https://artofproblemsolving.com/wiki/index.php?title=1992_AJHSME_Problems/Problem_10&feed=atom&action=history
1992 AJHSME Problems/Problem 10 - Revision history
2024-03-29T01:07:25Z
Revision history for this page on the wiki
MediaWiki 1.31.1
https://artofproblemsolving.com/wiki/index.php?title=1992_AJHSME_Problems/Problem_10&diff=55630&oldid=prev
Nathan wailes at 04:09, 5 July 2013
2013-07-05T04:09:18Z
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 04:09, 5 July 2013</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{AJHSME box|year=1992|num-b=9|num-a=11}}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{AJHSME box|year=1992|num-b=9|num-a=11}}</div></td></tr>
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Nathan wailes
https://artofproblemsolving.com/wiki/index.php?title=1992_AJHSME_Problems/Problem_10&diff=49997&oldid=prev
Gina at 01:32, 23 December 2012
2012-12-23T01:32:30Z
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 01:32, 23 December 2012</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>=== Solution 2 ===</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>=== Solution 2 ===</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Each of the triangle has side length of <math> \frac{1}{4} \times 8=2 </math>, so the area is <math> \frac{1}{2} \times 2 \times 2=2 </math>. Because there are <math> 10 </math> triangles is the shaded area, its area is <math> 2 \times 10 =\boxed{\text{(B)}\ 20} </math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Each of the triangle has side length of <math> \frac{1}{4} \times 8=2 </math>, so the area is <math> \frac{1}{2} \times 2 \times 2=2 </math>. Because there are <math> 10 </math> triangles is the shaded area, its area is <math> 2 \times 10 =\boxed{\text{(B)}\ 20} </math>.</div></td></tr>
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<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">==See Also==</ins></div></td></tr>
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Gina
https://artofproblemsolving.com/wiki/index.php?title=1992_AJHSME_Problems/Problem_10&diff=41158&oldid=prev
Math Kirby: Created page with "== Problem == An isosceles right triangle with legs of length <math>8</math> is partitioned into <math>16</math> congruent triangles as shown. The shaded area is <asy> for (in..."
2011-08-02T20:00:22Z
<p>Created page with "== Problem == An isosceles right triangle with legs of length <math>8</math> is partitioned into <math>16</math> congruent triangles as shown. The shaded area is <asy> for (in..."</p>
<p><b>New page</b></p><div>== Problem ==<br />
<br />
An isosceles right triangle with legs of length <math>8</math> is partitioned into <math>16</math> congruent triangles as shown. The shaded area is<br />
<br />
<asy><br />
for (int a=0; a <= 3; ++a)<br />
{<br />
for (int b=0; b <= 3-a; ++b)<br />
{<br />
fill((a,b)--(a,b+1)--(a+1,b)--cycle,grey);<br />
}<br />
}<br />
for (int c=0; c <= 3; ++c)<br />
{<br />
draw((c,0)--(c,4-c),linewidth(1));<br />
draw((0,c)--(4-c,c),linewidth(1));<br />
draw((c+1,0)--(0,c+1),linewidth(1));<br />
}<br />
<br />
label("$8$",(2,0),S); <br />
label("$8$",(0,2),W);<br />
</asy><br />
<br />
<math>\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 64</math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
Because the smaller triangles are congruent, the shaded area take <math> \frac{10}{16} </math> of the largest triangles area, which is <math> \frac{8 \times 8}{2}=32 </math>, so the shaded area is <math> \frac{10}{16} \times 32= \boxed{\text{(B)}\ 20} </math>.<br />
<br />
=== Solution 2 ===<br />
Each of the triangle has side length of <math> \frac{1}{4} \times 8=2 </math>, so the area is <math> \frac{1}{2} \times 2 \times 2=2 </math>. Because there are <math> 10 </math> triangles is the shaded area, its area is <math> 2 \times 10 =\boxed{\text{(B)}\ 20} </math>.</div>
Math Kirby