Question

The length, breadth and height of a cuboid are in the ratio 7:6:5. If the surface area of the
cuboid is 1926 cm2, find
(i) volume
(ii) length of a diagonal of the cuboid.
V = 1906 cm: Diagonal = 2V100cm
V = 1926cm Diagonal = 3v100cm
V = 1906cm; Diagonal = 3v100cm
O V = 1926cm: Diagonal = v100cm
Deselect
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Answer 1

Answer :

• Volume of the Cuboid, v = 5670 cm³.

• Diagonal of the cuboid, d = 3√110 cm.

Explanation :

Given :

• Ratio of the length , breadth and height of the cuboid = 7 : 6 : 5

• Total surface of the cuboid, TSA = 1926 cm².

To find :

• Volume of the cuboid , v = ?

• Diagonal of the cuboid , d = ?

Solution :

Let the sides of the cuboid be 7x , 6x and 5x.

So here ,

• Length = 7x
• Breadth = 6x
• Height = 5x

Formula for surface area of a cuboid :

⠀⠀⠀⠀⠀⠀⠀⠀TSA = 2(lb + lh + bh)

Where :

• TSA = Total surface area of the cuboid.
• l = Length of the cuboid.
• b = Breadth of the cuboid.
• h = Height of the cuboid.

Now ,

By using the formula for total surface of a cuboid and substituting the values in it, we get :

==> TSA = 2(lb + lh + bh)

==> 1926 = 2[(7x × 6x) + (7x × 5x) + (6x × 5x)]

==> 1926 = 2(42x² + 35x² + 30x²)

==> 1926/2 = 42x² + 35x² + 30x²

==> 963 = 42x² + 35x² + 30x²

==> 963 = (42 + 35 + 30)x²

==> 963 = 107x²

==> 963/107 = x²

==> 9 = x²

==> √9 = x

==> ±3 = x

[Note : Since the side of the figure , here Cuboid can't be negative , we will neglect the value of x as -3 ]

∴ x = 3 cm.

Hence the value of x is 3 cm.

Now by substituting the value of x in the sides of the cuboid, we get :

• Length of the cuboid = 7x

==> l = 7 × 3

==> l = 21

∴ l = 21 cm

Hence the length of the cuboid is 21 cm.

• Breadth of the cuboid = 6x

==> b = 6 × 3

==> b = 18

∴ b = 18 cm

Hence the breadth of the cuboid is 18 cm.

• Height of the cuboid = 5x

==> b = 5 × 3

==> b = 15

∴ b = 15 cm

Hence the hieght of the cuboid is 15 cm.

Volume of the cuboid :

We know the formula for volume of a cuboid,

⠀⠀⠀⠀⠀⠀⠀⠀V = l × b × h

Where :

• V = Volume of the cuboid.
• l = Length of the cuboid.
• b = Breadth of the cuboid.
• h = Height of the cuboid.

By using the formula for volume of a Cuboid and substituting the values in it, we get :

==> V = l × b × h

==> V = 21 × 18 × 15

==> V = 5670

∴ V = 5670 cm³.

Diagonal of the cuboid :

We know the formula for diagonal of a cuboid i.e,

⠀⠀⠀⠀⠀⠀⠀⠀d = √(l² × b² × h²)

Where :

• d = Diagonal of the cuboid.
• l = Length of the cuboid.
• b = Breadth of the cuboid.
• h = Height of the cuboid.

By using the formula for diagonal of a cuboid and substituting the values in it, we get :

==> d = √(l² + b² + h²)

==> d = √(21² + 18² + 15²)

==> d = √(441 + 324 + 225)

==> d = √990

==> d = √(3 × 3 × 110)

==> d = 3√110

∴ d = 3√110 cm

Therefore,

• Volume of the cuboid , v = 5670 cm³.

• Diagonal of the cuboid , d = 3√110 cm.

