Question

Tsa and volume of frustum of a cone , CUBOID and cube

Answer 1

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$\bf\Huge\red{\mid{\overline{\underline{ ANSWER }}}\mid }$

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$\Large\fbox{\color{purple}{QUESTION}}$

SURFACE AREA VOLUME FORMULAS

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$\Large\fbox{\color{purple}{ SOLUTION }}$

$\Large\mathcal\green{FRUSTUM}$

$\implies \: tsa = \pi \: l(r1 + r2) + \pi \: {r1}^{2} + \pi {r2}^{2}$

$\implies volume = \frac{1}{3}\pi \: h( {r1}^{2} + r1.r2 + {r2}^{2} )$

$\Large\mathcal\purple{CUBOID}$

$\implies \: lsa = 2(l + b)h \\ \\ \: \implies \: tsa = 2(lb + bl + hl) \\ \\ \implies \: volume \: = l \times b \times h$

$\Large\mathcal\blue{CUBE}$

$\implies \: lsa = {4a}^{2} \\ \\ \implies \: tsa = {6a}^{2} \\ \\ \implies \: volume = {a}^{3}$

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Answer 2

$\pi l(r1 + r2) + \pi r12 + \pi r22 \\ volume \: of \: frustam \: of \: a \: cone = (1|3)\pi h(r12 + r22 + r1 \: r2) \\ cube = 6 {a}^{2} \\ cuboid = 2(lb + bh + hl)</p><p></p><p>$