Ответы
Ответ дал:
1
1) ΔАОВ, <BAO=90, <ABO=43 ---> <AOB=90-43=47
2)
![V_1=16,\; \; V_1=\pi R_1^2\cdot H_1\\\\R_2=3R_1,\; \; H_2=\frac{H_1}{4}\\\\V_2=\pi R_2^2\cdot H_2=\pi \cdot 9R_1^2\cdot \frac{H_1}{4}=\frac{9 }{4}\cdot \pi R_1^2\cdot H_1=\frac{9}{4}\cdot V_1=\frac{9}{4}\cdot 16=36 V_1=16,\; \; V_1=\pi R_1^2\cdot H_1\\\\R_2=3R_1,\; \; H_2=\frac{H_1}{4}\\\\V_2=\pi R_2^2\cdot H_2=\pi \cdot 9R_1^2\cdot \frac{H_1}{4}=\frac{9 }{4}\cdot \pi R_1^2\cdot H_1=\frac{9}{4}\cdot V_1=\frac{9}{4}\cdot 16=36](https://tex.z-dn.net/?f=V_1%3D16%2C%5C%3B+%5C%3B+V_1%3D%5Cpi+R_1%5E2%5Ccdot+H_1%5C%5C%5C%5CR_2%3D3R_1%2C%5C%3B+%5C%3B+H_2%3D%5Cfrac%7BH_1%7D%7B4%7D%5C%5C%5C%5CV_2%3D%5Cpi+R_2%5E2%5Ccdot+H_2%3D%5Cpi+%5Ccdot+9R_1%5E2%5Ccdot+%5Cfrac%7BH_1%7D%7B4%7D%3D%5Cfrac%7B9+%7D%7B4%7D%5Ccdot+%5Cpi+R_1%5E2%5Ccdot+H_1%3D%5Cfrac%7B9%7D%7B4%7D%5Ccdot+V_1%3D%5Cfrac%7B9%7D%7B4%7D%5Ccdot+16%3D36)
![3)\; V=\frac{1}{3}\cdot S_{osnov}\cdot H\\\\S_{osnov}=S_{ABC}=\frac{1}{2}\cdot 6\cdot 6\cdot sin60^0=\frac{36}{2}\cdot \frac{\sqrt3}{2}=9\sqrt3\\\\\Delta AOD:\; AO=\frac{2}{3}\cdot h_{ABC}=\frac{2}{3}\cdot \sqrt{6^2-3^2}=\frac{2}{3}\cdot \sqrt{27}=\frac{2}{3}\cdot 3\sqrt3=2\sqrt3\\\\DO=H_{piramidu}=\sqrt{AD^2-AO^2}=\sqrt{(6\sqrt3)^2-(2\sqrt3)^2}=\\\\=\sqrt{36\cdot 3-4\cdot 3}=\sqrt{108-12}=\sqrt{96}=4\sqrt6\\\\V=\frac{1}{3}\cdot 9\sqrt3\cdot 4\sqrt6=\frac{9}{3}\sqrt3\cdot 4\sqrt{3\cdot 2}=\frac{9}{3}\cdot 3\cdot 4\cdot \sqrt2 3)\; V=\frac{1}{3}\cdot S_{osnov}\cdot H\\\\S_{osnov}=S_{ABC}=\frac{1}{2}\cdot 6\cdot 6\cdot sin60^0=\frac{36}{2}\cdot \frac{\sqrt3}{2}=9\sqrt3\\\\\Delta AOD:\; AO=\frac{2}{3}\cdot h_{ABC}=\frac{2}{3}\cdot \sqrt{6^2-3^2}=\frac{2}{3}\cdot \sqrt{27}=\frac{2}{3}\cdot 3\sqrt3=2\sqrt3\\\\DO=H_{piramidu}=\sqrt{AD^2-AO^2}=\sqrt{(6\sqrt3)^2-(2\sqrt3)^2}=\\\\=\sqrt{36\cdot 3-4\cdot 3}=\sqrt{108-12}=\sqrt{96}=4\sqrt6\\\\V=\frac{1}{3}\cdot 9\sqrt3\cdot 4\sqrt6=\frac{9}{3}\sqrt3\cdot 4\sqrt{3\cdot 2}=\frac{9}{3}\cdot 3\cdot 4\cdot \sqrt2](https://tex.z-dn.net/?f=3%29%5C%3B+V%3D%5Cfrac%7B1%7D%7B3%7D%5Ccdot+S_%7Bosnov%7D%5Ccdot+H%5C%5C%5C%5CS_%7Bosnov%7D%3DS_%7BABC%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ccdot+6%5Ccdot+6%5Ccdot+sin60%5E0%3D%5Cfrac%7B36%7D%7B2%7D%5Ccdot+%5Cfrac%7B%5Csqrt3%7D%7B2%7D%3D9%5Csqrt3%5C%5C%5C%5C%5CDelta+AOD%3A%5C%3B+AO%3D%5Cfrac%7B2%7D%7B3%7D%5Ccdot+h_%7BABC%7D%3D%5Cfrac%7B2%7D%7B3%7D%5Ccdot+%5Csqrt%7B6%5E2-3%5E2%7D%3D%5Cfrac%7B2%7D%7B3%7D%5Ccdot+%5Csqrt%7B27%7D%3D%5Cfrac%7B2%7D%7B3%7D%5Ccdot+3%5Csqrt3%3D2%5Csqrt3%5C%5C%5C%5CDO%3DH_%7Bpiramidu%7D%3D%5Csqrt%7BAD%5E2-AO%5E2%7D%3D%5Csqrt%7B%286%5Csqrt3%29%5E2-%282%5Csqrt3%29%5E2%7D%3D%5C%5C%5C%5C%3D%5Csqrt%7B36%5Ccdot+3-4%5Ccdot+3%7D%3D%5Csqrt%7B108-12%7D%3D%5Csqrt%7B96%7D%3D4%5Csqrt6%5C%5C%5C%5CV%3D%5Cfrac%7B1%7D%7B3%7D%5Ccdot+9%5Csqrt3%5Ccdot+4%5Csqrt6%3D%5Cfrac%7B9%7D%7B3%7D%5Csqrt3%5Ccdot+4%5Csqrt%7B3%5Ccdot+2%7D%3D%5Cfrac%7B9%7D%7B3%7D%5Ccdot+3%5Ccdot+4%5Ccdot+%5Csqrt2)
2)
oganesbagoyan:
V =1/3*9√3 *4√6 = 36√2 ||маленькое упущение_ множитель 1/3 ||
Ответ дал:
1
<BOA =90° - <ABO = 90° - 43° = 47°. Это и есть центральный угол.
Градусная мера дуги = 47°.
-------------------------
V =πR²H =16 ;
V₁=πR₁²H₁ =π(3R)²*H/4 =πR²H*(9/4) =16*9/4 =36 .
------------------------
Правильная пирамида SABC: AB = BC =CA =a =6 ;SA=SB =SC =L =6√3.
V =1/3*S*H ;
S =a²(√3)/4 =6²(√3)/4= 9√3 .
H =√(L² - (2/3*h)²) ,где h _высота основания: h =a(√3)/2 =6(√3)/2 =3√3.
H =√((6√3)² - (2/3*3√3)²) =√((6√3)² - (2√3)²) =4√6.
V =1/3*S*H =1/3*9√3 *4√6=12√(3*6) =36√2.
Градусная мера дуги = 47°.
-------------------------
V =πR²H =16 ;
V₁=πR₁²H₁ =π(3R)²*H/4 =πR²H*(9/4) =16*9/4 =36 .
------------------------
Правильная пирамида SABC: AB = BC =CA =a =6 ;SA=SB =SC =L =6√3.
V =1/3*S*H ;
S =a²(√3)/4 =6²(√3)/4= 9√3 .
H =√(L² - (2/3*h)²) ,где h _высота основания: h =a(√3)/2 =6(√3)/2 =3√3.
H =√((6√3)² - (2/3*3√3)²) =√((6√3)² - (2√3)²) =4√6.
V =1/3*S*H =1/3*9√3 *4√6=12√(3*6) =36√2.
Вас заинтересует
1 год назад
1 год назад
1 год назад
1 год назад
6 лет назад
6 лет назад
7 лет назад
7 лет назад