In a triangle the sum of two sides is x and the product of the same sides is y. If x² - c² = y, wher

Q: Why did we use the Cosine Rule instead of Heron's formula?

Heron's formula involves the semi-perimeter $s$ and is often computationally heavy when you don't have the explicit values of the sides. The Cosine Rule directly links the sides to the angle $C$, which is exactly what we need to simplify the area expression and find the angle.

Q: How do we know $C = 120^\circ$ and not $60^\circ$?

The Cosine Rule gave us $\cos C = -\frac{1}{2}$. Since the cosine function is negative in the second quadrant, the angle must be obtuse. Thus, $C = 120^\circ$ is the only valid solution in a triangle.

Q: Is there a faster way to find $\frac{r}{R}$?

While there are various identities for $\frac{r}{R}$, deriving it from the fundamental definitions is the most robust method for JEE. It ensures you don't miss any factors and helps you understand the underlying geometry, which is crucial for solving more complex problems.

Question

In a triangle the sum of two sides is $x$ and the product of the same sides is $y$. If $x^2 - c^2 = y$, where $c$ is the third side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is

Options:

(A) $\frac{3y}{2x(x+c)}$

(B) $\frac{3y}{2c(x+c)}$

(C) $\frac{3y}{4x(x+c)}$

(D) $\frac{3y}{4c(x+c)}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery! Today, we are not just solving a problem; we are peeling back the layers of a geometric mystery.

Imagine you are standing in a field, looking at a triangle with sides $a$, $b$, and $c$. You are given two simple clues: the sum of two sides is $x$, and their product is $y$.

Then, a cryptic equation appears: $x^2 - c^2 = y$. It feels like a puzzle, doesn't it? Let's solve it together.

### The Algebraic Trap

First, let's translate the language of the problem into the language of algebra. We know $a + b = x$ and $ab = y$.

The given equation is $x^2 - c^2 = y$. If we substitute our variables, we get $(a + b)^2 - c^2 = ab$.

Now, let's expand this. The square of $(a + b)$ is $a^2 + b^2 + 2ab$. So, our equation becomes:

$$a^2 + b^2 + 2ab - c^2 = ab$$

Subtracting $2ab$ from both sides, we arrive at a beautiful, clean expression:

$$a^2 + b^2 - c^2 = -ab$$

This is the moment where the algebra starts to whisper secrets about the geometry of the triangle.

### The Cosine Rule Revelation

Does the expression $a^2 + b^2 - c^2$ ring a bell? It should! It is the heart of the Cosine Rule.

Recall that:

$$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$

By substituting our result $a^2 + b^2 - c^2 = -ab$ into this formula, we get:

$$\cos C = \frac{-ab}{2ab}$$

The $ab$ terms vanish, leaving us with $\cos C = -\frac{1}{2}$. This is a profound moment.

In the world of triangles, an angle with a cosine of $-\frac{1}{2}$ can only be one thing: $120^\circ$. We have just discovered that our triangle is obtuse, with angle $C = 120^\circ$.

### The Ratio Bridge

The question asks for the ratio of the in-radius $r$ to the circum-radius $R$. We know $r = \frac{\Delta}{s}$ and $R = \frac{abc}{4\Delta}$.

Dividing $r$ by $R$, we get:

$$\frac{r}{R} = \frac{\Delta}{s} \cdot \frac{4\Delta}{abc} = \frac{4\Delta^2}{s \cdot abc}$$

Now, we need to express $\Delta$ and $s$ in terms of our known variables. The area $\Delta$ is $\frac{1}{2}ab \sin C$.

With $C = 120^\circ$, $\sin 120^\circ = \frac{\sqrt{3}}{2}$, so $\Delta = \frac{\sqrt{3}}{4}ab$. Squaring this, we get:

$$\Delta^2 = \frac{3}{16}a^2b^2$$

The semi-perimeter $s$ is defined as:

$$s = \frac{a + b + c}{2} = \frac{x + c}{2}$$

### The Final Synthesis

Now, let's bring it all together. Substituting $\Delta^2$ and $s$ into our ratio formula, we have:

$$\frac{r}{R} = \frac{4(\frac{3}{16}a^2b^2)}{\frac{x + c}{2} \cdot abc}$$

Simplifying the numerator, we get $\frac{3}{4}a^2b^2$. The denominator is $\frac{c(x+c)ab}{2}$.

When we divide, the $ab$ terms cancel, and the constants simplify to:

$$\frac{r}{R} = \frac{3ab}{2c(x+c)}$$

Finally, substituting $ab = y$, we reach our destination:

$$\frac{r}{R} = \frac{3y}{2c(x+c)}$$

We have successfully navigated the algebraic and geometric landscape to find the elegant solution. Keep this spirit of curiosity alive, and you will conquer any problem JEE throws your way!

Animated Solution for Mathematics - Trigonometry: In a triangle the sum of two sides is $x$ and the product of the same sides is $y$ . If $x^{2} - c^{2} = y$ , where $c$ is the third side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is

Select Answer:

Visualized Solution

$D e f inin g t h e T r ian g l e$

$T h e G i v e n E q u a t i o n$

$S u b s t i t u t in g V a r iab l es$

$E x p an d in g t h e E q u a t i o n$

$T h e C os in e R u l e$

$F in d in g A n g l e C$

$V a l u eo f A n g l e C$

$I n - r a d i u s an d C i r c u m - r a d i u s$

$F or m u l a t in g t h e R a t i o \frac{r}{R}$

$C a l c u l a t in g A r e a Δ$

$S q u a r in g t h e A r e a$

$S e mi - p er im e t er s$

$S u b s t i t u t in g in t o t h e R a t i o$

$S im pl i f y in g t h e R a t i o$

$F ina l A n s w er$

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Algebraic Trap

The Cosine Rule Revelation

The Ratio Bridge

The Final Synthesis

Similar Questions

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In a triangle $A B C$ , $2 a c sin \frac{1}{2} (A - B + C) =$

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } DefiningtheTriangle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TheGivenEquation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SubstitutingVariables

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ExpandingtheEquation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TheCosineRule

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FindingAngleC

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ValueofAngleC

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In−radiusandCircum−radius

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FormulatingtheRatioRr​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CalculatingAreaΔ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SquaringtheArea

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Semi−perimeters

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SubstitutingintotheRatio

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SimplifyingtheRatio

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FinalAnswer