Question

The sum of the length, breadth and height of a cuboid is $5\sqrt{3}$ cm and length of its diagonal is $3\sqrt{5}$ cm, then its total surface area is ___________.

Answer 1

Let the length, breadth and height of the cuboid be l, b and h, respectively.

Given:

$l+b+h=5\sqrt{3}\mathrm{cm}$ .....(1)

Length of diagonal of cuboid = $\sqrt{l^2+b^2+h^2}=3\sqrt{5}\mathrm{cm}$

$\Rightarrow l^2+b^2+h^2=45{\mathrm{cm}}^2$ .....(2)

Now,

$$\begin{gathered} {\left(l+b+h\right)}^2=l^2+b^2+h^2+2lb+2bh+2hl \\ \Rightarrow 2\left(lb+bh+hl\right)={\left(5\sqrt{3}\mathrm{cm}\right)}^2-45{\mathrm{cm}}^2\left[\mathrm{Using}\left(1\right)\mathrm{and}\left(2\right)\right] \\ \Rightarrow 2\left(lb+bh+hl\right)=75-45=30{\mathrm{cm}}^2 \end{gathered}$$

Total surface area of the cuboid = $2\left(lb+bh+hl\right)=30{\mathrm{cm}}^2$

The sum of the length, breadth and height of a cuboid is $5\sqrt{3}$ cm and length of its diagonal is $3\sqrt{5}$ cm, then its total surface area is __30 cm³_.