Answer:

[tex]\qquad\qquad\boxed{\sf \: \sf \: Height \: of \: room = 4.5 \: m \: } \\ \\ [/tex]

Step-by-step explanation:

Given that, the volume of a room is 378 m³ and the area of its floor is 84 m².

We know,

[tex]\sf \: Volume = Area \times Height \\ \\ [/tex]

So,

[tex]\sf \: Height = \dfrac{Volume}{Area} \\ \\ [/tex]

[tex]\sf \: Height = \dfrac{378}{84} \\ \\ [/tex]

[tex]\sf \: Height = \dfrac{54}{12} \: \: \: \{cancel \: by \: 7 \} \\ \\ [/tex]

[tex]\sf \: Height = \dfrac{9}{2} \: \: \: \{cancel \: by \: 6 \} \\ \\ [/tex]

[tex]\sf\implies \sf \: Height = 4.5 \: m \\ \\ [/tex]

Hence,

[tex]\sf\implies \sf \: Height \: of \: room = 4.5 \: m \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\begin{gathered}\: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]