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Answer 1

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Answer 2

Salut!

35. Calculati aria laterala a unei prisme triunghiulare regulate cu aria bazei de $\frac{25\sqrt3}{4}$ si inaltimea=5.

$A_{b}=\frac{l^2\sqrt3}{4}\\\\\frac{25\sqrt3}{4} =\frac{l^2\sqrt3}{4} \\\\100\sqrt3=4\sqrt3*l^2|:4\sqrt3\\\\25=l^2\\\\l=\sqrt{25}\\\\l=5$

$A_{l}=P_{b}*h$

$P_{b}$=3l=3*5=15

$A_{l}=P_{b}*h\\\\A_{l}=15*5\\\\A_{l}=75$

36. Volumul unei prisme triunghiulare regulate este egal cu 144√3, iar inaltimea prismei este egala cu 12. Calculati aria totala a prismei.

$V_{prismei}=A_{b}*h\\\\144\sqrt3=A_{b}*12|:12\\\\12\sqrt3=A_{b}$

$A_{b}=\frac{l^2\sqrt3}{4}\\\\\frac{12\sqrt3}{1}=\frac{l^2\sqrt3}{4} \\\\48\sqrt3=l^2\sqrt3|:\sqrt3\\\\48=l^2\\\\l=\sqrt{48} \\\\l=4\sqrt3$

$A_{t}=A_{l}+2*A_{b}$

$A_{l}=P_{b}*h\\\\A_{l}=(4\sqrt3*3)*12\\\\A_{l}=12\sqrt3*12\\\\A_{l}=144\sqrt3$

$A_{b}=\frac{l^2\sqrt3}{4} \\\\A_{b}=\frac{(4\sqrt3)^2*\sqrt3}{4} \\\\A_{b}=\frac{48\sqrt3}{4} \\\\A_{b}=12\sqrt3$

$A_{t}=A_{l}+2*A_{b}\\\\A_{t}=144\sqrt3+2*12\sqrt3\\\\A_{t}=144\sqrt3+24\sqrt3\\\\A_{t}=164\sqrt3$

Succes!