Помогите,решить! Задачи 4,5,6,7. Created by Veronira. geometriya-ru

Помогите,решить! Задачи 4,5,6,7...

Veronira @Veronira

September 2021 1 4 Report

Помогите,решить!
Задачи 4,5,6,7

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oganesbagoyan

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4) S(ABCD) =H*(AD+BC)/2 ;

Проведем BE⊥AD , H = BE ; AE =(AD -BC)/2 ⇒AD =BC +2AE ;
S(ABCD) =BE*(AD+BC)/2 =BE*(BC +2AE+BC)/2 =(BC+AE)*BE .
∠ABE =90° -∠A =90° -60°=30°. AE =AB/2(как катет против угла =30°). AE =4/2 =2 .
BE =√(AB² -AE²) =√(4² -2²)=2√3. * или сразу BE=AB*sin∠A =4*sin60° =4*(√3)/2 =2√3 * .
S(ABCD) =(BC+AE)*BE =(5+2)*2√3=14√3.
5) S(BOC) /S(DOA) =(BC/DA)² ( как подобные треугольники).
S(BOC) /S(DOA) =(4/8)² =1/4 . * * * 0,25 * * *
6) ME - KM =AD/2 -BC/2=(AD-BC)/2 ;
|ME - KM|=|(AD-BC)/2| =|(8-6)/2| =1.
7) NP = NK - PK = AD/2 -BC/2 =(AD - BC)/2 =(10 -6)/2 =2.
* * * средние линии треугольников ACD и ABC * * *
ΔAOD ~ ΔNOP :
AO/NO =AD/NP ⇔ (AN+NO)/NO =10/2⇔ (AC/2+NO)/NO =5 ;
(6+NO)/NO =5 ⇔6+NO =5NO ⇔6=4NO ⇒NO=1,5.

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