2024 AMC 8 第 11 题

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11.

ABC\bigtriangleup ABC 的三个顶点为 A(5,7)A(5, 7)B(11,7)B(11, 7)C(3,y)C(3, y),其中 y>7y > 7ABC\bigtriangleup ABC 的面积为 1212yy 的值是多少?

The coordinates of ABC\bigtriangleup ABC are A(5,7),A(5, 7), B(11,7),B(11, 7), and C(3,y),C(3, y), with y>7.y > 7. The area of ABC\bigtriangleup ABC is 12.12. What is the value of y?y?

88

99

1010

1111

1212

答案:D
知识点:坐标几何三角形面积一次方程
难度评级:960
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文字解答:

AB\overline{\rm AB} 是水平线段,长度为 115=611-5=6。三角形面积为 1212,所以高为 2(12)6=4\dfrac{2(12)}{6}=4。因为 y>7y>7,所以 yy 必须是 7+4=117+4=11

正确答案是 D

Consider the base of the triangle to be AB\overline{\rm AB} which has length 115=611-5=6. Given that the area of the triangle is 1212, its height must be of length 2(12)6=4\dfrac{2(12)}{6}=4. Since y>7y>7, then yy must be 7+4=117+4=11.

Thus, D is the correct answer.

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