Answer
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Hint: Take length, breadth and depth as l, b and h. Thus we get the relation connecting them. Expand \[{{\left( l+b+h \right)}^{2}}\] and substitute the values. Simplify and get the total surface area as cuboid.
Complete step-by-step solution -
Let us consider the length of cuboid as ‘l’, breadth as ‘b’ and height of cuboid as ‘h’. From the figure you can find the length, breadth and height.
The diagonal of the cuboid is marked as ‘d’. It is given the sum of length, breadth and depth of cuboid is 19 cm, i.e. l + b + h = 19…….(1)
Now, the length of diagonal, d = 11 cm.
We know that the length of diagonal of cuboid \[=\sqrt{{{l}^{2}}+{{b}^{2}}+{{h}^{2}}}=11cm.......(2)\]
The total surface area of a cube is the area of all 6faces of the cuboid \[=2\left( lb+bh+lh \right)......(3)\]
Now from equation (2), \[\sqrt{{{l}^{2}}+{{b}^{2}}+{{h}^{2}}}=11\]
Now square both sides of the equation (2).
\[{{l}^{2}}+{{b}^{2}}+{{h}^{2}}={{(11)}^{2}}=121\].
Now by basic trigonometric formula we know that,
\[\begin{align}
& {{\left( l+b+h \right)}^{2}}={{l}^{2}}+{{b}^{2}}+{{h}^{2}}+2lb+2bh+2lh \\
& {{\left( l+b+h \right)}^{2}}={{l}^{2}}+{{b}^{2}}+{{h}^{2}}+2\left( lb+bh+hl \right).....(4) \\
\end{align}\]
Now let us substitute all values from equation (1), (2) to equation (4).
\[l+b+h=19\] and \[l+b+h=121\]
\[\begin{align}
& \therefore {{\left( 19 \right)}^{2}}=121+2\left( lb+bh+lh \right) \\
& \therefore 2\left( lb+bh+lh \right)=361-121=240 \\
\end{align}\]
From equation (3) we know that TSA of cuboid = \[2\left( lb+bh+lh \right)\]
\[\therefore 2\left( lb+bh+lh \right)=240\]
TSA of cuboid = 240
i.e. the area of the cuboid = 240.
Thus we got the area of the cuboid as 240.
Note: We were only asked to find the area of the cuboid. So leave the answers as \[2\left( lb+bh+lh \right)\]. Don’t simplify again and make it complex. We haven’t been given sufficient data to find length, breadth and height. So get the area. Remember the formula for finding the length of diagonal of a cuboid.
Complete step-by-step solution -
Let us consider the length of cuboid as ‘l’, breadth as ‘b’ and height of cuboid as ‘h’. From the figure you can find the length, breadth and height.
The diagonal of the cuboid is marked as ‘d’. It is given the sum of length, breadth and depth of cuboid is 19 cm, i.e. l + b + h = 19…….(1)
Now, the length of diagonal, d = 11 cm.
We know that the length of diagonal of cuboid \[=\sqrt{{{l}^{2}}+{{b}^{2}}+{{h}^{2}}}=11cm.......(2)\]
The total surface area of a cube is the area of all 6faces of the cuboid \[=2\left( lb+bh+lh \right)......(3)\]
Now from equation (2), \[\sqrt{{{l}^{2}}+{{b}^{2}}+{{h}^{2}}}=11\]
Now square both sides of the equation (2).
\[{{l}^{2}}+{{b}^{2}}+{{h}^{2}}={{(11)}^{2}}=121\].
Now by basic trigonometric formula we know that,
\[\begin{align}
& {{\left( l+b+h \right)}^{2}}={{l}^{2}}+{{b}^{2}}+{{h}^{2}}+2lb+2bh+2lh \\
& {{\left( l+b+h \right)}^{2}}={{l}^{2}}+{{b}^{2}}+{{h}^{2}}+2\left( lb+bh+hl \right).....(4) \\
\end{align}\]
Now let us substitute all values from equation (1), (2) to equation (4).
\[l+b+h=19\] and \[l+b+h=121\]
\[\begin{align}
& \therefore {{\left( 19 \right)}^{2}}=121+2\left( lb+bh+lh \right) \\
& \therefore 2\left( lb+bh+lh \right)=361-121=240 \\
\end{align}\]
From equation (3) we know that TSA of cuboid = \[2\left( lb+bh+lh \right)\]
\[\therefore 2\left( lb+bh+lh \right)=240\]
TSA of cuboid = 240
i.e. the area of the cuboid = 240.
Thus we got the area of the cuboid as 240.
Note: We were only asked to find the area of the cuboid. So leave the answers as \[2\left( lb+bh+lh \right)\]. Don’t simplify again and make it complex. We haven’t been given sufficient data to find length, breadth and height. So get the area. Remember the formula for finding the length of diagonal of a cuboid.
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