Difference between revisions of "1993 AJHSME Problems/Problem 17"

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==Problem==
 
==Problem==
  
Square corners, <math>5</math> units on a side, are removed from a <math>20</math> unit by <math>30</math> unit rectangular sheet of cardboard.  The sides are then folded to form an open box.  The surface area, in square units, of the interior of the box is
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Square corners, 5 units on a side, are removed from a <math>20</math> unit by <math>30</math> unit rectangular sheet of cardboard.  The sides are then folded to form an open box.  The surface area, in square units, of the interior of the box is
  
 
<asy>
 
<asy>
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==Solution==
 
==Solution==
If the sides of the open box are folded down so that a flat sheet with four corners cut out remains, then the revealed surface would have the same area as the interior of the box. This is equal to the area of the four corners subtracted from the area of the original sheet, which is <math>(20)(30)-4(5)(5) = 600-100 = \boxed{\text{(B)}\ 500}</math>.
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If the sides of the open box are folded down so that a flat sheet with four corners cut out remains, then the revealed surface would have the same area as the interior of the box. This is equal to the area of the four corners subtracted from the area of the original sheet, which is <math>((20)(30)-4(5)(5)) = 600-100 = \boxed{\text{(B)}\ 500}</math>.
  
 
==See Also==
 
==See Also==
 
{{AJHSME box|year=1993|num-b=16|num-a=18}}
 
{{AJHSME box|year=1993|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 08:35, 1 December 2020

Problem

Square corners, 5 units on a side, are removed from a $20$ unit by $30$ unit rectangular sheet of cardboard. The sides are then folded to form an open box. The surface area, in square units, of the interior of the box is

[asy] fill((0,0)--(20,0)--(20,5)--(0,5)--cycle,lightgray); fill((20,0)--(20+5*sqrt(2),5*sqrt(2))--(20+5*sqrt(2),5+5*sqrt(2))--(20,5)--cycle,lightgray); draw((0,0)--(20,0)--(20,5)--(0,5)--cycle); draw((0,5)--(5*sqrt(2),5+5*sqrt(2))--(20+5*sqrt(2),5+5*sqrt(2))--(20,5)); draw((20+5*sqrt(2),5+5*sqrt(2))--(20+5*sqrt(2),5*sqrt(2))--(20,0)); draw((5*sqrt(2),5+5*sqrt(2))--(5*sqrt(2),5*sqrt(2))--(5,5),dashed); draw((5*sqrt(2),5*sqrt(2))--(15+5*sqrt(2),5*sqrt(2)),dashed); [/asy]

$\text{(A)}\ 300 \qquad \text{(B)}\ 500 \qquad \text{(C)}\ 550 \qquad \text{(D)}\ 600 \qquad \text{(E)}\ 1000$

Solution

If the sides of the open box are folded down so that a flat sheet with four corners cut out remains, then the revealed surface would have the same area as the interior of the box. This is equal to the area of the four corners subtracted from the area of the original sheet, which is $((20)(30)-4(5)(5)) = 600-100 = \boxed{\text{(B)}\ 500}$.

See Also

1993 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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 AJHSME/AMC 8 Problems and Solutions

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