Difference between revisions of "2002 AMC 12P Problems/Problem 8"

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[[2002 AMC 12P Problems/Problem 8|Solution]]
 
  
== Solution 1==
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== Solution==
 
We can solve this with some simple coordinate geometry. Let <math>A</math> be the origin at let <math>AB</math> be located on the positive <math>x-</math>axis. The equation of semi-circle <math>AB</math> is <math>(x-13)^2+y^2=13^2, y \geq 0.</math> Since <math>E</math> and <math>F</math> are both perpendicular to <math>C</math> and <math>D</math> respectively, they must have the same <math>x -</math> coordinate. Plugging in <math>1</math> and <math>8</math> into our semi-circle equation gives us <math>y=5</math> and <math>y=12</math> respectively. The distance formula on <math>(1, 5)</math> and <math>(8, 12)</math> gives us our answer of <math>\sqrt{(1-8)^2 + (5-12)^2}=\sqrt{2(7^2)}=\boxed{\textbf{(D) } 7\sqrt{2}}.</math>
 
We can solve this with some simple coordinate geometry. Let <math>A</math> be the origin at let <math>AB</math> be located on the positive <math>x-</math>axis. The equation of semi-circle <math>AB</math> is <math>(x-13)^2+y^2=13^2, y \geq 0.</math> Since <math>E</math> and <math>F</math> are both perpendicular to <math>C</math> and <math>D</math> respectively, they must have the same <math>x -</math> coordinate. Plugging in <math>1</math> and <math>8</math> into our semi-circle equation gives us <math>y=5</math> and <math>y=12</math> respectively. The distance formula on <math>(1, 5)</math> and <math>(8, 12)</math> gives us our answer of <math>\sqrt{(1-8)^2 + (5-12)^2}=\sqrt{2(7^2)}=\boxed{\textbf{(D) } 7\sqrt{2}}.</math>
  

Latest revision as of 20:24, 19 May 2024

Problem

Let $AB$ be a segment of length $26$, and let points $C$ and $D$ be located on $AB$ such that $AC=1$ and $AD=8$. Let $E$ and $F$ be points on one of the semicircles with diameter $AB$ for which $EC$ and $FD$ are perpendicular to $AB$. Find $EF.$

$\text{(A) }5 \qquad \text{(B) }5 \sqrt{2}  \qquad \text{(C) }7 \qquad \text{(D) }7 \sqrt{2} \qquad \text{(E) }12$


Solution

We can solve this with some simple coordinate geometry. Let $A$ be the origin at let $AB$ be located on the positive $x-$axis. The equation of semi-circle $AB$ is $(x-13)^2+y^2=13^2, y \geq 0.$ Since $E$ and $F$ are both perpendicular to $C$ and $D$ respectively, they must have the same $x -$ coordinate. Plugging in $1$ and $8$ into our semi-circle equation gives us $y=5$ and $y=12$ respectively. The distance formula on $(1, 5)$ and $(8, 12)$ gives us our answer of $\sqrt{(1-8)^2 + (5-12)^2}=\sqrt{2(7^2)}=\boxed{\textbf{(D) } 7\sqrt{2}}.$

See also

2002 AMC 12P (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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 12 Problems and Solutions

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