r₁ = 10 cm
r₂ = 5 cm
t₁ = 24 cm
t₂ = 12 cm
[tex]\sf s_{1} =\sqrt{r_{1}^{2}+t_{1}^{2}} =\sqrt{10^{2}+24^{2} } =\sqrt{100+576} =\sqrt{676} =26\:cm\\\\s_{2} =\sqrt{r_{2}^{2}+t_{2}^{2}} =\sqrt{5^{2}+12^{2} } =\sqrt{25+144} =\sqrt{169}=13 \:cm[/tex]
Luas permukaan:
Lp = L selimut kerucut bsr + L selimut kerucut kcl + L alas kerucut bsr - L alas kerucut kcl
= πr₁s₁ + πr₂s₂ + πr₁² - πr₂²
= (3,14 x 10 x 26) + (3,14 x 5 x 13) + (3,14 x 10²) - (3,14 x 5²)
= 816,4 + 204,1 + 314 - 78,5
= 1256 cm²
Volume:
V = V kerucut besar - V kerucut kecil
= πr₁²t₁ - πr₂²t₂
= (3,14 x 10² x 24) - (3,14 x 5² x 12)
= 7536 - 942
= 6594 cm³
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
r₁ = 10 cm
r₂ = 5 cm
t₁ = 24 cm
t₂ = 12 cm
[tex]\sf s_{1} =\sqrt{r_{1}^{2}+t_{1}^{2}} =\sqrt{10^{2}+24^{2} } =\sqrt{100+576} =\sqrt{676} =26\:cm\\\\s_{2} =\sqrt{r_{2}^{2}+t_{2}^{2}} =\sqrt{5^{2}+12^{2} } =\sqrt{25+144} =\sqrt{169}=13 \:cm[/tex]
Luas permukaan:
Lp = L selimut kerucut bsr + L selimut kerucut kcl + L alas kerucut bsr - L alas kerucut kcl
= πr₁s₁ + πr₂s₂ + πr₁² - πr₂²
= (3,14 x 10 x 26) + (3,14 x 5 x 13) + (3,14 x 10²) - (3,14 x 5²)
= 816,4 + 204,1 + 314 - 78,5
= 1256 cm²
Volume:
V = V kerucut besar - V kerucut kecil
= πr₁²t₁ - πr₂²t₂
= (3,14 x 10² x 24) - (3,14 x 5² x 12)
= 7536 - 942
= 6594 cm³