Difference between revisions of "1962 AHSME Problems/Problem 16"
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==Solution== | ==Solution== | ||
| − | + | Clearly, the other side of <math>R_1</math> has length <math>\frac{12}2=6</math>. | |
| + | Now call the sides of <math>R_2</math> a and b, with <math>b>a</math>. We know that <math>\frac{b}a=\frac31=3</math>, because the two rectangles are similar. We also know that <math>a^2+b^2=15^2=225</math>. But <math>b=3a</math>, so substituting gives | ||
| + | <cmath>a^2+(3a)^2=225</cmath> | ||
| + | <cmath>10a^2=225</cmath> | ||
| + | <cmath>a^2=\frac{45}2</cmath> | ||
| + | <cmath>a=\frac{3\sqrt{10}}2</cmath> | ||
| + | <cmath>b=\frac{9\sqrt{10}}2</cmath> | ||
| + | <cmath>[R_2]=ab=\boxed{\frac{135}2\textbf{ (C)}}</cmath> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1962|before=Problem 15|num-a=17}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 21:16, 3 October 2014
Problem
Given rectangle 
with one side 
inches and area 
square inches. Rectangle 
with diagonal 
inches is similar to . Expressed in square inches the area of is:
Solution
Clearly, the other side of has length 
.
Now call the sides of a and b, with 
. We know that 
, because the two rectangles are similar. We also know that 
. But 
, so substituting gives
![\[a^2+(3a)^2=225\]](http://latex.artofproblemsolving.com/1/2/2/122c79a916eef044962f7a765a99ec70d6fafe61.png)
![\[10a^2=225\]](http://latex.artofproblemsolving.com/c/1/5/c158aaad5a5178d713583e34cf6a260a9988299c.png)
![\[a^2=\frac{45}2\]](http://latex.artofproblemsolving.com/c/a/6/ca6ddfd7f836ccbace7c6e3f92da80e6bcee48ab.png)
![\[a=\frac{3\sqrt{10}}2\]](http://latex.artofproblemsolving.com/9/f/b/9fbbaf3bdadb6848aa58dadcc8d921316317046a.png)
![\[b=\frac{9\sqrt{10}}2\]](http://latex.artofproblemsolving.com/5/0/f/50f664b0695ea25a59c3489c95ceb7d60b5ebccd.png)
![\[[R_2]=ab=\boxed{\frac{135}2\textbf{ (C)}}\]](http://latex.artofproblemsolving.com/1/9/0/1900ccd39fce190d04007f39ea3fef4cdc0f9ae3.png)
See Also
| 1962 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
