Difference between revisions of "2003 AMC 10A Problems/Problem 22"
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<math>\angle BHA = \angle HAD</math> (Alt. Interior Angles are congruent). | <math>\angle BHA = \angle HAD</math> (Alt. Interior Angles are congruent). | ||
− | Therefore <math> | + | Therefore <math>\triangle GFA</math> and <math>\triangle ABH</math> are similar. |
− | <math>GCH</math> and <math>GEA</math> are also similar. | + | <math>\triangle GCH</math> and <math>\triangle GEA</math> are also similar. |
<math>DA</math> is 9, therefore <math>EA</math> must equal 5. Similarly, <math>CH</math> must equal 3. | <math>DA</math> is 9, therefore <math>EA</math> must equal 5. Similarly, <math>CH</math> must equal 3. | ||
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<math>\frac{25}{10}\: =\: \frac{GF}{8}</math>. | <math>\frac{25}{10}\: =\: \frac{GF}{8}</math>. | ||
− | Therefore <math>GF | + | Therefore <math>GF= \boxed{\mathrm{(B)}\ 20}</math>. |
=== Solution 2 === | === Solution 2 === | ||
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<math>FD=6</math> | <math>FD=6</math> | ||
− | <math>GF=2 \cdot FD+8=2\cdot6+8= | + | <math>GF=2 \cdot FD+8=2\cdot6+8=\boxed{\mathrm{(B)}\ 20}</math> |
=== Solution 3 === | === Solution 3 === | ||
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<cmath>\dfrac{8}{4\sqrt{5}}=\dfrac{GF}{10\sqrt{5}}</cmath> | <cmath>\dfrac{8}{4\sqrt{5}}=\dfrac{GF}{10\sqrt{5}}</cmath> | ||
− | We can multiply both sides by <math>\sqrt{5}</math> to get that <math>GF</math> is twice of 10, or <math> | + | We can multiply both sides by <math>\sqrt{5}</math> to get that <math>GF</math> is twice of 10, or <math>\boxed{\mathrm{(B)}\ 20}</math> |
=== Solution 4 === | === Solution 4 === | ||
− | We extend BC such that it intersects GF at X. Since ABCD is a rectangle, it follows that CD=8, therefore, XF=8. Let GX=y. From the similarity of triangles GCH and GEA, we have the ratio 3:5 (as CH=9-6=3, and EA=9-4=5). GX and GF are the altitudes of GCH and GEA, respectively. Thus, y:y+8 = 3:5, from which we have y=12, thus GF=y+8=12+8=20 | + | We extend <math>BC</math> such that it intersects <math>GF</math> at <math>X</math>. Since <math>ABCD</math> is a rectangle, it follows that <math>CD=8</math>, therefore, <math>XF=8</math>. Let <math>GX=y</math>. From the similarity of triangles <math>GCH</math> and <math>GEA</math>, we have the ratio <math>3:5</math> (as <math>CH=9-6=3</math>, and <math>EA=9-4=5</math>). <math>GX</math> and <math>GF</math> are the altitudes of <math>GCH</math> and <math>GEA</math>, respectively. Thus, <math>y:y+8 = 3:5</math>, from which we have <math>y=12</math>, thus <math>GF=y+8=12+8=\boxed{\mathrm{(B)}\ 20}</math> |
== See Also == | == See Also == |
Revision as of 13:55, 1 August 2011
Contents
Problem
In rectangle , we have , , is on with , is on with , line intersects line at , and is on line with . Find the length of .
Solution
Solution 1
(Opposite angles are equal).
(Both are 90 degrees).
(Alt. Interior Angles are congruent).
Therefore and are similar. and are also similar.
is 9, therefore must equal 5. Similarly, must equal 3.
Because and are similar, the ratio of and , must also hold true for and . , so is of . By Pythagorean theorem, .
.
So .
.
Therefore .
Solution 2
Since is a rectangle, .
Since is a rectangle and , .
Since is a rectangle, .
So, is a transversal, and .
This is sufficient to prove that and .
Using ratios:
Since can't have 2 different lengths, both expressions for must be equal.
Solution 3
Since is a rectangle, , , and . From the Pythagorean Theorem, .
Lemma
Statement:
Proof: , obviously.
$\begin{eqnarray} \angle HCE=180^{\circ}-\angle CHG\\ \angle DCE=\angle CHG-90^{\circ}\\ \angle CEED=180-\angle CHG\\ \angle GEA=\angle GCH \end{eqnarray}$ (Error compiling LaTeX. ! Missing \endgroup inserted.)
Since two angles of the triangles are equal, the third angles must equal each other. Therefore, the triangles are similar.
Let .
Also, , therefore
We can multiply both sides by to get that is twice of 10, or
Solution 4
We extend such that it intersects at . Since is a rectangle, it follows that , therefore, . Let . From the similarity of triangles and , we have the ratio (as , and ). and are the altitudes of and , respectively. Thus, , from which we have , thus
See Also
2003 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |