The sum of length,  breadth and depth of a cuboid is 19 cm and the length of its diagonal is 11 cm. Find the surface area of the cuboid.

Hi!

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

Given, l + b + h = 19cm ...(1)

Length of diagonal of the cuboid =

Squaring on both sides, we get

l 2 + b 2 = h 2 = 121cm2 ...(2)

Now, (l + b + h)2 = l 2 + b 2 + h 2 + 2lb + 2bh + 2hl

∴ 2(lb + bh + hl ) = ( l + b + h)2 – (l 2 + b 2 + h 2)

2(lb + bh + hl ) = (19cm)2 – 121cm2 = 361cm2 – 121cm2 = 361cm2 – 121cm2 = 240cm2 (Using(1) and (2))

Surface area of the cuboid = 2(lb + bh + hl) = 240cm2

∴ Surface area of the cuboid is 240cm2.

Cheers!

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Diagonal of a cuboid- root l?+b?+h?
L+b+h -19
Diagonal-11 squaring on both sides
We get 121
L?+b?+h?- 121
According to formula,of( a+b+c)?
It is l?+b?+h?+2(lb+bh+hl)
-19?
2(lb+bh+hl)-361-121
So,the answer is 240
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Length of diagonal cuboid=l^2+b^2+h^2
Surface area of cuboid=2lb+bh+hl
Given equation 1 l+b+h=19
Equation 2 length of diagonal =11
l^2+ b^2+ h^2 = 11
Simplyfying equation 2
l^2+ b^2+ h^2= 121
Taking l^2+b^2+h^2= l^2+ b^2 h^2+ 2lb+ 2bh+ 2lh
Substituting values
(19)^2= 121+2(lb+ bh+ hl)
361-121=2(lb+ bh+ hl)
240= 2(lb+ bh+ hl)
Surface area of cuboid=240
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Denoting length by l, breadth by b and height by h..

You gave me, l+b+h = 19

l^2+b^2+h^2 = 121.

Squaring the first equation, we have,

l^2+b^2+h^2+2(l+b)h+2lb = 361

→121+2(lb+bh+lh) = 361 →2(lb+bh+lh) = 240

So, the total surface area is 240 sq.cm

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