Problem 68 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

In a circle of radius \(r\), an isosceles triangle \(A B C\) is inscribed with \(A B=A C\). If the \(\Delta A B C\) has perimeter \(P=\) \(2\left[\sqrt{2 h r-h^{2}}+\sqrt{2 h r}\right]\) and area \(A=h \sqrt{2 h r-h^{2}}\), where \(h\) is the altitude from \(A\) to \(B C\), then \(\lim _{h \rightarrow 0^{+}} \frac{A}{P^{3}}\) is equal to \(\begin{array}{ll}\text { (A) } 128 r & \text { (B) } \frac{1}{128 r}\end{array}\) (C) \(\frac{1}{64 r}\) (D) None of these

Step-by-Step Solution

Answer

(B) \( \frac{1}{128r} \)

Step 1 - Write the Expressions for Perimeter and Area

The problem gives us two formulas: the perimeter \( P \) of the triangle is \( P = 2 \left[ \sqrt{2hr-h^2} + \sqrt{2hr} \right] \), and the area \( A \) of the triangle is \( A = h \sqrt{2hr-h^2} \). We need these expressions to find the limit.

Step 2 - Substitute h Approaching 0

As \( h \to 0^+ \), both \( \sqrt{2hr-h^2} \) and \( \sqrt{2hr} \) simplify to \( \sqrt{2hr} \) because \( h^2 \) becomes negligible. Therefore, the perimeter \( P \) approaches \( 2 \cdot 2\sqrt{2hr} = 4\sqrt{2hr} \), and the area \( A \to h\sqrt{2hr} \).

Step 3 - Evaluate the Limit Expression

The limit expression \( \lim_{h \to 0^+} \frac{A}{P^3} \) becomes \( \frac{h\sqrt{2hr}}{(4\sqrt{2hr})^3} = \frac{h\sqrt{2hr}}{64hr^{3/2}(2hr)^{3/2}} \). Simplify this to \( \frac{h}{128h^{3/2}r} \).

Step 4 - Simplify and Take Limit as h Approaches 0

Notice the \( h \) in the numerator. When \( h \to 0 \), \( h^{1/2} \to 0 \), hence \( \frac{h}{128h^{3/2}r} \to \frac{1}{128r} \).

Step 5 - Conclude the Calculation

We find that \( \lim_{h \to 0^+} \frac{A}{P^3} = \frac{1}{128r} \). This corresponds to option (B) in the choices.

Key Concepts

calculus

Calculus is all about understanding change and motion, a concept that plays a pivotal role in solving limit problems. In this exercise, we look into the behavior of an expression as a variable approaches a specific value—in this case, as the height ( h) of the isosceles triangle approaches zero.When evaluating limits, calculus provides tools such as the Squeeze Theorem, L'Hôpital's Rule, and algebraic simplifications to find the resultant value of an expression. Here, the focus is on how expressions simplify as a dependent variable becomes negligible. For example, when h becomes very small, terms involving h^2 disappear because they become so tiny that their effect is negligible in comparison to the dominant terms.Through calculus, we assess this negligible impact and apply concepts of continuity and limits to refine and simplify our mathematical models, allowing us to find values like \(\lim_{h \to 0^+} \frac{A}{P^3}\) in a structured way.

geometry

Geometry encompasses the study of shapes, sizes, and properties of space, key elements in understanding figures like triangles inscribed in circles. In this exercise, the geometry of the circle and the inscribed triangle form the basis of our calculations.An isosceles triangle is a geometrical shape characterized by having two equal sides. In this particular problem, triangle \( \Delta ABC \) is isosceles, with its base BC inside the circle. The two equal sides are AB and AC, originating from the vertex A and meeting the circle at B and C respectively.Understanding these properties allows us to use given formulas for perimeter and area. By relating the triangle's elements with the fundamental properties of the circle, the exercise combines geometry's visual aspect with calculus’s analytical approach.

isosceles triangle

The isosceles triangle holds unique properties making it quite interesting in geometric problems. As previously mentioned, an isosceles triangle has two equal sides. This symmetry is used in numerous mathematical applications and problems due to its straightforwardness and elegant properties.In the context of the circle, the apex of the isosceles triangle is at point A, and the equal sides reach out to touch the circle at points B and C. The altitude h, which acts as the perpendicular line dropped from the apex A to the base BC, becomes crucial in calculations of area and perimeter. The symmetry of an isosceles triangle simplifies calculations. It ensures equal distribution of lengths and angles, enabling the use of specific formulae for easy calculations. Calculating the area with \( A = h \sqrt{2hr-h^2} \) and understanding how adjustments affect properties are all grounded in these unique geometrical principles.