Question

The areas of the three of faces of a cuboid are in the ratio 1:3:4 and its volume is 144 cu cm,
The length of its diagonal is​

Answer 1

Given:

The ratio of the areas of the 3 faces of a cuboid are 1:3:4

Volume of the cuboid = 144 cm³

To find:

The length of its diagonal

Formula to be used:

$\boxed{Volume \:of\: a\: cuboid\: = length\: *\: breadth\:*\: height}\\\\and\\\\\boxed{Diagonal\: of\: a\: cuboid = \sqrt{length^2\:+\:breadth^2\:+\:height^2} }$

Solution:

Let's assume the dimensions of the cuboid as:

"l" → length

"b" → breadth

"h" → height

The ratio of the areas of the faces of the cuboid will be ⇒ l×b : b×h : h×l = 1:3:4

So, we have

$\boxed{l*b = k},\; \boxed{b*h = 3k} \:and\: \boxed{h*l = 4k}$ ....... where k > 0

We will multiply the above three equations:

l × b × b × h × h × l = k × 3k × 4k

⇒ l² × b² × h² = 12k³ ...... (i)

Using the formula of the volume of a cuboid, we have

Volume = l × b × h = 144 cm³ ..... (ii)

On squaring the equation (ii), we get

[l × b × h]² = [144]²

⇒ l² × b² × h² = 144 × 144

substituting the value of l² × b² × h² from eq. (i)

⇒ 12k³ = 144 × 144

⇒ k³ = $\frac{144\:*\:144}{12}$

⇒ k³ = 1728

⇒ k = $\sqrt[3]{1728}$

⇒ k = 12 cm

Therefore,

l × b = 12 .... (iii)

b × h = 3k = 3 × 12 = 36 ...... (iv)

h × l = 4k = 4 × 12 = 48 ....... (v)

We will be substituting the values from (iii), (iv) & (v) in (ii) to find l, b & h

l = $\frac{l\:*\:b\:*\:h}{b\:*\:h} = \frac{144}{36} = 4\: cm$

b = $\frac{l\:*\:b\:*\:h}{h\:*\:l} = \frac{144}{48} = 3\:cm$

h = $\frac{l\:*\:b\:*\:h}{l\:*\:b} = \frac{144}{12} = 12\:cm$

Now,

Using the formula of the diagonal of a cuboid we will find the length of the diagonal:

Diagonal of the cuboid is,

$= \sqrt{l^2\:+\:b^2\:+\:h^2} \\\\= \sqrt{4^2\:+\:3^2\:+\:12^2} \\\\= \sqrt{16\:+\:9\:+\:144} \\\\= \sqrt{169} \\\\ = 13 \:cm$

Thus, the length of its diagonal is 13 cm.

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