Question

直线 $y=5$ 、 $y=1+x$ 、 $y=1-x$ 围成的三角形面积是多少？

What is the area of the triangle formed by the lines $y=5,$ $y=1+x,$ and $y=1-x?$

$4$

$8$

$10$

$12$

$16$

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先求三角形的顶点。直线 $y = 5$ 与另外两条直线的交点为 $(4, 5)$ 和 $(-4, 5)$ 。

求第三个交点时，令两条斜线相等，得到 $1 + x = 1 - x \Rightarrow x = 0.$

所以第三个顶点是 $(0, 1)$ 。

前两个交点关于 $y$ 轴对称，因此这个三角形是等腰三角形。

底边长为 $2 \cdot 4 = 8$ ，高为 $5 - 1 = 4$ 。三角形面积为 $\dfrac{1}{2} \cdot 8 \cdot 4 = 16.$

正确答案是 E。

First, let us find the vertices of the triangle. The intersections between $y = 5$ and the other two lines are $(4, 5)$ and $(-4, 5)$ .

To find the third intersection, we can equate the two equations to get $1 + x = 1 - x \Rightarrow x = 0.$

This gives us the point $(0, 1).$

Note that the first two intersection points are mirrored across the $y$ -axis. This tells us that the triangle is isosceles.

The base is therefore $2 \cdot 4 = 8,$ and the height is $5 - 1 = 4.$ The area of the triangle is therefore $\dfrac{1}{2} \cdot 8 \cdot 4 = 16.$

Thus, the correct answer is E.