Question

. The length, breadth and height of a cuboidal room be
a unit, b unit, and c unit respectively and a+b+c = 25, ab+bc+ca = 240.5 then
find the length of the longest rod to be kept inside thé room.
...thot of the base of the​

Answer 1

The maximum length is 12 units.

Step-by-step explanation:

The maximum length of a rod that can be kept inside the cuboidal room having a length, breadth and height a units, b units, and c units respectively is the diagonal of the cuboidal i.e. $\sqrt{a^{2} + b^{2} + c^{2}}$ units.

Now, we know the formula that,

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

⇒ (25)² = (a² + b² + c²) + 2(240.5)

{Since, (a + b + c) = 25 and (ab + bc + ca) = 240.5. given}

⇒ (a² + b² + c²) = 144

⇒ $\sqrt{a^{2} + b^{2} + c^{2}} = \sqrt{144} = 12$ units. (Answer)

Answer 2

Answer:

Length, breadth and height of a cuboidal shaped room are a units , b units , and c units respectively.

Given that , a + b + c = 25

ab + bc + ca = 240.5

Now the formula is ,

$( {a}^{2} + {b}^{2} + {c}^{2} ) = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca)$

Putting the values ,

$= > ( 25) {2} = {a}^{2 } + {b}^{2} + {c}^{2} + 2(240.5)$

$= > {a}^{2} + {b}^{2} + {c}^{2} = 625 - 481 = 144$

$\sqrt{ {a}^{2} } + {b}^{2} + {c}^{2} = \sqrt{144} = 12units$

Thank you ...