Question

find the total surface area of a cubiod
shaped solid of length 8m,breadth 6m and height 5cm
​

Answer 1

${\huge{\star{\underbrace{\tt{\red{Answer}}}}}}{\huge{\star}}⋆ </p><p>Answer</p><p> </p><p> ⋆</p><p></p><p>\Large{\blue{\tt{\underline{\underline{Given \ :-}}}}} </p><p>Given :−</p><p> </p><p> </p><p> </p><p> </p><p></p><p>{\sf{❥ \ Area \ of \ square \ = \ 81cm²}}❥ Area of square = 81cm²</p><p></p><p>\Large{\green{\tt{\underline{\underline{To \ find \ :-}}}}} </p><p>To find :−</p><p> </p><p> </p><p> </p><p> </p><p></p><p>{\sf{✿ \ Circumference \ of \ circle}}✿ Circumference of circle</p><p></p><p>\Large{\orange{\tt{\underline{\underline{Solution \ :-}}}}} </p><p>Solution :−</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}{\tt{The \ side \ of \ square \ will \ be \ equal \ to }}\\ {\tt{diameter \ of \ circle}}\end{gathered} </p><p>The side of square will be equal to</p><p>diameter of circle</p><p> </p><p> </p><p></p><p>{\bf{\large{✯Area \ of \ square \ = \ Side ²}}}✯Area of square = Side²</p><p></p><p>{\sf{↣ \ 81 \ = \ Side²}}↣ 81 = Side²</p><p></p><p>{\sf{↣ \ Side \ = \ \sqrt{81}}}↣ Side = </p><p>81</p><p> </p><p> </p><p></p><p>{\sf{↣ \ Side \ = \ 9cm}}↣ Side = 9cm</p><p></p><p>________________________________________</p><p></p><p>Now,</p><p></p><p>Diameter of circle = 9cm</p><p></p><p>{\boxed{\sf \ ✯ \ Circumference \ of \ circle \ = \ \pi × diameter}} </p><p> ✯ Circumference of circle = π×diameter</p><p> </p><p> </p><p></p><p>{\sf{\longmapsto \ Circumference \ = \ \frac{22}{7} × 9}}⟼ Circumference = </p><p>7</p><p>22</p><p> </p><p> ×9</p><p></p><p>{\sf{\longmapsto \ Circumference \ = \ 28.28cm}}⟼ Circumference = 28.28cm</p><p></p><p>\begin{gathered} \\ \end{gathered} </p><p> </p><p> </p><p></p><p>▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬</p><p></p><p>\Large\mathcal{\fcolorbox{lime}{black}{\red{Hope it's help ⚓⚓}}} </p><p>Hope it’s help ⚓⚓</p><p> </p><p> </p><p></p><p>\large\mathcal{\fcolorbox{yellow}{teal}{\orange{Please mark me as brainliest (. ❛ ᴗ ❛.)}}} </p><p>Please mark me as brainliest (. ❛ ᴗ ❛.)</p><p>$