In a triangle ABC, let ∠C = π/2. If r is the inradius and R is the circumradius of the triangle ABC,

Q: Wait, why is the circumradius $R$ equal to $c/2$?

In a right-angled triangle, the hypotenuse subtends a $90^\circ$ angle at the vertex $C$. By the converse of Thales' Theorem, the hypotenuse must be the diameter of the circumcircle. Thus, the radius $R$ is half the diameter, which is $c/2$.

Q: Is the formula $r = \frac{a+b-c}{2}$ valid for all triangles?

No! That is a common trap. That specific, elegant formula is derived specifically for right-angled triangles using the property that the incircle touches the sides at distances related to the vertices. For a general triangle, you would use $r = \frac{\Delta}{s}$.

Q: Can I solve this using coordinate geometry?

Absolutely! Place vertex $C$ at $(0,0)$, $A$ at $(0,b)$, and $B$ at $(a,0)$. You can calculate the circumcenter and incenter coordinates and find the radii. It works, but it is much longer. Geometry is the elegant path here.

Question

In a triangle $ABC$, let $\angle C = \frac{\pi}{2}$. If $r$ is the inradius and $R$ is the circumradius of the triangle $ABC$, then $2(r+R)$ equals

Options:

(A) $b+c$

(B) $a+b$

(C) $a+b+c$

(D) $c+a$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of a right-angled triangle $ABC$, with the right angle at $C$. The sides are $a$, $b$, and the hypotenuse $c$. This is our canvas.

### The Circumcircle: The Guardian of the Hypotenuse

Let us first consider the circumcircle. This is the circle that perfectly encloses our triangle, touching all three vertices.

Because $\angle C = 90^{\circ}$, the hypotenuse $c$ acts as the diameter of this circle. This is a classic property: the angle subtended by a diameter at the circumference is always $90^{\circ}$.

Therefore, the circumcenter lies exactly at the midpoint of the hypotenuse. If the hypotenuse is the diameter, then the circumradius $R$ must be half of that length.

So, we have our first pillar of truth:

$$R = \frac{c}{2}$$

This implies that $2R = c$. Keep this in your pocket; it is going to be vital later.

### The Incircle: The Heart of the Triangle

Now, let us look inside. We want to fit the largest possible circle within this triangle. This is the incircle, with radius $r$.

For a right-angled triangle, there is a beautiful, specific formula for this radius. If you draw the tangents from the vertices to the incircle, you will find that the inradius $r$ is given by:

$$r = \frac{a+b-c}{2}$$

Why? Because the distance from the vertex $C$ to the points of contact on sides $a$ and $b$ creates a small square of side $r$ at the corner. The remaining lengths of the sides $a$ and $b$ are then tangent segments that sum up to the hypotenuse $c$.

When you derive this, you get $2r = a+b-c$. This formula is a lifesaver in competitive exams.

### The Synthesis: The Grand Cancellation

Now, we bring it all together. The problem asks us to evaluate $2(r+R)$. Let us expand this:

$$2(r+R) = 2r + 2R$$

We have our two pillars ready. We know $2r = a+b-c$ and we know $2R = c$. Let us substitute these into our expression:

$$2(r+R) = (a+b-c) + c$$

Look at that! The $-c$ and the $+c$ are staring at each other, waiting to vanish. They cancel out perfectly, leaving us with the elegant result:

**$$2(r+R) = a+b$$**

The hypotenuse, which seemed so important, has completely disappeared from the final answer. This is the elegance of geometry.

We started with a complex relationship involving the hypotenuse, and through the symmetry of the triangle, we arrived at a simple sum of the two perpendicular sides. Remember this result, not just as a formula, but as a testament to the harmony of mathematics.

Animated Solution for Mathematics - Trigonometry: In a triangle $A B C$ , let $∠ C = \frac{π}{2}$ . If $r$ is the inradius and $R$ is the circumradius of the triangle $A B C$ , then $2 (r + R)$ equals

Select Answer:

Visualized Solution

Visualizing $△ A B C$

Labeling the Sides

Circumcircle of a Right Triangle

Circumradius $R$ Property

Finding $2 R$

Incircle of the Triangle

Inradius $r$ Formula

Finding $2 r$

Setting up $2 (r + R)$

Substituting the Values

Final Simplification

Conclusion

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Circumcircle

The Incircle

The Synthesis

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In a triangle ABC, let ∠C=2π​. If r is the inradius and R is the circumradius of the triangle ABC, then 2(r+R) equals