Question

pirâmide reta de base quadrada= {ARESTA DA BASE 8cm}

(ARESTA LATERAL 10cm)


1°.Calcule

A) area lateral

B) area total

C) volume

Answer 1

${l}^{2}={ap}^{2}+{(\frac{a}{2})}^{2}$

${10}^{2}={ap}^{2}+{(\frac{8}{2})}^{2}$

$100={ap}^{2}+16 \\ {ap}^{2}=100-16$

${ap}^{2}=84 \\ ap=\sqrt{84} \\ap=2\sqrt{21}cm$

$m=\frac{8}{2} \\ m=4cm$

${ap}^{2}={h}^{2}+{m}^{2} \\ 84={h}^{2}+{4}^{2}$

${h}^{2}=84-16 \\{h}^{2}=68 \\ h=2\sqrt{17}cm$

a)

$Al=4.\frac{1}{2}.8.2\sqrt{21} \\ Al=32\sqrt{21}{cm}^{2}$

b)

$Ab={8}^{2} \\ Ab=64{cm}^{2}$

$At=Al+Ab \\ At=32\sqrt{21}+64 \\ At=32(\sqrt{21}+2)$

c)

$V=\frac{1}{3}.Ab.h \\ V=\frac{1}{3}.64.2\sqrt{17}$

$V=\frac{128\sqrt{17}}{3} {cm}^{3}$