Answer 2

$\large{\underline{\bf{Given:-}}}$

⟡Ratio of Length, Breadth and Height = 7:6:5

⟡Total Surface Area = 1926cm²

$\large{\underline{\bf{Find:-}}}$

⟡Volume of Cuboid

⟡Length of its Diagonal

$\large{\underline{\bf{Diagram:-}}}$

$\setlength{\unitlength}{0.74 cm}\begin{picture} (0,0) \thinlines\put(5.6,5.4){\bf A}\put(11.1,5.4){\bf B}\put(11.2,9){\bf C}\put(5.3,8.6){\bf D}\put(3.3,10.2){\bf E}\put(3.3,7){\bf F}\put(9.25,10.35){\bf H}\put(9.35,7.35){\bf G}\put(4.2,6.4){\bf 6x}\put(7.7,5.6){\bf 7x}\put(11.3,7.45){\bf 5x}\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}} \put(11,9){\line(-4,-1){7}}\put(7.8,7.4){\huge{?}} \end{picture}$

$\large{\underline{\bf{Solution:-}}}$

Let,

➮ Length = 7x

➮ Breadth = 6x

➮ Height = 5x

we, know that

$\huge{\underline{\boxed{\sf T.S.A \: of \: cuboid = 2(lb + bh + hl)}}}$

where,

• Total Surface Area, T.S.A = 1926cm²
• Length, l = 7x
• Breadth, b = 6x
• Height, h = 5x

So,

$\dashrightarrow\sf T.S.A \: of \: cuboid = 2(lb + bh + hl)$

$\dashrightarrow\sf 1926 = 2 \big\lgroup{(7x)(6x) + (6x)(5x) + (5x)(7x)} \big \rgroup$

$\dashrightarrow\sf 1926 = 2 \big\lgroup{42 {x}^{2} + 30 {x}^{2} + 35 {x}^{2} } \big \rgroup$

$\dashrightarrow\sf 1926 = 2 \big\lgroup{107 {x}^{2}} \big \rgroup$

$\dashrightarrow\sf 1926 = 214 {x}^{2}$

$\dashrightarrow\sf \dfrac{1926}{214} = {x}^{2}$

$\dashrightarrow\sf 9= {x}^{2}$

$\dashrightarrow\sf \sqrt{9} = x$

$\dashrightarrow\sf 3cm= x$

$\dashrightarrow\sf x = \pm 3cm$

$:\to\sf x = 3cm,- 3cm$

$\tiny{ \underline{\sf \: ignoring \: negative \: one \:as \: side \: can't \: be \: in \: negative }}$

$\small{ \therefore \underline{\sf value \: of \: x \: will \: be \: 3cm}}$

Now,

✒ Length, l = 7x = 7×3 = 21cm

✒ Breadth, b = 6x = 6×3 = 18cm

✒ Height, h = 5x = 5×3 = 15cm

Now, Using

$\huge{\underline{\boxed{\sf Volume \: of \: cuboid =l \times b \times h}}}$

where,

• Length, l = 21cm
• Breadth, b = 18cm
• Height, h = 15cm

So,

$\dashrightarrow\sf Volume \: of \: cuboid = l \times b \times h$

$\dashrightarrow\sf Volume \: of \: cuboid = 21\times 18 \times 15$

$\dashrightarrow\sf Volume \: of \: cuboid = 21\times 270$

$\dashrightarrow\sf Volume \: of \: cuboid = 5670 {cm}^{3}$

$\small{\therefore \underline{\sf Volume \: of \: cuboid = 5670 {cm}^{3}}}$

_____________________________

we, know that

$\huge{\underline{\boxed{\sf Diagonal \: of \: cuboid = \sqrt{l^2+b^2 + h^2}}}}$

where,

• Length, l = 21cm
• Breadth, b = 18cm
• Height, h = 15cm

So,

$\dashrightarrow\sf Diagonal \: of \: cuboid = 2(lb + bh + hl)$

$\dashrightarrow\sf Diagonal \: of \: cuboid = \sqrt{l^2+b^2 + h^2}$

$\dashrightarrow\sf Diagonal \: of \: cuboid = \sqrt{21^2+18^2 + 15^2}$

$\dashrightarrow\sf Diagonal \: of \: cuboid = \sqrt{441+324+225}$

$\dashrightarrow\sf Diagonal \: of \: cuboid = \sqrt{990}$

$\sf \bigstar \sqrt{990} \: can \: be \: written \: as \: \sqrt{9 \times 110} \bigstar$

$\dashrightarrow\sf Diagonal \: of \: cuboid = \sqrt{9 \times 110}$

$\dashrightarrow\sf Diagonal \: of \: cuboid = 3\sqrt{110}$

$\dashrightarrow\sf Diagonal \: of \: cuboid = 3\sqrt{110}cm$

$\small{\therefore \underline{\sf Diagonal \: of \: cuboid =3 \sqrt{110} cm}}$

_____________________________

Hence,

• Volume of Cuboid = 5670cm³
• Diagonal of Cuboid = 3√(110)cm