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A

$?\frac {4}{25}?$

6

1：16

$解：?(1)?周長之比等于相似比$

$∴?C_{五邊形ABCDE}：??C_{五邊形A'B'C'D'E'}=3：??2?$

$∵五邊形?ABCDE = 72\ \mathrm {cm}?$

$∴?C_{五邊形A'BC'D'E'} = 48\ \mathrm {cm}?$

$?(2)?面積之比等于相似比的平方$

$∴?S _{五邊形ABCDE}：??S_{ 五邊形A'B'C'D'E'}=9：??4?$

$∵?S_{ 五邊形A'B'C'D'E' }= 120\ \mathrm {cm}2?$

$∴?S _{五邊形ABCDE} = 270\ \mathrm {cm}2?$

$解：∵點?{A}_1、??{A}_2，?點?{B}_1、??{B}_2，?$

$點?{C}_1、??{C}_2?分別是?△ABC?的邊?BC、??CA、?? AB?的三等分點$

$∴?\frac {B{C}_2}{AB}=\frac {B{A}_1}{BC}=\frac {1}{3}?$

$∵?∠B=∠B?$

$∴?△B{C}_2{A}_1∽△BAC?$

$同理可得：?△A{B}_2{C}_1∽△ACB、??△C{B}_1{A}_2∽△ CAB，?且相似比均為?1 ： 3?$

$∴?{A}_1{C}_2={B}_1{B}_2=\frac {1}{3}AC?$

$?{A}_2{B}_1={C}_1{C}_2=\frac {1}{3}AB?$

$?{C}_1{B}_2={A}_1{A}_2=\frac {1}{3}BC?$

$∴?{C}_{六邊形}=\frac {2}{3}{C}_{△ABC}=\frac {2}{3}l?$

$解：由題意可知，?An，??Bn?是?AC、??BC?中最靠近點?C?的?{2}^{n}?等分點$

$則?{S}_{四邊形AnABBn}=\frac {3}{4}+\frac {3}{42}+\frac {3}{43}+···+\frac {3}{{4}^{n}}?$

$∵?An，??Bn?是?AC、??BC?中最靠近點?C?的?{2}^{n}?等分點$

$∴?△ABC∽△AnBnC，?相似比為?{2}^{n}：??1?$

$∴?S_{△ABC} ：?? S_{△AnBnC}={4}^{n}：??1?$

$∴?S_{△AnBnC} =\frac {1}{{4}^{n}}S_{△ABC} ?$

$又∵?S_{四邊形AnABB}= S_{△ABC}- S_{△AnBnC}?$

$∴?S_{四邊形AnABBn}=1-\frac {1}{{4}^{n}}?$

$∴?\frac {3}{4}+\frac {3}{42}+\frac {3}{43}+···+\frac {3}{{4}^{n}}=1-\frac {1}{{4}^{n}}?$

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