Problem 66 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length \(x\). The maximum area enclosed by the park is (A) \(\frac{3}{2} x^{2}\) (B) \(\sqrt{\frac{x^{3}}{8}}\) (C) \(\frac{1}{2} x^{2}\) (D) \(\pi x^{2}\)

Step-by-Step Solution

Answer

The maximum area is \(\frac{1}{2} x^{2}\), option (C).

Step 1: Understanding the Problem

We are given a triangular park with two sides of equal length \(x\) and need to find the maximum possible area of this triangle. Since it's isosceles with the third side being the river bank, the triangle must be an isosceles right triangle for maximum area.

Step 2: Determining the Triangle Type

For maximum area with two equal sides and a constant perimeter along a straight line, the triangle must be right-angled, as this makes optimal use of the given lengths \(x\). So we assume an isosceles right triangle.

Step 3: Finding Precise Area

In an isosceles right triangle with legs of length \(x\), the formula for area is \(A = \frac{1}{2} \times x \times x\). Hence, the area becomes \(A = \frac{1}{2}x^{2}\).

Step 4: Checking Options

The calculated maximum area \(\frac{1}{2}x^{2}\) matches with option \(C\). Therefore, the correct answer is option \(C\).

Key Concepts

Isosceles Right Triangle

An **isosceles right triangle** is a special kind of triangle. It has two sides of equal length and a right angle (90 degrees) between them, making the third side the hypotenuse. In our exercise, the park shaped in this manner uses two fence sides of equal length \(x\). The property of having equal legs simplifies calculations, especially when trying to find areas or other properties of the triangle.

• **Equal Lengths**: The two legs are equal, simplifying many calculations like perimeter and area.
• **Right Angle**: Offers advantages because trigonometric ratios and simple formulas are applicable.
• **Symmetrical**: The symmetry simplifies many geometric problems involving this triangle.

Understanding these points helps clarify why an isosceles right triangle is optimal for this problem. The symmetry and simplicity make it easier to calculate the maximum area of the park.

Maximum Area

The concept of **maximum area** in this context is linked to utilizing the lengths \(x\) most efficiently. To maximize a triangle's area with two sides \(x\) fixed, making it an isosceles right triangle optimizes the enclosed area due to its symmetrical properties.For an isosceles right triangle:

• The area is given by the formula \(A = \frac{1}{2} \times x \times x\), which simplifies to \(A = \frac{1}{2}x^{2}\).
• This formula arises because you multiply the base by the height and then take half, typical for any triangle area but straightforward here due to symmetry.

This equation shows that the maximum area for such a park configuration, with given side lengths \(x\), is \(\frac{1}{2}x^{2}\). Thus, choosing this shape ensures no wasted fence length, maximizing space efficiently.

Geometry Problem Solving

**Geometry problem solving** involves understanding the figures and applying known formulas and logical reasoning to find a solution. In the task of optimizing a triangle's area, critical steps include identifying constraints and selecting the best possible configuration.

• **Identifying Problem Type**: Recognize that the task entails maximizing an area with given side lengths.
• **Choosing Optimal Triangle Shape**: The isosceles right triangle was chosen based on inherent characteristics of maximizing space.
• **Applying Formulas Correctly**: Use relevant area and geometry formulas to calculate maximum possible values.

Effective problem solving in geometry often means choosing the simplest, most symmetric shape when conditions allow, as complexity usually doesn’t add value here. Understanding these principles aids not only in this exercise but in a wide range of geometric contexts.