2011 AMC 10B 第 17 题

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17.

在给定圆中,直径 EB\overline{EB} 平行于 DC\overline{DC},且 AB\overline{AB} 平行于 ED\overline{ED}。角 AEBAEBABEABE 的比为 4:54 : 5。角 BCDBCD 的度数是多少?

In the given circle, the diameter EB\overline{EB} is parallel to DC,\overline{DC}, and AB\overline{AB} is parallel to ED.\overline{ED}. The angles AEBAEB and ABEABE are in the ratio 4:5.4 : 5. What is the degree measure of angle BCD?BCD?

120120

125125

130130

135135

140140

答案:C
知识点:圆周角平行线导角
难度评级:1670
解答:

因为 EBEB 是直径,EAB=90\angle EAB=90^\circ。由 AEB:ABE=4:5\angle AEB:\angle ABE=4:5,得 AEB=40\angle AEB=40^\circABE=50\angle ABE=50^\circ

因为 ABEDAB\parallel ED,所以 DEB=50\angle DEB=50^\circ。又 EBDCEB\parallel DC,四边形 EBCDEBCD 是等腰梯形,所以 BCD=CDE\angle BCD=\angle CDE

DEBDEBCDECDE 互补,所以 CDE=18050=130\angle CDE=180^\circ-50^\circ=130^\circ,因此 BCD=130\angle BCD=130^\circ

所以正确答案是 C

Since EBEB is a diameter, EAB=90\angle EAB=90^\circ. The ratio AEB:ABE=4:5\angle AEB:\angle ABE=4:5 then gives AEB=40\angle AEB=40^\circ and ABE=50\angle ABE=50^\circ.

Because ABEDAB\parallel ED, DEB=50\angle DEB=50^\circ. Since EBDCEB\parallel DC, quadrilateral EBCDEBCD is an isosceles trapezoid, so BCD=CDE\angle BCD=\angle CDE.

Angles DEBDEB and CDECDE are supplementary, so CDE=18050=130\angle CDE=180^\circ-50^\circ=130^\circ. Hence BCD=130\angle BCD=130^\circ.

Thus, C is the correct answer.

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