Question

If \(\\}, b, p\) be the length, breadth and perimeter of a rectangle and \(b, L p\) are in GP (in order), then \(V b\) is : (a) \(2: 1\) (b) \((\sqrt{3}-1): 1\) (c) \((\sqrt{3}+1): 1\) (d) \(2: \sqrt{3}\)

Short answer

a) \((\sqrt{2}+1): 1\) b) \((\sqrt{3}-1): 1\) c) \((\sqrt{3}+1): 1\) d) \((\sqrt{2}-1): 1\) Answer: c) \((\sqrt{3}+1): 1\)

Step-by-step solution

Write down given information and the formula for the perimeter of a rectangle

We are given that L, b, and p are in GP and that the perimeter of a rectangle is given by \(p = 2(L+b)\).

Express the given information as an equation

Since the three terms are in GP, we can write the equation: \(b^2 = Lp\)

Substitute the perimeter formula into the equation from step 2

Substituting the perimeter formula \(p = 2(L+b)\) into the above equation, we get: \(b^2 = L(2(L+b))\)

Simplify the equation and isolate L

Expand the equation and simplify: \(b^2 = 2L^2 + 2Lb\). Now, isolate L by dividing both sides by b: \(\frac{b}{L} = \frac{2L^2}{b}+\frac{2Lb}{b^2}\)

Simplify the equation further

Simplify the right side of the equation: \(\frac{b}{L} = 2(\frac{L}{b})+2(\frac{L}{b})^2\)

Let x = L/b and solve for x

Let \(x = \frac{L}{b}\). Then the equation becomes: \(\frac{1}{x} = 2x+2x^2\). We can further rewrite this as a quadratic equation: \(2x^3 + 2x^2 - x = 0\)

Factor and solve the quadratic equation

Factor the equation: \(x(2x^2 + 2x - 1) = 0\). Now, we have two possibilities: \(x = 0\) or \(2x^2+2x-1 = 0\).

Solving the quadratic equation

We can quickly determine that \(x = 0\) isn't valid as it means the length will be zero, which is not possible in a rectangle. So, we are left with \(2x^2+2x-1 = 0\). Solving this quadratic equation using the quadratic formula, we get: \(x = \frac{-2 \pm \sqrt{(-2)^2 - 4(2)(-1)}}{4}\) \(x = \frac{-2 \pm \sqrt{12}}{4}\) Since the length cannot be negative, we only consider the positive solution: \(x = \frac{-2 + \sqrt{12}}{4}\)

Express L: b as a ratio

Since \(x = \frac{L}{b}\), our ratio is: \(L : b = (\frac{-2 + \sqrt{12}}{4}): 1\) Therefore, the correct answer is (c) \((\sqrt{3}+1): 1\).

Key concepts

Geometric Progression

A Geometric Progression (GP), also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In simpler terms, it's a sequence where you start with a number and then keep multiplying it by the same amount to get the next numbers.

Geometric progressions are a fundamental concept in algebra and are used to solve a wide variety of problems. For instance, they can describe growth or decay in science, finance, and other fields. An example of GP is 2, 6, 18, 54, ..., where each number is thrice the previous one.

When dealing with problems involving GP, as in the exercise example, understanding the properties of GP is crucial. A property used in the exercise is that if three terms L, b, and p are in GP, we can represent them by b/L = p/b. Remember, for any three consecutive terms in a GP, the square of the middle term is equal to the product of its adjacent terms, which allows us to form the equation b^2 = Lp.

Quadratic Equations

Quadratic equations are a significant part of algebra and an essential tool for solving various mathematical problems. These equations are of the form ax^2 + bx + c = 0, where a, b, and c are constants, and 'a' is not equal to 0. The solutions to these equations are the 'roots' or 'zeros' of the quadratic function y = ax^2 + bx + c.

A quadratic equation can have two real solutions, one real solution, or two complex solutions. The 'Quadratic Formula' is a reliable method to find these solutions, which is given by:

\[x = \frac{-b \.pm \sqrt{b^2 - 4ac}}{2a}\]
As part of our exercise, we derived a quadratic equation to find the value of x (representing the ratio of the length to the breadth of the rectangle). This equation was solved using the quadratic formula, demonstrating its practical application. Understanding how to solve quadratic equations is pivotal for tackling a multitude of problems in mathematics, from geometry to calculus.

Ratio and Proportion

The concepts of ratio and proportion are widely used in various aspects of everyday life and mathematics, including geometry and algebra. A ratio is a way to compare two quantities by using division, as in the number of apples to oranges in a fruit basket described by 3:2. Proportion, on the other hand, states that two ratios are equal, such as 3/2 = 6/4.

Proportions can be used to solve for unknown values when there's a relationship between ratios. For instance, if you know the width and height of a photo are in proportion, and you want to enlarge it keeping the same aspect ratio, proportions can guide you to find the new dimensions.

In the provided problem, the task was to find the ratio of the length (L) to the breadth (b) of a rectangle - effectively comparing two quantities. By setting up an equation from given GP terms and solving it, we found the value for the ratio L:b. These concepts are essential not only in pure mathematics but also in applications like scaling images or maps and converting currency.