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

(Created page with "==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 f...")
 
Line 14: Line 14:
  
 
<math>\text{(A)}\ 300 \qquad \text{(B)}\ 500 \qquad \text{(C)}\ 550 \qquad \text{(D)}\ 600 \qquad \text{(E)}\ 1000</math>
 
<math>\text{(A)}\ 300 \qquad \text{(B)}\ 500 \qquad \text{(C)}\ 550 \qquad \text{(D)}\ 600 \qquad \text{(E)}\ 1000</math>
 +
 +
==Solution==
 +
If the sides of the open box are folded down so that a flat sheet with four corners cur 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==
 +
{{AJHSME box|year=1993|num-b=16|num-a=18}}

Revision as of 22:12, 22 December 2012

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 cur 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
Invalid username
Login to AoPS