Question

Find the capacity of a closed rectangular cistern whose length is 8 m and breadth is 6 m and depth 2.5 m also find the area of the iron sheet required to make the cistern​

Answer 1

$volume \: of \: cuboid = 8 \times 6 \times 2.5 \\ = 120 {cm}^{3} = capacity \\ \\ area \: of \: iron \: sheet \: = tsa \\ = 2((8 \times 6) + (6 \times 2.5) + (2.5 \times 8)) \\ = 2 (48 + 15 + 20) \\ = 2(83) \\ = 166 {cm}^{2}$

hope it helps....

please mark me as brainliest answer....

Answer 2

$\Large{ \mathfrak{ \underline{ Question : }}} \\ \\ \textsf {Find the capacity of a closed rectangular } \\ \textsf{cistern whose length is 8 m, breadth 6 m } \\ \textsf{and depth 2.5 m. Also, find the area of the } \\ \textsf{iron sheet requires making the cistern.} \\ \\ \Large{ \mathfrak{ \underline{ Answer : }}} \\ \\ {\underline{\boxed{ \textsf {Capacity of Closed Rectangular Cistern = 120 m} \sf {}^{3} }}} \\ {\underline{\boxed{ \textsf {Area of the iron sheet required = 166} \sf \: {m}^{2} }}} \\ \\ \Large{ \mathfrak{ \underline{ Solution : }}} \\ \\ \sf Length \: of \: Cistern = 8 m \\ \sf Breadth \: of \: Cistern = 6 m \\ \sf Height \: of \: Cistern = 2.5 m \\ \\ \textsf{Therefore, the capacity of the cistern = Volume of the cistern} \\ \\ \sf Volume \: of \: cistern \\ \qquad \sf= l × b × h \\ \qquad \sf = 8 \times 6 \times 2.5 \\ \qquad \sf = 120 \: {m}^{3} \\ \\ \textsf{Area of the iron sheet required = Total surface area of the cistern} \\ \\ \sf ∴Total \: surface \: area \\ \qquad \sf = 2(lb + bh + hl) \\ \qquad \sf = 2 \{(8 × 6) + (6 × 2.5) + (2.5 × 8) \} \\ \qquad \sf = 2(48 + 15 + 20) \\ \qquad \sf = 2 \times 83 \\ \qquad \sf = 166 \: {m}^{2}$