CD at point E, which is (-2, 2 ). If the x intercepts of CD is (-7,0), what is the area of AEDO?
#3: The slope of AB is -3 and it intercepts with CD at point E, which is ( 5, 8 ). If the x intercepts of CD is (14,0), what is the area of AEDO? -- Andrew's question.
#4: The slope of AB is \(\dfrac {-3} {2}\)and it intercepts with CD at point E, which is ( 2, 7 ). If the x intercepts of CD is (16,0), what is the area of AEDO? --Daniel's question.
#5: The slope of AB is -4/5 and it intercepts with CD at point E, which is (5, 2 ). If the x intercepts of CD is (12,0), what is the area of AEDO?
Answer key:
#1: 33
#2: 13
#3: 113.5
#4: 66
#5: 27
This question is similar to 2010 #3 Mathcouts Chapter Team problem.
Solution I: Using the given slope, you can find the equation for AB ,
which is: y = -2 x + b, plug in (4, 7) and you have 7 = -2 x 4 + b so b = 15
The y intercept for AB is (0, 15)
Now you have four points, (0, 0), (0, 15), (4, 7) and (15, 0).
Using shoestring method to find area of any polygon, you have the area of polygon
AEDO = 82.5 square units.
Solution II:
After finding the y intercept for AB.
You break the area into three parts.
The side of the shaded region is 4 x 7 = 28.
The area of triangle ACE = 4 * (15-7) / 2 = 16
The area of triangle
EBD = (external height)7 * (15-4) / 2 = 38.5
So the total area is 82.5 square units.
Applicable problems: Answer key below.
#1: The slope of AB is -3 and it intercepts with CD at point E, which is (3, 3). If the x intercepts of CD is (10,0), what is the area of AEDO?
#2: The slope of AB is 2 and it intercepts with CD at point E, which is (-2, 2 ). If the x intercepts of CD is (-7,0), what is the area of AEDO?
#3: The slope of AB is -3 and it intercepts with CD at point E, which is ( 5, 8 ). If the x intercepts of CD is (14,0), what is the area of AEDO? -- Andrew's question.
#4: The slope of AB is \(\dfrac {-3} {2}\)and it intercepts with CD at point E, which is ( 2, 7 ). If the x intercepts of CD is (16,0), what is the area of AEDO? --Daniel's question.
#5: The slope of AB is -4/5 and it intercepts with CD at point E, which is (5, 2 ). If the x intercepts of CD is (12,0), what is the area of AEDO?
Answer key:
#1: 33
#2: 13
#3: 113.5
#4: 66
#5: 27
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 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.]
Download this year's Mathcounts handbook here.
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.]
Thursday, November 22, 2012
Similar Triangles I
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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.
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.
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.
Sunday, November 18, 2012
2007 State Harder Algebra Questions
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Download this year's Mathcounts handbook here.
#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:
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.)
Download this year's Mathcounts handbook here.
#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 2Equation 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.)
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