Question

can anyone answer the question​

Answer 1

$\large\textsf{ }$

$\maltese\LARGE\textsf{\underline{ GiVeN :-}}$

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{Base of the tile = 24cm}$

$\qquad\tt{:}\longrightarrow\large\textsf{Height of the tile = 0.1 m = 10cm}$

$\qquad\tt{:}\longrightarrow\large{\sf{Area\;\; of\;\; the\;\; floor\;\; = 1080 m^2 = 10800000 cm^2}}$

$\large\textsf{ }$

$\maltese\LARGE\textsf{\underline{ To FiNd :-}}$

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{No. of tiles required to cover the floor = ?}$

$\large\textsf{ }$

$\maltese\LARGE\textsf{\underline{ SoLuTioN :-}}$

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{ {\large\textsf{Area}}_{\small\textsf{( \; Parallelogram \; )}} = \large\sf{Base \times Height} }$

$\large\textsf{ }$

$\maltese\LARGE\textsf{\underline{ SoLuTiOn :-}}$

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\sf{Area \;\;of\;\; tiles = 24 \times 10}$

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\boxed{\large\sf{= 240\;\; cm^2}}$

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$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{No. of tiles = \cfrac{\large\textsf{Area of the floor}}{\large\textsf{Area of the tiles}} }\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{ = \cancel\cfrac{\large\textsf{10800000}}{\large\textsf{240}} }\\\\\\\qquad\tt{:}\longrightarrow\boxed{\therefore\large\textsf{ \; \; No. of tiles = 45000}}$

$\large\textsf{ }$

$\maltese\LARGE\textsf{\underline{ MoRe fOrMuLaS :-}}$

$\large\textsf{ }$

$\large\textsf{ {\large\textsf{L.S.A.}}_{\large\textsf{( \; Cuboid \; )}} = \large\textsf{2h ( l + b )} }$

$\large\textsf{ {\large\textsf{T.S.A.}}_{\large\textsf{( \; Cuboid \; )}} = \large\textsf{2 ( lb + bh + hl )} }$

$\large\textsf{ {\large\textsf{Volume}}_{\large\textsf{( \; Cuboid \; )}} = \large\sf{l\times b\times h} }$

$\large\textsf{ }$

$\large\textsf{ {\large\textsf{L.S.A.}}_{\large\textsf{( \; Cube \; )}} = \large\sf{4\times l^2} }$

$\large\textsf{ {\large\textsf{T.S.A.}}_{\large\textsf{( \; Cube \; )}} = \large\sf{6 \times l^2} }$

$\large\textsf{ {\large\textsf{Volume}}_{\large\textsf{( \; Cube \; )}} = \large\sf{l^2} }$

$\large\textsf{ }$

$\large\textsf{ {\large\textsf{C.S.A.}}_{\large\textsf{( \; Cylinder \; )}} = \large\sf{2 \times \pi \; rh} }$

$\large\textsf{ {\large\textsf{T.S.A.}}_{\large\textsf{( \; Cylinder \; )}} = \large\sf{2\pi r \times ( r + h )} }$

$\large\textsf{ {\large\textsf{Volume}}_{\large\textsf{( \; Cylinder \; )}} = \large\sf{\pi r^2h} }$

$\large\textsf{ }$

$\large\textsf{ {\large\textsf{C.S.A.}}_{\large\textsf{( \; Cone \; )}} = \large\sf{\pi rl} }$

$\large\textsf{ {\large\textsf{T.S.A.}}_{\large\textsf{( \; Cone \; )}} = \large\sf{\pi r \times ( r + l )} }$

$\large\textsf{ {\large\textsf{Volume}}_{\large\textsf{( \; Cone \; )}} } \large\textsf{ = \cfrac{\large\textsf{1}}{\large\textsf{3}} }\large\sf{\times \pi r^2h}$

$\large\textsf{ }$

$\large\textsf{ {\large\textsf{T.S.A.}}_{\large\textsf{( \; Sphere \; )}} = \large\sf{4\pi r^2} }$

$\large\textsf{ {\large\textsf{Volume}}_{\large\textsf{( \; Sphere \; )}} } \large\textsf{ = \cfrac{\large\textsf{4}}{\large\textsf{3}} }\large\sf{\times \pi r^3}$

$\large\textsf{ }$

Answer 2

therefore 45000 tiles are needed

hope it helps you