Bardzo proszę o pomoc w rozwiązaniu tych zadan, nie mam pojęcia jak je zrobic
zad1. Punkt E lezy na ramieniu BC trapezu ABCD, w którym AB||CD. Udowodnij, że kątAED=kątBAE+kątCDE.
zad2. Dany jest prostokąt o bokach a i b oraz prostokąt o bokach c i d. Długość boku c to 90% długości boku a. Długość boku d to 120% długości boku b. Oblicz, ile procent pola prostokąta o bokach a i b stanowi pole prostokata o bokach c i d.
zad3. Wysokość ostrosłupa prawidłowego czworokątnego jest równa 8. Krawędź boczna jest nachylona do płaszczyzny podstawy pod kątem 40(stopni). Oblicz objętość tego ostrosłupa.
Bardzo proszę o pomoc, zalezy mi na tych rozwiązaniach.
trapez, prostokąt..
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JednaZWieluxp
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trapez, prostokąt..
Zad 1
|\(\displaystyle{ \sphericalangle}\)EAD|=180\(\displaystyle{ ^{o}}\)-a-b-c
|\(\displaystyle{ \sphericalangle}\)AED|=180\(\displaystyle{ ^{o}}\)-c-(180\(\displaystyle{ ^{o}}\)-a-b-c)
|\(\displaystyle{ \sphericalangle}\)AED|=a+b
|\(\displaystyle{ \sphericalangle}\)AED|=|\(\displaystyle{ \sphericalangle}\)BAE|+|\(\displaystyle{ \sphericalangle}\)CDE|
-- 26 mar 2010, o 21:07 --
Zad 2
P\(\displaystyle{ _{1}}\) - pole prostokąta o bokach a i b
P\(\displaystyle{ _{2}}\) - pole prostokąta o bokach c i d
c=0,9a d=1,2b
P\(\displaystyle{ _{2}}\)=a\(\displaystyle{ \cdot}\)b
P\(\displaystyle{ _{2}}\)=c\(\displaystyle{ \cdot}\)d=0,9a\(\displaystyle{ \cdot}\)1,2b=1,08\(\displaystyle{ \cdot}\)a\(\displaystyle{ \cdot}\)b=108\(\displaystyle{ \%}\)P\(\displaystyle{ _{2}}\)-- 26 mar 2010, o 22:02 --H=8
\(\displaystyle{ \frac{0,5d}{H}}\)=ctg 40\(\displaystyle{ ^{o}}\)
0,5d=H\(\displaystyle{ \cdot}\)ctg 40\(\displaystyle{ ^{o}}\)
d=2H\(\displaystyle{ \cdot}\)ctg 40\(\displaystyle{ ^{o}}\)
d=a\(\displaystyle{ \sqrt{2}}\)
a\(\displaystyle{ \sqrt{2}}\)=2H\(\displaystyle{ \cdot}\)ctg 40\(\displaystyle{ ^{o}}\)
2a\(\displaystyle{ ^{2}}\)=4H\(\displaystyle{ ^{2}}\)\(\displaystyle{ \cdot}\)ctg\(\displaystyle{ ^{2}}\) 40\(\displaystyle{ ^{o}}\)
a\(\displaystyle{ ^{2}}\)=4H\(\displaystyle{ ^{2}}\)\(\displaystyle{ \cdot}\)ctg\(\displaystyle{ ^{2}}\) 40\(\displaystyle{ ^{o}}\)\(\displaystyle{ \cdot}\)\(\displaystyle{ \frac{1}{2}}\)
V=\(\displaystyle{ \frac{1}{3}}\)\(\displaystyle{ \cdot}\)a\(\displaystyle{ ^{2}}\)\(\displaystyle{ \cdot}\)H
V=\(\displaystyle{ \frac{1}{3}}\)\(\displaystyle{ \cdot}\)4H\(\displaystyle{ ^{2}}\)\(\displaystyle{ \cdot}\)ctg\(\displaystyle{ ^{2}}\) 40\(\displaystyle{ ^{o}}\)\(\displaystyle{ \cdot}\)\(\displaystyle{ \frac{1}{2}}\)\(\displaystyle{ \cdot}\)H
