Difference between revisions of "AoPS Wiki:Sandbox"
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We want to find the area of this figure: | We want to find the area of this figure: | ||
| + | |||
<asy> | <asy> | ||
| − | path rt,tt; | + | path rt,tt, tri; |
| − | + | real x, y; | |
| − | + | y = 1+sqrt(2); | |
| − | tt=(0,0)..( | + | x = y+(6/1.7); |
| − | rt=(1 | + | tt=(0,0)..(y,1)--(y,-1)..cycle; |
| + | rt=(y,-1)--(x,-1)--(x,1)--(y,1); | ||
| + | tri=(y,-1)--(y-1,0)--(y,1); | ||
draw(rt); | draw(rt); | ||
draw(tt); | draw(tt); | ||
| + | draw(tri); | ||
label("1.7", (x, 0), E); | label("1.7", (x, 0), E); | ||
| + | label("3", (y+(3/1.7), -1), S); | ||
| + | label("C", (y-1, -0.1), S); | ||
</asy> | </asy> | ||
| + | |||
| + | 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. | ||
| + | |||
| + | <asy> | ||
| + | path rt; | ||
| + | real x, y; | ||
| + | y = 1+sqrt(2); | ||
| + | x = y+(6/1.7); | ||
| + | rt=(y,-1)--(x,-1)--(x,1)--(y,1)--cycle; | ||
| + | draw(rt); | ||
| + | label("1.7", (x, 0), E); | ||
| + | label("3", (y+(3/1.7), -1), S); | ||
| + | </asy> | ||
| + | |||
| + | It has an area of <math> 3 * 1.7 = 5.1</math> . | ||
| + | |||
| + | |||
| + | Lets now take a look at the triangle, after drawing the height. | ||
| + | |||
| + | <asy> | ||
| + | unitsize(0.8inch); | ||
| + | path tri, lin; | ||
| + | real x, y; | ||
| + | y = 1+sqrt(2); | ||
| + | x = y+(6/1.7); | ||
| + | tri=(y,-1)--(y-1,0)--(y,1)--cycle; | ||
| + | lin=(y-1, 0)--(y,0); | ||
| + | draw(tri); | ||
| + | draw(lin); | ||
| + | label("1.7", (y, 0), E); | ||
| + | label("C", (y-1, -0.1), S); | ||
| + | </asy> | ||
| + | |||
| + | 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 <math>\frac{1.7 * \frac{1.7}{2}}{2} = 0.7225</math> . | ||
| + | |||
| + | 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. | ||
| + | |||
| + | <asy> | ||
| + | unitsize(0.8inch); | ||
| + | path tt, tri; | ||
| + | real x, y; | ||
| + | y = 1+sqrt(2); | ||
| + | x = y+(6/1.7); | ||
| + | tt=(0,0)..(y,1)--(y,-1)..cycle; | ||
| + | tri=(y,-1)--(y-1,0)--(y,1); | ||
| + | draw(tt); | ||
| + | draw(tri); | ||
| + | label("1.7", (y, 0), E); | ||
| + | label("C", (y-1, -0.1), S); | ||
| + | </asy> | ||
| + | |||
| + | We find the radius using the 45-45-90 triangle. Since the ratios of the sides are 1:1:<math>\sqrt{2}</math>, we can find the radius to be <math>\frac{1.7}{\sqrt{2}} = \frac{1.7 \sqrt{2}}{2}</math> . | ||
Revision as of 18:16, 2 April 2014
Welcome to the sandbox, a location to test your newfound wiki-editing abilities.
Please note that all contributions here may be deleted periodically and without warning.
In the computer world, a sandbox is a place to test and experiment -- essentially, it's a place to play.
This is the AoPSWiki Sandbox. Feel free to experiment here.
Warning: anything you place here is subject to deletion without notice.
[This was deleted due to its inappropriateness.]
test
test---experimenthere-----
Please do not delete from this point on until 5:00 PST on 4/2!
We want to find the area of this figure:
![[asy] path rt,tt, tri; real x, y; y = 1+sqrt(2); x = y+(6/1.7); tt=(0,0)..(y,1)--(y,-1)..cycle; rt=(y,-1)--(x,-1)--(x,1)--(y,1); tri=(y,-1)--(y-1,0)--(y,1); draw(rt); draw(tt); draw(tri); label("1.7", (x, 0), E); label("3", (y+(3/1.7), -1), S); label("C", (y-1, -0.1), S); [/asy]](http://latex.artofproblemsolving.com/d/2/f/d2f5c70186cf1e51f1e65a13ef95d95335aa1f17.png)
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.
![[asy] path rt; real x, y; y = 1+sqrt(2); x = y+(6/1.7); rt=(y,-1)--(x,-1)--(x,1)--(y,1)--cycle; draw(rt); label("1.7", (x, 0), E); label("3", (y+(3/1.7), -1), S); [/asy]](http://latex.artofproblemsolving.com/2/c/e/2ced4c211717f2579089a230c942cda1c78649d9.png)
It has an area of 
.
Lets now take a look at the triangle, after drawing the height.
![[asy] unitsize(0.8inch); path tri, lin; real x, y; y = 1+sqrt(2); x = y+(6/1.7); tri=(y,-1)--(y-1,0)--(y,1)--cycle; lin=(y-1, 0)--(y,0); draw(tri); draw(lin); label("1.7", (y, 0), E); label("C", (y-1, -0.1), S); [/asy]](http://latex.artofproblemsolving.com/7/4/4/7440fa82cbbf77c5c1bbfe5ab26d521732e44dd6.png)
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.
![[asy] unitsize(0.8inch); path tt, tri; real x, y; y = 1+sqrt(2); x = y+(6/1.7); tt=(0,0)..(y,1)--(y,-1)..cycle; tri=(y,-1)--(y-1,0)--(y,1); draw(tt); draw(tri); label("1.7", (y, 0), E); label("C", (y-1, -0.1), S); [/asy]](http://latex.artofproblemsolving.com/5/b/4/5b4ff3a004442e5d420758ec69d7a80112b9b6f1.png)
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 
.
