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 

The Largest Rectangle Inscribed in Any Triangle

From Mathcounts Mini : Maximum area of inscribed rectangles and triangles

\(\Delta EHI\sim\Delta EFG\) \(\rightarrow\) \(\dfrac {a} {c}=\dfrac {d-b} {d}\)\(\rightarrow\) \(a=\dfrac {c\left( d-b\right) } {d}=\dfrac {-c\left( b-d\right) } {d}\)

We are going to find out what the largest area of a rectangle is with the side length a and b.
It can be shown that by substituting the side length "a" with the previous equation + completing the square that the largest area is half of the area of the triangle the rectangle is embedded.

\(a\times b=\dfrac {-c\left( b-d\right) \times b} {d}=\dfrac {-c\left( b^{2}-bd\right)} {d}= \dfrac {-c\left( b-\dfrac {1} {2}d\right) ^{2}+\dfrac {1} {4}dc} {d}\).

From there, you know that when \(b= \dfrac {1} {2}d\), it will give you the largest area, which is \(\dfrac {1} {4}dc\).

\(a=\dfrac {-c\left( b-d\right) } {d}= \dfrac {-c\left( \dfrac {1} {2}d-d\right) } {d}=\dfrac {c\left( d-\dfrac {1} {2}d\right) } {d}=\dfrac {1} {2}c\).

Thus, the maximum rectangle area occurs when the midpoints of two of the sides of the triangle were joined to make a side of the rectangle and its area is thus 50% or half of the area of the triangle or 1/4 of the base times height.

Proof without words from Mr. Rusczyk 

Try using different types of triangles to experiment and see for yourself.
Paper folding is fun !!!!!
It's very cool :D

Dimensional Change questions I:

Questions written by Willie, a volunteer.  Answer key and detailed solutions below.

1a. There is a regular cylinder, which has a height equal to its radius. If the radius and height are both increased by 50%, by what % does the total volume of the cylinder increase?

1b. If the radius and height are both decreased by 10%, by what % does the total volume of the cylinder decrease?

1c. If the radius is increased by 20% and the height is decreased by 40%, what % of the volume of the original cylinder does the volume of the new cylinder represent?

1d. If the radius is increased by 40% and the height is decreased by 20%, what % of the volume of the original cylinder does the volume of the new cylinder represent?

1e. If the height is increased by 125%, what % does the radius need to be decreased by for the volume to remain the same?

2. If the side of a cube is increased by 50%, by what % does the total surface area of the cube increase?

3a. If the volume of a cube increases by 72.8%, by what % does the total surface area of the cube increase?

3b. By what % did the side length of the cube increase?

4. You have a collection of cylinders, all having a radius of 5. The first cylinder has a height of 2, the second has a height of 4, the third a height of 6, etc. The last cylinder has a height of 50. What is the sum of the volumes of all the cylinders (express your answer in terms of pi)?

Answer key: (Each question should not take you more than 30 seconds to solve if you really understand the concepts involved.)

 

 

1a.  The volume of a cylinder is πr²x h (height). The radius itself will be squared and the height stays at constant ratio. The volume will increased thus (1.5)³ - 1³ -- the original 100% of the volume = 2.375

=237.5%

1b.  Like the previous question: 1³ - 0.9³ [when it's discount/percentage decrease, you use the 100% or 1 - the discount/decrease percentage] = 0.271 =  27.1% decrease

1c.  1.2² [100% + 20% increase = 1.2] x 0.6 [100% -40% = 0.6] = 0.864  or  

86.4% of the original volume

1d.  1.4² [100% + 40% increase = 1.4] x 0.8 [100% -20% = 0.8] = 1.568 = 156.8% of the original volume

1e.  When the height of a cylinder is increased 125%, the total volume is is 225% of the original cylinder, or 9/4.

Since the radius is used two times (or squared), it has to decrease 4/9^1/2 = 2/3 for the new cylinder to have the same volume as the old one. [9/4 times 4/9 = 1 or the original volume.]

1 - (2/3) = 1/3 = 0.3 = 33.3%

2. Surface area is 2-D so 1.5² - 1 = 1.25 = 125% increase

3a. If a volume of a cube is increased by 72.8 percent, it's 172.8% or 1.728 of the original volume. Now you are going from 3-D (volume) to 2-D (surface area). 1.728^2/3 = 1.44 or 44% increase. [Don't forget to minus 1 (the original volume) since it is asking you the percentage increase.]

3b. From surface area, you can get the side increase by using 1.44^1/2 = 1.2, so 20% increase.

Or you can also use 1.728^1/3 = 1.2;  1.2 - 1 = 20%

4. The volume of a cylinder is πr²x h . (2 + 4 + 6 + ...50) x 5²π = (25 x 26) x 25π =16250π

2011 Mathcounts Chapter Sprint Round solutions

#22: The answer is 2674.
 See left for explanations.














#23: Let the two consecutive positive integers be x and x + 1.
( x + 1 ) / x = 1.02, x + 1 = 1.02x, 0.02x = 1, x = 1 divided by 0.02 = 1 times 100/2 = 50
The two numbers are 50 and 51 and their sum is 50 + 51 = 101.

#24: The two x-intercepts when y is "0" are 10 or -10; the two y-intercepts when x is "0" are 5 or -5.
Area of a rhombus is D1 x D2 / 2 so the answer is [10-(-10)] x [5 -(-5)] = 100 square units.

#25: The area ratio of the two similar triangle is 150/6 so the line ratio is √ 150/6  or 5:1.
the length of the hypotenuse of the smaller triangle is 5 inches, so the other two legs are 3 and 4. 
(a Pythagorean triple)
The sum of the lengths of the legs of the larger triangle is (3 + 4) * 5 = 35.

#26: To have same number of boys and girls, the committee needs to consist of 3 boys and 3 girls. 
(6C3 x 4C3)/ 10C6 = 80/210 = 8/21

#27: When the point (3, 4) is reflected over the x-axis to B, B would = (3, -4). 
When B is reflected over the line y = x to C, C would = (-4, 3).

The area of the triangle is [4 -(-4)] x [ 3 - (-4)]/ 2 = 28 square units. 


#28: Tonisha is 45 miles ahead Sheila when Sheila leaves Maryville at 8: 15 a.m. 
Each hour Sheila will be 15 miles closer to Tonisha. 45/15 = 3, which means that 3 hours 
later Sheila will pass Tonisha. 
8:15 + 3 hours = 11: 15 a.m.


#29: Using 30-60-90 degree angle ratio, you can make the radius be √ 3 and half of the side of the hexagon would be 1 so each side of the hexagon
is 2.

The area of the hexagon is (√ 3/4) times 2² times 6 = 6√ 3.
The area of the circle is 3Π.
The fraction is 3Π/6√ 3 = √ 3 / 6
a = 3 and b = 6, ab = 18

#30:  Area of triangle KDC is easy to find once you realize the height is just the right triangle with a hypotenuse 6 and a leg 4. (half of the length of CD),

Using Pythagorean theorem, you get the height to be 2√ 5 .

8 x 2√ 5 /2 = 8 √ 5 .

Original Problem from a Student Problem Solver on Similar Triangles

An original problem on similar triangles from Varun (from FL)