V=\(\displaystyle{ \frac{1}{6}}\)\(\displaystyle{ \cdot}\)4H\(\displaystyle{ ^{3}}\)\(\displaystyle{ \cdot}\)ctg\(\displaystyle{ ^{2}}\) 40\(\displaystyle{ ^{o}}\)
V=\(\displaystyle{ \frac{1}{6}}\)\(\displaystyle{ \cdot}\)4\(\displaystyle{ \cdot}\)8\(\displaystyle{ ^{3}}\)\(\displaystyle{ \cdot}\)ctg\(\displaystyle{ ^{2}}\) 40\(\displaystyle{ ^{o}}\)
V=85\(\displaystyle{ \frac{1}{3}}\)\(\displaystyle{ \cdot}\)ctg\(\displaystyle{ ^{2}}\) 40\(\displaystyle{ ^{o}}\)
|\(\displaystyle{ \sphericalangle}\)EAD|=180\(\displaystyle{ ^{o}}\)-a-b-c
|\(\displaystyle{ \sphericalangle}\)AED|=180\(\displaystyle{ ^{o}}\)-c-(180\(\displaystyle{ ^{o}}\)-a-b-c)
|\(\displaystyle{ \sphericalangle}\)AED|=a+b
|\(\displaystyle{ \sphericalangle}\)AED|=|\(\displaystyle{ \sphericalangle}\)BAE|+|\(\displaystyle{ \sphericalangle}\)CDE|
-- 26 mar 2010, o 21:07 --
Zad 2
P\(\displaystyle{ _{1}}\) - pole prostokąta o bokach a i b
P\(\displaystyle{ _{2}}\) - pole prostokąta o bokach c i d
c=0,9a d=1,2b
P\(\displaystyle{ _{2}}\)=a\(\displaystyle{ \cdot}\)b
P\(\displaystyle{ _{2}}\)=c\(\displaystyle{ \cdot}\)d=0,9a\(\displaystyle{ \cdot}\)1,2b=1,08\(\displaystyle{ \cdot}\)a\(\displaystyle{ \cdot}\)b=108\(\displaystyle{ \%}\)P\(\displaystyle{ _{2}}\)-- 26 mar 2010, o 22:02 --H=8
\(\displaystyle{ \frac{0,5d}{H}}\)=ctg 40\(\displaystyle{ ^{o}}\)
0,5d=H\(\displaystyle{ \cdot}\)ctg 40\(\displaystyle{ ^{o}}\)
d=2H\(\displaystyle{ \cdot}\)ctg 40\(\displaystyle{ ^{o}}\)
d=a\(\displaystyle{ \sqrt{2}}\)
a\(\displaystyle{ \sqrt{2}}\)=2H\(\displaystyle{ \cdot}\)ctg 40\(\displaystyle{ ^{o}}\)
2a\(\displaystyle{ ^{2}}\)=4H\(\displaystyle{ ^{2}}\)\(\displaystyle{ \cdot}\)ctg\(\displaystyle{ ^{2}}\) 40\(\displaystyle{ ^{o}}\)
a\(\displaystyle{ ^{2}}\)=4H\(\displaystyle{ ^{2}}\)\(\displaystyle{ \cdot}\)ctg\(\displaystyle{ ^{2}}\) 40\(\displaystyle{ ^{o}}\)\(\displaystyle{ \cdot}\)\(\displaystyle{ \frac{1}{2}}\)
V=\(\displaystyle{ \frac{1}{3}}\)\(\displaystyle{ \cdot}\)a\(\displaystyle{ ^{2}}\)\(\displaystyle{ \cdot}\)H
V=\(\displaystyle{ \frac{1}{3}}\)\(\displaystyle{ \cdot}\)4H\(\displaystyle{ ^{2}}\)\(\displaystyle{ \cdot}\)ctg\(\displaystyle{ ^{2}}\) 40\(\displaystyle{ ^{o}}\)\(\displaystyle{ \cdot}\)\(\displaystyle{ \frac{1}{2}}\)\(\displaystyle{ \cdot}\)H
V=\(\displaystyle{ \frac{1}{6}}\)\(\displaystyle{ \cdot}\)4H\(\displaystyle{ ^{3}}\)\(\displaystyle{ \cdot}\)ctg\(\displaystyle{ ^{2}}\) 40\(\displaystyle{ ^{o}}\)
V=\(\displaystyle{ \frac{1}{6}}\)\(\displaystyle{ \cdot}\)4\(\displaystyle{ \cdot}\)8\(\displaystyle{ ^{3}}\)\(\displaystyle{ \cdot}\)ctg\(\displaystyle{ ^{2}}\) 40\(\displaystyle{ ^{o}}\)
V=85\(\displaystyle{ \frac{1}{3}}\)\(\displaystyle{ \cdot}\)ctg\(\displaystyle{ ^{2}}\) 40\(\displaystyle{ ^{o}}\)