Question

volume of a cuboidal box is 750 cubic metre if the dimensions of cuboid are in the ratio of 3:2:1 find its original Length breadth and height.​

Answer 1

Step-by-step explanation:

As per the provided information in the given question, we have :

• Volume of cuboidal box = 750 m³
• The dimensions of cuboid are in the ratio of 3:2:1.

We've been asked to calculate its length, breadth and height.

As the dimensions of cuboid are in the ratio of 3:2:1, so let's suppose the length , breadth and the height of the cuboid as 3x , 2x and 1x. Thus,

1. Length = 3x
2. Breadth = 2x
3. Height = x

― According to the question, the volume of the cuboid is 750 m³. Here, the formula to calculate the volume of the cuboid will act as the linear equation to find the value of x. As we know that,

$\bigstar \quad \underline{\boxed{ \pmb{\mathfrak{Volume}}_{\pmb{\mathfrak{(Cuboid)}}} = \pmb{\mathfrak{l \times b \times h}} }}$

• l denotes length
• b denotes base
• h denotes height

So, as per the question,

$\longrightarrow \sf{\quad {750 = 3x \times 2x \times x }}$

Multiplying the terms in the RHS.

$\longrightarrow \sf{\quad {750 = 6x^3 }}$

Transposing 6 from RHS to LHS.

$\longrightarrow \sf{\quad {\cancel{\dfrac{750}{6}} = x^3 }}$

Dividing 750 by 6 in LHS.

$\longrightarrow \sf{\quad { 125= x^3 }}$

Now, put the sign of cube root in both sides.

$\longrightarrow \sf{\quad { \sqrt[3]{125} = \sqrt[3]{ {x}^{3} } }}$

Writing the cube roots of the terms in LHS and RHS.

$\longrightarrow \quad \boxed{\sf { \textbf{\textsf{5 = x}} }}$

We have to calculated the value of x. We'll be calculating the dimensions of the cuboid.

• Length = 3x = 3(5) = 15 m
• Breadth = 2x = 2(5) = 10 m
• Height = x = 5 m

Answer 2

Answer:

Given :-

• The volume of a cuboidal box is 750 m³. The dimensions of cuboid are in the ratio of 3 : 2 : 1.

To Find :-

• What is the original length, breadth and height of a cuboidal box.

Formula Used :-

$\clubsuit$ Volume Of Cuboid Formula :

$\footnotesize\mapsto \sf\boxed{\bold{\pink{Volume_{(Cuboid)} =\: Length \times Breadth \times Height}}}$

Solution :-

Let,

$\mapsto \bf Length_{(Cuboidal\: Box)} =\: 3a\: m$

$\mapsto \bf Breadth_{(Cuboidal\: Box)} =\: 2a\: m$

$\mapsto \bf Height_{(Cuboidal\: Box)} =\: a\: m$

Given :

$\leadsto \sf Volume_{(Cuboidal\: Box)} =\: 750\: m^3$

According to the question by using the formula we get,

$\bigstar\: \: \bold{\purple{Volume_{(Cuboid)} =\: 3a \times 2a \times a}}$

$\longrightarrow \sf 3a \times 2a \times a =\: 750$

$\longrightarrow \sf 6a^2 \times a =\: 750$

$\longrightarrow \sf 6a^3 =\: 750$

$\longrightarrow \sf a^3 =\: \dfrac{\cancel{750}}{\cancel{6}}$

$\longrightarrow \sf a^3 =\: \dfrac{125}{1}$

$\longrightarrow \sf a^3 =\: 125$

$\longrightarrow \sf a =\: \sqrt[3]{125}$

$\longrightarrow \sf a =\: \sqrt[3]{\underline{5 \times 5 \times 5}}$

$\longrightarrow \sf\bold{\green{a =\: 5}}$

Hence, the required original length, breadth and height are ;

❒ Length Of Cuboidal Box :

$\longrightarrow \sf Length_{(Cuboidal\: Box)} =\: 3a\: m$

$\longrightarrow \sf Length_{(Cuboidal\: Box)} =\: (3 \times 5)\: m$

$\longrightarrow \sf\bold{\red{Length_{(Cuboidal\: Box)} =\: 15\: m}}$

❒ Breadth Of Cuboidal Box :

$\longrightarrow \sf Breadth_{(Cuboidal\: Box)} =\: 2a\: m$

$\longrightarrow \sf Breadth_{(Cuboidal\: Box)} =\: (2 \times 5)\: m$

$\longrightarrow \sf\bold{\red{Breadth_{(Cuboidal\: Box)} =\: 10\: m}}$

❒ Height Of Cuboidal Box :

$\longrightarrow \sf Height_{(Cuboidal\: Box)} =\: a$

$\longrightarrow \sf\bold{\red{Height_{(Cuboidal\: Box)} =\: 5\: m}}$

${\footnotesize{\bold{\underline{\therefore\: The\: original\: length\: breadth\: and\: height\: of\: cuboidal\: box\: is\: 15\: m\: , 10\: m\: and\: 5\: m\: respectively\: .}}}}$

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★ VERIFICATION ★

$\implies \sf 3a \times 2a \times a =\: 750$

By putting a = 5 we get,

$\implies \sf 3(5) \times 2(5) \times 5 =\: 750$

$\implies \sf 15 \times 10 \times 5 =\: 750$

$\implies \bf 750 =\: 750$

Hence, Verified.