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Question

The sum of the lengths of the edges of a cube is equal to half the perimeter of a square. If the numerical value of the volume of the cube is equal to one-sixth of the numerical value of the area of the square, then the length of one side of the square is:

This question was previously asked in
RRB ALP 2018 CBT 2 Fitter Question Paper (21-Jan-2019) (Shift 3)
The correct answer is

36 units

Understanding the Geometry Problem

This problem involves two common geometric shapes: a cube and a square. We are given two relationships connecting properties of the cube (specifically its edges and volume) with properties of the square (its perimeter and area). Our goal is to use these relationships to find the length of one side of the square.

Let's define variables for the dimensions of the shapes:

  • Let 'a' be the length of one edge of the cube.
  • Let 's' be the length of one side of the square.

Setting Up Equations from Given Relationships

The problem provides two key pieces of information that we can translate into mathematical equations.

Relationship 1: Cube Edges and Square Perimeter

The first relationship states that the sum of the lengths of the edges of a cube is equal to half the perimeter of a square.

  • A cube has 12 edges, and all edges have the same length 'a'. The sum of the lengths of the edges is \(12a\).
  • The perimeter of a square with side length 's' is \(4s\).

So, the equation for Relationship 1 is:

\(12a = \frac{1}{2} \times 4s\)

\(12a = 2s\)

We can simplify this equation:

\(s = 6a \quad (Equation 1)\)

This equation tells us that the side length of the square is 6 times the edge length of the cube.

Relationship 2: Cube Volume and Square Area

The second relationship states that the numerical value of the volume of the cube is equal to one-sixth of the numerical value of the area of the square.

  • The volume of a cube with edge length 'a' is \(a^3\).
  • The area of a square with side length 's' is \(s^2\).

So, the equation for Relationship 2 is:

\(a^3 = \frac{1}{6} \times s^2 \quad (Equation 2)\)

Solving for the Side Length of the Square

Now we have a system of two equations with two variables ('a' and 's'):

  1. \(s = 6a\)
  2. \(a^3 = \frac{1}{6} s^2\)

We can use substitution to solve this system. Substitute Equation 1 into Equation 2:

\(a^3 = \frac{1}{6} (6a)^2\)

Simplify the right side of the equation:

\(a^3 = \frac{1}{6} (36a^2)\)

\(a^3 = 6a^2\)

To solve for 'a', move all terms to one side:

\(a^3 - 6a^2 = 0\)

Factor out \(a^2\):

\(a^2(a - 6) = 0\)

This equation holds true if either \(a^2 = 0\) or \(a - 6 = 0\).

  • If \(a^2 = 0\), then \(a = 0\). An edge length of 0 is not possible for a physical cube.
  • If \(a - 6 = 0\), then \(a = 6\). This is a valid length.

So, the edge length of the cube is 6 units.

Now that we have the value of 'a', we can find the value of 's' using Equation 1 (\(s = 6a\)):

\(s = 6 \times 6\)

\(s = 36\)

The length of one side of the square is 36 units.

Conclusion

Based on the given relationships between the cube and the square, the calculated length of one side of the square is 36 units.

Shape Dimension Formula Value (Calculated)
Cube Edge Length (a) - 6 units
Cube Sum of Edge Lengths \(12a\) \(12 \times 6 = 72\) units
Cube Volume \(a^3\) \(6^3 = 216\) cubic units
Square Side Length (s) - 36 units
Square Perimeter \(4s\) \(4 \times 36 = 144\) units
Square Area \(s^2\) \(36^2 = 1296\) square units

Let's verify the original conditions using these values:

  • Sum of edges of cube = 72. Half the perimeter of the square = \(0.5 \times 144 = 72\). Condition 1 holds.
  • Volume of cube = 216. One-sixth of the area of the square = \(\frac{1}{6} \times 1296 = 216\). Condition 2 holds.

The calculated side length of the square, 36 units, satisfies both conditions.

Revision Table: Key Geometric Formulas

Shape Property Formula
Cube (edge 'a') Sum of Edge Lengths \(12a\)
Cube (edge 'a') Volume \(a^3\)
Square (side 's') Perimeter \(4s\)
Square (side 's') Area \(s^2\)

Additional Information: Properties of Cubes and Squares

Let's review some basic properties of cubes and squares that are relevant to this problem.

Cube Properties

  • A cube is a three-dimensional solid object bounded by six square faces, facets or sides, with three meeting at each vertex.
  • It has 6 faces, 12 edges, and 8 vertices.
  • All edges of a perfect cube are equal in length.
  • All angles between faces are right angles (90 degrees).

Square Properties

  • A square is a regular quadrilateral, which means it has four equal sides and four equal angles (90 degrees or right angles).
  • It is a special type of rectangle (since all angles are 90 degrees) and also a special type of rhombus (since all sides are equal).
  • The perimeter is the total length of its boundary (sum of all sides).
  • The area is the space enclosed within its sides.

Understanding these basic properties is crucial for setting up the correct equations when solving geometry problems like this one.

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Similar Questions

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  2. The two sides holding the right-angle in a right-angled triangle are 3 cm and 4 cm long. The area of its circumcircle will be:

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Important Questions from Plane Figures

  1. If the area of a square is 625 cm 2, then what is the perimeter of the square?

  2. The area and the perimeter of a sheet of paper are 240 cm 2and 68 cm, respectively. What would be its length and breadth?

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  4. The bisector of ∠B in ΔABC meets AC at D. If AB = 12 cm, BC = 18 cm and AC = 15 cm, then the length of AD (in cm) is:

  5. The perimeter and the length of one of the diagonals of a rhombus is 26 cm and 5 cm respectively. Find the length of its other diagonal (in cm).

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