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| − | {{AoPSWiki:Sandbox/header}} <!-- Please do not delete this line --> | + | <noinclude>{{shortcut|[[A:SAND]]}} {{Template:Sandbox}} [[Category:AoPS Wiki]] <!-- Please do not delete this line --></noinclude> |
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| − | ==Solution==
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| − | We want to find the area of this figure:
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| − | <asy>
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| − | //import graph;
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| − | draw(circle((0,0), 1));
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| − | draw(circle((1,0), 1));
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| − | </asy>
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| − | -------------------------
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| − | We want to find the area of this figure:
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| − | <asy>
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| − | path rt,tt, tri;
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| − | real x, y;
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| − | y = 1+sqrt(2);
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| − | x = y+(6/1.7);
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| − | tt=(0,0)..(y,1)--(y,-1)..cycle;
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| − | rt=(y,-1)--(x,-1)--(x,1)--(y,1);
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| − | tri=(y,-1)--(y-1,0)--(y,1);
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| − | draw(rt);
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| − | draw(tt);
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| − | draw(tri);
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| − | label("1.7", (x, 0), E);
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| − | label("3", (y+(3/1.7), -1), S);
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| − | label("C", (y-1, -0.1), S);
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| − | </asy>
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| − | | |
| − | 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.
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| − | Lets first take a look at the rectangle.
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| − | <asy>
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| − | path rt;
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| − | real x, y;
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| − | y = 1+sqrt(2);
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| − | x = y+(6/1.7);
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| − | rt=(y,-1)--(x,-1)--(x,1)--(y,1)--cycle;
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| − | draw(rt);
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| − | label("1.7", (x, 0), E);
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| − | label("3", (y+(3/1.7), -1), S);
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| − | </asy>
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| − | It has an area of <math> 3 * 1.7 = 5.1</math> .
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| − | Lets now take a look at the triangle, after drawing the height.
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| − | <asy>
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| − | unitsize(0.8inch);
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| − | path tri, lin;
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| − | real x, y;
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| − | y = 1+sqrt(2);
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| − | x = y+(6/1.7);
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| − | tri=(y,-1)--(y-1,0)--(y,1)--cycle;
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| − | lin=(y-1, 0)--(y,0);
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| − | draw(tri);
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| − | draw(lin);
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| − | label("1.7", (y, 0), E);
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| − | label("C", (y-1, -0.1), S);
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| − | </asy>
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| − | We see that both the radii are the two shorter sides of the triangle, making this a isosceles 45-45-90 triangle.
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| − | We also see that the height that we drew is half the hypotenuse(note the two smaller 45-45-90 isosceles triangles).
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| − | Hence, the area of the triangle is <math>\frac{1.7 * \frac{1.7}{2}}{2} = 0.7225</math> .
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| − | | |
| − | 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.
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| − | <asy>
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| − | unitsize(0.6inch);
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| − | path tt, tri;
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| − | real x, y;
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| − | y = 1+sqrt(2);
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| − | x = y+(6/1.7);
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| − | tt=(0,0)..(y,1)--(y-1, 0)--(y,-1)..cycle;
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| − | tri=(y, 1)--(y,-1);
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| − | draw(tt);
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| − | fill(tt, gray(0.4));
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| − | label("1.7", (y, 0), E);
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| − | label("C", (y-1, -0.1), S);
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| − | </asy>
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| − | 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> .
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| − | Hence, the area of the whole circle is <math>\pi r^2</math>, and the area of the 3/4 circle is <math>\frac{3}{4} * \pi * (\frac{1.7 \sqrt{2}}{2})^2 </math>.
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| − | Adding it all up, we find the answer to be <math>\frac{3}{4} \pi (\frac{1.7 \sqrt{2}}{2})^2 + 0.7225 + 5.1</math> .
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| − | Just plug it into your calculator. And please understand where all the numbers came from before writing the answer down.
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