Problem 5 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A string of length 12 units is bent first into a square \(P Q R S\) and then into a right-angled \(\Delta P Q T\) by keeping the side \(P Q\) of the square fixed and other is one more than its side. Then, the area of \(P Q R S\) equals (A) ar \((\Delta P Q T)\) (B) \(\frac{3}{2} \cdot \operatorname{ar}(\Delta P Q T)\) (C) \(2 \cdot \operatorname{ar}(\Delta P Q T)\) (D) none of these

Step-by-Step Solution

Answer

The correct answer is (B), \(\frac{3}{2} \cdot \operatorname{ar}(\Delta P Q T)\).

Step 1: Determine the side length of the square

Since the string is bent into a square, the perimeter of the square equals the length of the string, which is 12 units. Hence, the perimeter is 4 times the side length of the square. Let the side length of the square be \( s \). Then, \( 4s = 12 \), so \( s = 3 \) units.

Step 2: Calculate the area of the square

Using the side length from Step 1, the area of square \( P Q R S \) is \( s^2 \). Substituting the side length, we have \( 3^2 = 9 \) square units.

Step 3: Determine the shape and dimensions of \( \Delta P Q T \)

The right-angled triangle \( \Delta P Q T \) involves the side \( P Q \, = 3 \) units, which is fixed, and the other side is given to be one more than this, so it is \( 4 \) units.

Step 4: Calculate the area of \( \Delta P Q T \)

The area of a right-angled triangle is given by \( \frac{1}{2} \times \text{base} \times \text{height} \). Here, the base is \( 3 \) and the height is \( 4 \). So the area is \( \frac{1}{2} \times 3 \times 4 = 6 \) square units.

Step 5: Compare the areas

The area of \( \Delta P Q T \) is \( 6 \) square units while the area of the square \( P Q R S \) is \( 9 \) square units. Since \( 9 = \frac{3}{2} \times 6 \), the area of \( P Q R S \) equals \( \frac{3}{2} \times \text{area of } \Delta P Q T \).

Key Concepts

Square geometry

The concept of square geometry primarily revolves around understanding the properties of a square, which is a four-sided polygon (quadrilateral) with all sides of equal length and every interior angle measuring 90 degrees.
When we talk about forming a square from a string, we're examining how its perimeter can translate to a square's layout. This involves ensuring that the entire length of the string is used to frame the square's borders.

In our exercise, a string of 12 units is formed into a square. Hence, the string's length represents the square's perimeter. To find the side length of the square, we use the formula for the perimeter of a square:

• Perimeter = 4 times the side length

If the perimeter is 12 units, we set up the equation:

• 4s = 12

Solving for s (the side length), we find that each side of the square must be 3 units.

Once we've established the side length, calculating the area becomes straightforward, as the area of a square is given by the formula:

• \[\text{Area} = s^2\]

Substitute in our value for side length, so the area of this square is:

• 3 squared = 9 square units

Right-angled triangle

A right-angled triangle is a type of triangle that has one angle measuring exactly 90 degrees. Recognizing and calculating properties related to such triangles is crucial in geometry.

In our case, after bending the 12-unit string into a square, it's then reconfigured into a right-angled triangle, while maintaining one side — the side PQ of the square, which is 3 units — as one of its legs. The problem also states that the other leg of the triangle is one unit longer than PQ, making it 4 units.

To find the area of this triangle, we use the formula for the area of a right-angled triangle:

• \[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\]

For our triangle, the base is 3 units and the height is 4 units, thus:

• Area = \( \frac{1}{2} \times 3 \times 4 = 6 \) square units

This calculation confirms the right-angled triangle’s area of 6 square units.

Area comparison

The final step in our exercise involves comparing the areas of the square and the right-angled triangle that were constructed from the string. This comparison is a fascinating aspect of geometry, as it allows us to see how shapes utilizing the same perimeter can enclose different areas.

As calculated earlier, the square has an area of 9 square units, while the right-angled triangle has an area of 6 square units. By comparing these, we aim to express one area in terms of the other.

Notice how the area of the square can be compared to the area of the triangle using the relationship:

• \[9 = \frac{3}{2} \times 6\]

This implies that the area of the square is \( \frac{3}{2} \) times the area of the triangle. Comparison like this not only enhances our understanding of the spatial distribution but also broadens our perspective on how various geometric shapes behave in terms of area, given a set perimeter. Such exercises can significantly help in grasping area-related problems in geometry.