AMC 8 2019 Problem 21

AMC8 2019 年第 21 题

坐标几何★★★☆☆
What is the area of the triangle formed by the lines y=5 , y=1+x , and y=1-x ?
由三条直线 y=5,y=1+x 和 y=1-x 所包围的三角形的面积是多少?
  1. A.4
  2. B.8
  3. C.10
  4. D.12
  5. E.16

正确答案 · Correct Answer

E

解析 · Solution

Solution 1

First, we need to find the coordinates where the graphs intersect. We want the points x and y to be the same. Thus, we set 5=x+1, and get x=4. Plugging this into the equation, y=1-x, y=5, and y=1+x intersect at (4,5), we call this line x. Doing the same thing, we get x=-4. Thus, y=5. Also, y=5 and y=1-x intersect at (-4,5), and we call this line y. It's apparent the only solution to 1-x=1+x is 0. Thus, y=1. y=1-x and y=1+x intersect at (0,1), we call this line z. Using the Shoelace Theorem we get:
((20−4)−(−20+4)2)=322=16\left( \frac{(20-4)-(-20+4)}{2} \right)= \frac{32}{2} = 16
So, our answer is 16. We might also see that the lines yy and xx are mirror images of each other. This is because, when rewritten, their slopes can be multiplied by -1 to get the other. As the base is horizontal, this is an isosceles triangle with base 8, as the intersection points have a distance of 8. The height is 5-1=4, so 4⋅82=16.\displaystyle \frac{4\cdot 8}{2} = 16. by WrenMath Warning: Do not use the distance formula for the base then use Heron's formula. It will take you half of the time you have left!

继续练习

真题版权归属 MAA(AMC)。本站仅供学习用途。版权方如有异议,请联系本站予以下架。

Problems © MAA (AMC). This site is for study purposes only. Rightsholders may contact us for removal.