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考题13-14 2014 AMC 8

2018-08-04 重点归纳

AMC 8数学竞赛专为8年级及以下的初中学生设计，但近年来的数据显示，越来越多小学4-6年级的考生加入到AMC 8级别的考试行列中，而当这些学生能在成绩中取得“A”类标签，则是对孩子数学天赋的优势证明，不管是对美高申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC 8数学竞赛试题及答案吧：

Problem 13

If $n$ and $m$ are integers and $n^2+m^2$ is even, which of the following is impossible?

$\textbf{(A) }$ $n$ and $m$ are even $\qquad\textbf{(B) }$ $n$ and $m$ are odd $\qquad\textbf{(C) }$ $考题13-14 2014 AMC 8$ is even $\qquad\textbf{(D) }$ $n+m$ is odd $\qquad \textbf{(E) }$ none of these are impossible

Solution

Since $n^2+m^2$ is even, either both $n^2$ and $m^2$ are even, or they are both odd. Therefore, $n$ and $m$ are either both even or both odd, since the square of an even number is even and the square of an odd number is odd. As a result, $n+m$ must be even. The answer, then, is $\boxed{D}$ .

Problem 14

Rectangle $ABCD$ and right triangle $DCE$ have the same area. They are joined to form a trapezoid, as shown. What is $DE$ ?

AMC 8数学竞赛试题及答案

$考题13-14 2014 AMC 8$

Solution

The area of $\bigtriangleup CDE$ is $\frac{DC\cdot CE}{2}$ . The area of $ABCD$ is $AB\cdot AD=5\cdot 6=30$ , which also must be equal to the area of $\bigtriangleup CDE$ , which, since $DC=5$ , must in turn equal $\frac{5\cdot CE}{2}$ . Through transitivity, then, $\frac{5\cdot CE}{2}=30$ , and $CE=12$ . Then, using the Pythagorean Theorem, you should be able to figure out that $AMC 8数学竞赛试题及答案$ is a $5-12-13$ triangle, so $DE=\boxed{13}$ , or $\boxed{B}$ .

Solution 2

The area of the rectangle is $5\times6=30.$ Since the parallel line pairs are identical, $DC=5$ . Let $CE$ be $x$ . $\dfrac{5x}{2}=30$ is the area of the right triangle. Solving for $x$ , we get $x=12.$ According to the Pythagorean Theorem, we have a 5-12-13 triangle. So, the hypotenuse $DE$ has to be $\boxed{B}$ .