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2005 AMC 12B Problems: Difference between revisions

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== Problem 3 ==
== Problem 3 ==
A rectangle with a diagonal of length <math>x</math> is twice as long as it is wide.  What is the area of the rectangle?
A rectangle with a diagonal of length <math>x</math> is twice as long as it is wide.  What is the area of the rectangle?
<math>
\mathrm{(A)}\ \frac14x^2      \qquad
\mathrm{(B)}\ \frac25x^2      \qquad
\mathrm{(C)}\ \frac12x^2      \qquad
\mathrm{(D)}\ x^2      \qquad
\mathrm{(E)}\ \frac32x^2
</math>


[[2005 AMC 12B Problems/Problem 3|Solution]]
[[2005 AMC 12B Problems/Problem 3|Solution]]

Revision as of 20:38, 17 April 2009

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Problem 1

Two is $10\%$ of $x$ and $20\%$ of $y$. What is $x-y$?

$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 5 \qquad \mathrm{(D)}\ 10 \qquad \mathrm{(E)}\ 20$

Solution

Problem 2

The equations $2x+7=3$ and $bx-10=-2$ have the same solution for $x$. What is the value of $b$?

$\mathrm{(A)}\ -8 \qquad \mathrm{(B)}\ -4 \qquad \mathrm{(C)}\ -2 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 8$

Solution

Problem 3

A rectangle with a diagonal of length $x$ is twice as long as it is wide. What is the area of the rectangle?

$\mathrm{(A)}\ \frac14x^2 \qquad \mathrm{(B)}\ \frac25x^2 \qquad \mathrm{(C)}\ \frac12x^2 \qquad \mathrm{(D)}\ x^2 \qquad \mathrm{(E)}\ \frac32x^2$

Solution

Problem 4

Solution

Problem 5

Solution

Problem 6

Solution

Problem 7

Solution

Problem 8

Solution

Problem 9

Solution

Problem 10

Solution

Problem 11

Solution

Problem 12

Solution

Problem 13

Solution

Problem 14

Solution

Problem 15

Solution

Problem 16

Solution

Problem 17

Solution

Problem 18

Solution

Problem 19

Solution

Problem 20

Solution

Problem 21

Solution

Problem 22

Solution

Problem 23

Solution

Problem 24

Solution

Problem 25

Solution

See also

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