【基础级练习题69】1 p1 y2 }/ |( D$ p4 m
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求:! j7 E! M8 k( `# z
1. a之距离为何?
* `5 T5 _& ~$ i, q, B (A) 51.4023 (B) 51.4230 (C) 51.4302
4 J1 t; ^% f( U6 Z7 Z2. b之距离为何?, U9 `2 i4 K( }
(A) 58.4176 (B) 58.4617 (C) 58.4761
% s: A: h& q4 y8 o, s0 s1. c之半径值为何?& N( {2 U% I6 y( w" i1 y
(A) 15.6275 (B) 15.6527 (C) 15.6752
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; D6 L) ~- a5 A7 O3 J2 h" v. m【基础级练习题69】标准答案:0 n. u+ {2 k' v
1. a之距离 = 51.4230,就是(B)
, k7 {: g5 i$ A2 A3 G3 i T2. b之距离 = 58.4761,就是(C)
" g* ` S/ x4 H- {3. c之半径值 = 15.6275,就是(A)/ Y$ S2 t9 T" k. u+ N; e
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【基础级练习题69】解题思路:$ a O A9 W4 j2 H5 `
本题比较简单,画三边形,大家都会。关键是三边形内那个长度; k1 H& X1 E+ s+ X
为40的线段,线段到顶角的距离要相等,这里用了角平分线并往
* @( Q" J6 F! S两边各偏移20的画法来解决。7 W+ q4 n% o+ z _1 u0 W
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