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Difference between revisions of "1967 AHSME Problems/Problem 11"

(Created page with "== Problem == If the perimeter of rectangle <math>ABCD</math> is <math>20</math> inches, the least value of diagonal <math>\overline{AC}</math>, in inches, is: <math>\textbf{(A)...")
 
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== Solution ==
 
== Solution ==
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<math>\fbox{B}</math>
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For rectangle <math>ABCD</math> with perimeter 20, the diagonal <math>AC</math> is given by:
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<cmath> AC = \sqrt{l^2 + w^2} </cmath>
 +
To minimize <math>AC</math>, <math>l</math> and <math>w</math> should be equal (i.e., the rectangle is a square). Thus, <math>l = w = 5</math>.
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So, the minimum <math>AC</math> is:
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<cmath> AC = \sqrt{5^2 + 5^2} = \boxed{\textbf{(B) } \sqrt{50}} </cmath>
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~ proloto
  
 
== See also ==
 
== See also ==
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{{AHSME box|year=1967|num-b=10|num-a=12}}   
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{{AHSME 40p box|year=1967|num-b=10|num-a=12}}   
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:00, 28 September 2023

Problem

If the perimeter of rectangle $ABCD$ is $20$ inches, the least value of diagonal $\overline{AC}$, in inches, is:

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ \sqrt{50}\qquad \textbf{(C)}\ 10\qquad \textbf{(D)}\ \sqrt{200}\qquad \textbf{(E)}\ \text{none of these}$

Solution

For rectangle $ABCD$ with perimeter 20, the diagonal $AC$ is given by: \[AC = \sqrt{l^2 + w^2}\] To minimize $AC$, $l$ and $w$ should be equal (i.e., the rectangle is a square). Thus, $l = w = 5$. So, the minimum $AC$ is: \[AC = \sqrt{5^2 + 5^2} = \boxed{\textbf{(B) } \sqrt{50}}\] ~ proloto

See also

1967 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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