Question

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24. 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.
​

Answer 1

Answer:

Surface area of the cuboid 240 square cm.

Step-by-step explanation:

Surface area of cuboid = 2(lb+bh+hl) square units

Length of the diagonal of cuboid= $\sqrt{l^{2}+b^{2}+h^{2} }$

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

Given: l+b+h= 19 cm

Also ,

Length of diagonal = 11 cm

$\sqrt{l^{2}+b^{2}+h^{2} }=11$

squaring on both sides

$(\sqrt{l^{2}+b^{2}+h^{2} })^{2} =(11)^{2}$

= $l^{2}+b^{2}+h^{2} =121$

Now,

$(l+b+h)^{2} = l^{2}+b^{2}+h^{2} +2(lb+bh+hl)$

⇒$19^{2} = 121+2(lb+bh+hl)$

⇒$361-121= 2(lb+bh+hl)$

⇒$2(lb+bh+hl)=240$

$surface$  $area= 240$$cm^2$

Answer 2

✦ Question :-

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.

Given:

Length + breadth + depth = 19 cm { Cuboid }

Length of diagonal = 11 cm

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To find:

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What is Surface area of the cuboid ?

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Solution:

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We consider the length, breadth and height of the cube be a cm, b cm and c cm .

Then from this we get :-

⠀⠀⠀⠀⠀⠀⠀⠀

a + b + c = 19 cm eqn 1.

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Length of the diagonal = 11 cm

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Diagonal of a cuboid = √ l² + b² + h²

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{ l² + b² + h² = 121 } .....eqn 2

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Then ,

a + b + c = 19

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(l²+ b²+ h²) = 121

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l$^{2}$ + b$^{2}$ + h$^{2}$ + 2(lb + bh + lh) = 361

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

2(lb + bh + lh) = 240

Hence ,

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Surface area of the cuboid is 240 cm².

Additional Information :-

❥ Perimeter of Rectangle = 2( L + B )

❥ Perimeter of square = 4 × Side

❥ Perimeter of triangle = AB + BC + CA

❥ Area of Rectangle = L × B

❥ Area of Square = ( side ) ²

❥ Area of Rhombus = Product of Diagonal/2.

❥ Area of Parallelogram = Base × Height.

❥ Area of triangle = 1/2 × base × height .