Question

the length of the diagonal of a cube of side
$3 \sqrt{3}$
meter is​

Answer 1

❏ Question:-

@ The length of the diagonal of a cube of side $3 \sqrt{3}$ meter is ?

❏ Solution:-

➝ FIG:-

$\setlength{\unitlength}{0.74 cm}\begin{picture}(12,4)\thicklines\put(5.6,5.6){ A }\put(11.1,5.8){ B }\put(11.08,8.9){ C }\put(5.46,8.7){ D }\put(3.55,10.15){ E }\put(3.55,7.15){ F }\put(9.14,10.235){ H }\put(9.14,7.3){ G }\put(7.75,6.2){ 3\sqrt{3}\:cm }\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\put(6,6){\line(2,3){2.86}}\end{picture}$

➝ Given:-

• Side of the cube = $\sf 3 \sqrt{3}$ m.

➝ Explanation:-

$\sf\longrightarrow Diagonal=\sqrt{3}\times 3\sqrt{3}\:m$

$\sf\longrightarrow \boxed{\large{\red{Diagonal=9\:m}}}$

➝ Answer:-

Diagonal of the cube is 9 m.

$\setlength{\unitlength}{0.8cm}\begin{picture}(10,5)\thicklines\qbezier(1,1)(1,1)(9,1)\end{picture}$

$\setlength{\unitlength}{0.8cm}\begin{picture}(10,5)\thicklines\qbezier(1,1)(1,1)(9,1)\end{picture}$

➝ Used Formula:-

✦ For a cuboid of length l , breadth b and height h .

$\sf\longrightarrow\boxed{ Diagonal=\sqrt{l^{2}+b^{2}+h^{2}}}$

$\sf\longrightarrow\boxed{T.S.A.=2\times(lb+bh+hl)}$

$\sf\longrightarrow\boxed{ Volume=l\times b\times h}$

Where, • T.S.A.=Total Surface area,

✦For a Cube of Side a ,

$\sf\longrightarrow\boxed{ Diagonal=\sqrt{3}\times side}$

$\sf\longrightarrow\boxed{T.S.A.=6\times side{}^{2}}$

$\sf\longrightarrow\boxed{ Volume=Side^{3}}$

Where, •T.S.A.=Total Surface area.

$\setlength{\unitlength}{0.8cm}\begin{picture}(10,5)\thicklines\qbezier(1,1)(1,1)(9,1)\end{picture}$

$\setlength{\unitlength}{0.8cm}\begin{picture}(10,5)\thicklines\qbezier(1,1)(1,1)(9,1)\end{picture}$