Question

The length, breadth and height of a box are
75 cm, 85 cm and 95 cm respectively. What
will be the greatest length of tape which can
measure the three dimensions of the box
exactly? ​

Answer 1

$\huge{\mathbb{\red{ANSWER:-}}}$

For a Cuboid Box :-

Given :-

$\sf{length (L) = 75 \: cm}$

$\sf{breadth (b) = 85 \: cm}$

$\sf{height (h) = 95 \: cm}$

To Find :-

$\sf{The \: greatest \: length \: of \: this \: cuboid \: box = ?}$

Using Formula :-

$\sf{Diagonal \: of \: Cuboid =\sqrt{L^{2} + b^{2} + h^{2}}}$

Solution :-

On putting the values of length , breadth and height -

$\sf{The \: greatest \: length \: of \: tape-}$

$\sf{D =\sqrt{(75)^{2} + (85)^{2} + (95)^{2}}}$

$\sf{D =\sqrt{5^{2}(15^{2} + 17^{2} + 19^{2})}}$

$\sf{D =5\sqrt{(15^{2} + 17^{2} + 19^{2})}}$

$\sf{D =5\sqrt{(225 + 289 + 361)}}$

$\sf{D =5\sqrt{(225 + 650)}}$

$\sf{D =5\sqrt{875}}$

$\sf{D =5\sqrt{25*35}}$

$\sf{D =5\times 5\sqrt{35}}$

$\sf{D =25\sqrt{35}}$

$\sf{D =25\times 5.92}$

$\sf{D = 25\times \dfrac{592}{100}}$

$\sf{D =\dfrac{592}{4}}$

$\sf{D = 148 \: cm}$

As We got :-

$\sf{We \: need \: 148 \: cm \: tape \: which \: can \: measure \: all}$

$\sf{the \: three \: dimensions \: of \: the \: box \: exactly.}$

Here ,

$\sf{\sqrt{35}}$

$\sf{\sqrt{(36-1)}}$

$\sf{\sf\boxed{\sqrt{x-y}=\sqrt{x}-\dfrac{y}{2\sqrt{x}}}}$

So ,

$\sf{\sqrt{36} - \dfrac{1}{2\sqrt{36}}}$

$\sf{6 - \dfrac{1}{2\times 6}}$

$\sf{6 - \dfrac{1}{12}}$

$\sf{6 - 0.08333}$

$\sf{5.917}$

$\sf{5.92}$