Math, asked by sharmasakshi3939, 4 months ago

solve it quickly with step by step explanation ​

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Answers

Answered by kavinkarthik81
0

Answer:

I don't know

please marke as brainlist

Answered by dansudakshina060
1

Answer:

16.15 cm

Step-by-step explanation:

(For understanding this question and my solution, you have to draw a cuboid and ABCDEFGH; then only you can understand, because I have done the solution according to my prospective of drawing the cuboid, but the answer will be ofcourse same in all the cases)

The dimensions of the box are in the ratio of 2:3:4

So, the dimensions will be 2x, 3x and 4x

We know TSA of a cuboid = 2(lb+bh+hl)

Here, TSA of the cuboid = 468cm^2

So, 2(lb+bh+hl) = 468cm^2

=> 2[(2x*3x)+(3x*4x)+(4x*2x)] = 468cm^2

=> 2[6x^2 + 12x^2 + 8x^2] = 468cm^2

=> 2[26x^2] = 468cm^2

=> 52x^2 = 468cm^2

=> x^2 = 468/52

=> x^2 = 9

=> x = √9

=> x = 3

Now, Let AE be the diagonal of the largest surface

So, (AE)^2 = (4x)^2 + (3x)^2

=>(AE)^2 = (4*3)^2 + (3*3)^2

=>(AE)^2 = 12^2 + 9^2

=>(AE)^2= 144 + 81

=>(AE)^2 = 225

=>AE = √225

=>AE = 15 cm

Now, DE(or you can take any of the diagonals of the cuboid) is the longest rod which we can take

So, (DE)^2 = (DA)^2 + (AE)^2

=>(DE)^2 = (2x)^2 + (15)^2

=>(DE)^2 = (2*3)^2 + 225

=>(DE)^2 = 6^2 + 225

=>(DE)^2 = 36 + 225

=>(DE)^2 = 261

=>DE = √261

=>DE = 16.15 cm

Therefore, the longest rod which can be kept in the box should be of approx. 16.15 cm.

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