HMMT 十一月 2012 · GEN 赛 · 第 6 题
HMMT November 2012 — GEN Round — Problem 6
题目详情
英文原题
[5]ABCD is a parallelogram satisfying AB = 7, BC = 2, and ∠DAB = 120◦. Parallelogram
ECFA is contained in ABCD and is similar to it. Find the ratio of the area of ECFA to the area of
ABCD .
解析
英文解析
- [ 5 ] ABCD is a parallelogram satisfying AB = 7, BC = 2, and ∠ DAB = 120 . Parallelogram ECF Ais contained in ABCD and is similar to it. Find the ratio of the area of ECF A to the area of ABCD .
Answer: First, note that BD is the long diagonal of ABCD , and AC is the long diagonal of 39
ECF A . Because the ratio of the areas of similar figures is equal to the square of the ratio of their side 67
AC2
lengths, we know that the ratio of the area of ECF A to the area of ABCD is equal to the ratio .
BD2
2 2 2 ° 2
Using law of cosines on triangle ABD , we have BD = AD + AB − 2( AD )( AB ) cos(120 ) = 2 +
7 − 2(2)(7)( − ) = 67.21
2 2 2 ° 2 22
Using law of cosines on triangle ABC , we have AC = AB + BC − 2( AB )( BC ) cos(60 ) = 7 + 2 −
2(7)(2)( ) = 39.1
AC 3922
Finally, = .
BD 672
General Test