Difference between revisions of "1990 AJHSME Problems/Problem 15"

(Created page with '==Problem== The area of this figure is <math>100\text{ cm}^2</math>. Its perimeter is <asy> draw((0,2)--(2,2)--(2,1)--(3,1)--(3,0)--(1,0)--(1,1)--(0,1)--cycle,linewidth(1)); …')
 
 
Line 22: Line 22:
 
{{AJHSME box|year=1990|num-b=14|num-a=16}}
 
{{AJHSME box|year=1990|num-b=14|num-a=16}}
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 23:06, 4 July 2013

Problem

The area of this figure is $100\text{ cm}^2$. Its perimeter is

[asy] draw((0,2)--(2,2)--(2,1)--(3,1)--(3,0)--(1,0)--(1,1)--(0,1)--cycle,linewidth(1)); draw((1,2)--(1,1)--(2,1)--(2,0),dashed); [/asy]

[figure consists of four identical squares]

$\text{(A)}\ \text{20 cm} \qquad \text{(B)}\ \text{25 cm} \qquad \text{(C)}\ \text{30 cm} \qquad \text{(D)}\ \text{40 cm} \qquad \text{(E)}\ \text{50 cm}$

Solution

Since the area of the whole figure is $100$, each square has an area of $25$ and the side length is $5$.

There are $10$ sides of this length, so the perimeter is $10(5)=50\rightarrow \boxed{\text{E}}$.

See Also

1990 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png