Question

the surface are of cuboid is 758cm2. if lenght and breath are 11cm respectivly find its height

Answer 1

$\qquad\quad\boxed{\bf{\mid{\overline{\underline{\purple{\bigstar\: Correct~Question:-}}}}}\mid}$⠀

• The surface area of a cuboid is 758cm2 it's length and breadth are 14 cm and 11cm find its height

$\qquad\quad\underline{\bf{\mid{\overline{\underline{\purple{\bigstar\: Answer:-}}}}}\mid}$⠀

Given :-

• Surface area of Cuboid = 758 sq. cm

• Given Length of Cuboid = 14 cm

• Given Breadth of Cuboid = 11 cm

To Find : -

• Height of the Cuboid.

Let height of the cuboid be 'h'

As we know that,

• Total Surface Area of Cuboid = 2(lb + bh +hl)

where, l stands for length, b stands for breadth and h stands for height of the cuboid.

Therefore, we can say that --

2(lb + bh + hl) = 758

2([14 × 11]+ [11 × h] + [h × 14]) = 758

2(154 + 11h + 14h) = 758

2(154 + 25h) = 758

154 + 25h = 758 ÷ 2

154 + 25h = 379

25h = 379 - 154

25h = 225

h = 225 ÷ 25

h = 9 cm

$\bf\blue{Therefore,~ Height~ of ~the ~Cuboid~ is~ 9 cm.}$

Answer 2

$\large\textsf{ }$

$\LARGE\mathfrak{\underline{✯\; Given :-}}$

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{Surface Area of the Cuboid = 758 cm²}$

$\qquad\tt{:}\longrightarrow\large\textsf{Length of the Cuboid ( l ) = 14 cm}$

$\qquad\tt{:}\longrightarrow\large\textsf{Breadth of the Cuboid ( b ) = 11 cm}$

$\large\textsf{ }$

$\LARGE\mathfrak{\underline{✯\; To \; \; Find :-}}$

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{Height of the Cuboid ( h ) = ?}$

$\large\textsf{ }$

$\LARGE\mathfrak{\underline{✯\; Formula :-}}$

$\large\textsf{ }$

• As it is not given that which Surface Area it is so I will calculate with both the Total Surface Area and Lateral Surface Area of the cuboid .

$\boxed{\large\textsf\textcolor{purple}{ {\large\textsf{T.S.A.}}_{\large\textsf{( \; Cuboid \; )}} = \large\textsf{2 ( lb + bh + hl )} }}\\\\\boxed{\large\textsf\textcolor{purple}{ {\large\textsf{L.S.A.}}_{\large\textsf{( \; Cuboid \; )}} = \large\textsf{2h ( l + b )} }}$

$\large\textsf{ }$

$\LARGE\mathfrak{\underline{✯\; Solution :-}}$

$\large\textsf{ }$

$\large\textsf{ {\large\textsf{T.S.A.}}_{\large\textsf{( \; Cuboid \; )}} = \large\textsf{2 ( lb + bh + hl )} }\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{758 = 2 [ ( 14 × 11 ) + ( 11 × h ) + ( h × 14 )}\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{758 = 2 ( 154 + 11h + 14h )}\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{758 = 2 ( 154 + 25h )}\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{758 = 308 + 50h}\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{758 - 308 = 50h}\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{450 = 50h}\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{ \cfrac{\large\textsf{450}}{\large\textsf{50}} = \large\textsf{h} }\\\\\\\qquad\tt{:}\longrightarrow\boxed{\large\textsf\textcolor{red}{9 = h}}$

$\boxed{\large\textsf\textcolor{orange}{∴ Height of the Cuboid = 9 cm}}$

__________________________________________________________

$\large\textsf{ }$

$\large\textsf{ {\large\textsf{L.S.A.}}_{\large\textsf{( \; Cuboid \; )}} = \large\textsf{2h ( l + b )} }\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{758 = 2h ( 14 + 11 )}\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{758 = 2h ( 25 )}\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{758 = 50h}\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{ \cfrac{\large\textsf{758}}{\large\textsf{50}} = h }\\\\\\\qquad\tt{:}\longrightarrow\boxed{\large\textsf\textcolor{red}{15.16 = h }}$

$\boxed{\large\textsf\textcolor{orange}{∴ Height of the Cuboid = 15.16 cm}}$

$\large\textsf{ }$

➛ Now in both the above case's the height we have calculated is different so just check in your notebook that which Surface Area they have mentioned and just note down the answer of the respective Surface Area :)

__________________________________________________________

$\large\textsf{ }$

$\LARGE\mathfrak{\underline{✯\; More \; \; Formulas :-}}$

$\large\textsf{ }$

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

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

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

$\large\textsf{ {\large\textsf{L.S.A.}}_{\large\textsf\textcolor{red}{( \; Cube \; )}} = \large\textsf{4×l²} }$

$\large\textsf{ {\large\textsf{T.S.A.}}_{\large\textsf{( \; Cube \; )}} = \large\textsf{6 × l²} }$

$\large\textsf{ {\large\textsf{Volume}}_{\large\textsf{( \; Cube \; )}} = \large\textsf{l²} }$

$\large\textsf{ {\large\textsf{C.S.A.}}_{\large\textsf\textcolor{red}{( \; Cylinder \; )}} = \large\textsf{2 × πrh} }$

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

$\large\textsf{ {\large\textsf{Volume}}_{\large\textsf{( \; Cylinder \; )}} = \large\textsf{πr²h} }$

$\large\textsf{ {\large\textsf{C.S.A.}}_{\large\textsf\textcolor{red}{( \; Cone \; )}} = \large\textsf{πrl} }$

$\large\textsf{ {\large\textsf{T.S.A.}}_{\large\textsf{( \; Cone \; )}} = \large\textsf{πr × ( r + l )} }$

$\large\textsf{ {\large\textsf{Volume}}_{\large\textsf{( \; Cone \; )}} } \large\textsf{ = \cfrac{\large\textsf{1}}{\large\textsf{3}} }\large\textsf{× πr²h}$

$\large\textsf{ {\large\textsf{T.S.A.}}_{\large\textsf\textcolor{red}{( \; Sphere \; )}} = \large\textsf{4πr²} }$

$\large\textsf{ {\large\textsf{Volume}}_{\large\textsf{( \; Sphere \; )}} } \large\textsf{ = \cfrac{\large\textsf{4}}{\large\textsf{3}} }\large\textsf{× πr³}$

$\large\textsf{ }$

$\large\textsf\textcolor{purple}{ \; \; \; \; \; \; \; \; \; \; \; \; ◈ ━━━━━━━ ✪ ━━━━━━━ ◈}$