Difference between revisions of "2015 AMC 10A Problems/Problem 20"

Revision as of 00:16, 1 February 2016

Problem

A square with positive integer side lengths in $\text{cm}$ has area $A$ $\text{cm}^2$ and perimeter $P$ $\text{cm}$ . Which of the following numbers cannot equal $A+P$ ?

$\textbf{(A) }100\qquad\textbf{(B) }102\qquad\textbf{(C) }104\qquad\textbf{(D) }106\qquad\textbf{(E) }108$

Solution

Let the square's length and width be $a$ . Its area is $a^2$ and the perimeter is $4a$ .

Then $A + P = a^2+ 4a$ . Factoring, this is $(a + 2)(a + 2) - 4$ .

Looking at the answer choices, only $102$ cannot be written this way, because then $a$ would be $0$ .

So the answer is $\boxed{\textbf{(B) }102}$ .

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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All AMC 10 Problems and Solutions

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

@@ Line 7: / Line 7: @@
 ==Solution==
-Let the square's length and width be <math>a</math> . Its area is <math>a^2</math> and the perimeter is <math>2(a^2)</math>.
+Let the square's length and width be <math>a</math> . Its area is <math>a^2</math> and the perimeter is <math>4a</math>.
 Then <math>A + P = a^2+ 4a</math>. Factoring, this is <math>(a + 2)(a + 2) - 4</math>.

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Difference between revisions of "2015 AMC 10A Problems/Problem 20"

Revision as of 00:16, 1 February 2016

Problem

Solution

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