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How to solve the sixth problem of mathematics in 20 10 Hengyang senior high school entrance examination?
Analysis: In? In ABCD, AB=CD=6, AD=BC=9, and the bisector of ∠BAD intersects BC at point E, and it can be obtained that △ADF is an isosceles triangle, and AD = df = 9;; △ADF is an isosceles triangle, AB=BE=6, so CF = 3;; In △ABG, BG⊥AE, AB=6, BG=4 times the root number 2, which leads to AG=2, while △ADF is an isosceles triangle, BG⊥AE, so AE=2AG=4, so the circumference of △ABE is equal to 16, which is determined by? ABCD can get △CEF∽△BEA, and the similarity ratio is 1: 2, so the circumference of △CEF is 8, so A is chosen.

Solution: ∫In? In ABCD, AB=CD=6, AD=BC=9, and the bisector of ∠BAD intersects BC at point E,

∴△ADF is isosceles triangle, △ABE is isosceles triangle, and ad = df = 9;;

AB = BE = 6,

∴cf=3;

∴, BG⊥AE, AB=6, BG=4 in △ABG, multiplied by the radical number 2, can be obtained as follows: AG=2.

And BG⊥AE,

∴AE=2AG=4,

∴△∴△abe's circumference is equal to 16,

Again? Accelerated business collection and delivery system (adopted by the United States post office)

∴△CEF∽△BEA, the similarity ratio is 1: 2,

The circumference of ∴△CEF is 8.

So choose a.