Difference between revisions of "Geometry Solutions"
(→Probability) |
|||
Line 319: | Line 319: | ||
r = (0,0)--(4*sqrt(2), 0)--(4*sqrt(2), 3*sqrt(2))--(0, 3*sqrt(2))--cycle; | r = (0,0)--(4*sqrt(2), 0)--(4*sqrt(2), 3*sqrt(2))--(0, 3*sqrt(2))--cycle; | ||
draw(r); | draw(r); | ||
+ | |||
path sq; | path sq; | ||
sq = (0,0)--(2*sqrt(2), 2*sqrt(2))--(4*sqrt(2), 0); | sq = (0,0)--(2*sqrt(2), 2*sqrt(2))--(4*sqrt(2), 0); | ||
draw(sq); | draw(sq); | ||
+ | |||
path sq2; | path sq2; | ||
sq2 = (0,3*sqrt(2))--(2*sqrt(2), sqrt(2))--(4*sqrt(2), 3*sqrt(2)); | sq2 = (0,3*sqrt(2))--(2*sqrt(2), sqrt(2))--(4*sqrt(2), 3*sqrt(2)); | ||
draw(sq2); | draw(sq2); | ||
+ | |||
+ | path tot;//bottom right top left | ||
+ | tot = (2*sqrt(2),1*sqrt(2))--(2.5*sqrt(2), 1.5*sqrt(2))--(2*sqrt(2), 2*sqrt(2))--(1.5*sqrt(2),1.5*sqrt(2))--cycle; | ||
+ | fill(tot, gray(0.6)); | ||
+ | |||
+ | label("x", (2.25*sqrt(2),1.25*sqrt(2)), SE); | ||
+ | label("x", (1.75*sqrt(2),1.25*sqrt(2)), SW); | ||
+ | label("3x", (3*sqrt(2),2*sqrt(2)), SE); | ||
+ | label("3x", (1*sqrt(2),2*sqrt(2)), SW); | ||
+ | label("3x", (3*sqrt(2),1*sqrt(2)), NE); | ||
+ | label("3x", (1*sqrt(2),1*sqrt(2)), NW); | ||
+ | </asy> | ||
+ | |||
+ | |||
+ | We look at the the 45-45-90 triangle on the right. (label?) | ||
+ | |||
+ | <asy> | ||
+ | path r; | ||
+ | r = (0,0)--(4*sqrt(2), 0)--(4*sqrt(2), 3*sqrt(2))--(0, 3*sqrt(2))--cycle; | ||
+ | draw(r); | ||
+ | |||
+ | path sq; | ||
+ | sq = (0,0)--(2*sqrt(2), 2*sqrt(2))--(4*sqrt(2), 0); | ||
+ | draw(sq); | ||
+ | |||
+ | path sq2; | ||
+ | sq2 = (0,3*sqrt(2))--(2*sqrt(2), sqrt(2))--(4*sqrt(2), 3*sqrt(2)); | ||
+ | draw(sq2); | ||
+ | |||
+ | path tot;//bottom right top left | ||
+ | tot = (2*sqrt(2),1*sqrt(2))--(2.5*sqrt(2), 1.5*sqrt(2))--(2*sqrt(2), 2*sqrt(2))--(1.5*sqrt(2),1.5*sqrt(2))--cycle; | ||
+ | fill(tot, gray(0.8)); | ||
+ | |||
+ | path trir; | ||
+ | trir = (2.5*sqrt(2), 1.5*sqrt(2))--(4*sqrt(2), 3*sqrt(2))--(4*sqrt(2),0)--cycle; | ||
+ | fill(trir, gray(0.6)); | ||
+ | |||
+ | |||
+ | label("3x", (3*sqrt(2),1*sqrt(2)), NE); | ||
+ | label("3x", (3*sqrt(2),2*sqrt(2)), SE); | ||
</asy> | </asy> |
Revision as of 13:54, 4 April 2014
Geometry Explained
This is primarily meant for explaining problems in geometry, written by RMS students.
This page can be accessed by tinyurl.com/teachgeo.
Composite Figure
We want to find the area of this figure:
We label the circle as circle C. We can break the figure into three parts, shown as the 3/4 circle, the triangle, and the rectangle.
Lets first take a look at the rectangle.
It has an area of .
Lets now take a look at the triangle, after drawing the height.
We see that both the radii are the two shorter sides of the triangle, making this a isosceles 45-45-90 triangle.
We also see that the height that we drew is half the hypotenuse(note the two smaller 45-45-90 isosceles triangles).
Hence, the area of the triangle is .
Now, let's take a look at the 3/4 circle. We know it is 3/4 because there is a 90 degree triangle cut out of it.
We find the radius using the 45-45-90 triangle. Since the ratios of the sides are 1:1:, we can find the radius to be .
Hence, the area of the whole circle is , and the area of the 3/4 circle is .
Adding it all up, we find the answer to be .
Just plug it into your calculator.
Spotlight!
We want to find the area of the intersection of the circles in this figure:
Lets call the radius of each of the circles 1, because we are calculating probability. To do this, we need to divide the intersection area into smaller parts that we can find the area of.
Lets examine the shaded . Since each of its sides are radii, it is equilateral.
Hence, each of its angles measure 60 degrees. Namely, measures 60 degrees.
Important note: The area of an equilateral triangle is calculated by , where is the side length of a triangle. This is an EXTREMELY important formula to know, and it it also extremely simple to prove and memorize.
After this, we can deduce that the area of triangle ABC with side length 1 is
Let us now look at the encompassing sector.
We then find that the area of the entire sector is .
Subtracting the area of from sector gives us the shaded part below:
Plugging in the values we know, we find that this has area .
We see that there are 4 such small curved areas.
Multiplying the value we got for an individual small curved part by 4, we find the area of the above shaded region to be .
We then need to add back the areas of the equilateral triangles.
Recall that the area of each equilateral triangle was .
Hence the area of the above shaded region is .
We then look at the total.
Hence the area of the total shaded region is .
We plug this into the calculator to find the final answer.
In order to find the union(total) of the two circles, we examine the figure.
We want to find the area of the intersection of the circles in this figure:
Using , the area of the circle on the left is . We also know that the area of the circle on the right is . When we add these together, we are counting the entire area, but we are double-counting the overlap area. We thus need to subtract the overlap from the in order to find the union(total) of the figure.
Hence, the total is .
Probability
We want to find the probability that a point lies inside the shaded square below, given that it lies inside the larger rectangle.
We look at the the 45-45-90 triangle on the right. (label?)