In $\triangle ABC$, the lengths of sides AC and AB are 12 cm and 5 cm, respectively. If the area of $\triangle ABC$ is 30 cm$^2$ and R and r are respectively the radii of circumcircle and incircle of $\triangle ABC$ then the value of $2R + r$ (in cm ) is equal to

### Analyzing the Setup We are given a triangle $\triangle ABC$ with side lengths $AC = 12$ cm and $AB = 5$ cm, and an area of $30$ cm$^2$. Our objective is to determine the value of $2R + r$, where $R$ is the circumradius and $r$ is the inradius. ### The Hidden Angle To unlock the properties of this triangle, we utilize the sine formula for the area of a triangle: $$\text{Area} = \frac{1}{2} \cdot AB \cdot AC \cdot \sin A$$ Substituting the known values into this equation, we obtain: $$30 = \frac{1}{2} \cdot 5 \cdot 12 \cdot \sin A$$ Simplifying the expression yields $30 = 30 \sin A$, which implies $\sin A = 1$. Consequently, the angle $A$ must be $90^\circ$, revealing that $\triangle ABC$ is a **right-angled triangle**. ### The Right Triangle Revelation Since $\triangle ABC$ is a right-angled triangle at $A$, we apply the Pythagorean theorem to calculate the length of the hypotenuse $BC$: $$BC^2 = AB^2 + AC^2$$ $$BC^2 = 5^2 + 12^2 = 25 + 144 = 169$$ Taking the square root, we find $BC = 13$ cm. The sides of the triangle are therefore $5$, $12$, and $13$. ### The Circles Within and Without In a right-angled triangle, the circumcenter is located at the midpoint of the hypotenuse. Thus, the circumradius $R$ is: $$R = \frac{BC}{2} = \frac{13}{2} = 6.5 \text{ cm}$$ To find the inradius $r$, we first calculate the semi-perimeter $s$: $$s = \frac{5 + 12 + 13}{2} = \frac{30}{2} = 15 \text{ cm}$$ Using the relationship $r = \frac{\text{Area}}{s}$, we find: $$r = \frac{30}{15} = 2 \text{ cm}$$ ### Final Calculation We now compute the final expression $2R + r$ using our derived values: $$2R + r = 2(6.5) + 2$$ $$2R + r = 13 + 2 = 15$$ The final value is **15**.

JEE Main 2021 (March)

LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: In $△ A B C$ , the lengths of sides AC and AB are 12 cm and 5 cm, respectively. If the area of $△ A B C$ is 30 cm $^{2}$ and R and r are respectively the radii of circumcircle and incircle of $△ A B C$ then the value of $2 R + r$ (in cm ) is equal to

Q: Wait, why do we assume the angle is 90 degrees? Can't the area formula work for any angle?

That is a great question! The area formula $\text{Area} = \frac{1}{2} ab \sin C$ is indeed universal. However, when we plug in the given values, we find that $\sin A = 1$. Since the maximum value of the sine function is 1, the only possible angle is $90^\circ$. It is not an assumption; it is a mathematical necessity derived from the given data!

Enter Numerical Value:

Visualized Solution

Visualize the Triangle $A B C$

Given sides: $A C = 12$ cm and $A B = 5$ cm.
Given Area of $△ A B C = 30$ cm $^{2}$ .
We need to find the relationship between these sides and the area.

Apply the Area Formula

Area formula: $Area = \frac{1}{2} \cdot A B \cdot A C \cdot sin A$
We use this formula because it connects two sides, the included angle, and the area.

Substitute Known Values

Substitute the given values into the formula.
$30 = \frac{1}{2} \cdot 5 \cdot 12 \cdot sin A$

Solve for $sin A$

Simplifying the right side: $30 = 30 sin A$
Divide both sides by $30$ : $sin A = 1$
Therefore, $A = 9 0^{\circ}$

Identify the Right Triangle

Since $A = 9 0^{\circ}$ , $△ A B C$ is a right-angled triangle at $A$ .
This allows us to use the Pythagoras theorem to find the hypotenuse $B C$ .

Calculate the Hypotenuse $B C$

By Pythagoras theorem: $B C^{2} = A B^{2} + A C^{2}$
Substitute values: $B C^{2} = 5^{2} + 1 2^{2} = 25 + 144 = 169$
Taking square root: $B C = 169 = 13$ cm

Find the Circumradius $R$

For a right-angled triangle, the circumcenter lies at the midpoint of the hypotenuse.
Circumradius $R = \frac{hypotenuse}{2}$
$R = \frac{B C}{2} = \frac{13}{2}$ cm

Calculate Semi-perimeter $s$

Semi-perimeter $s = \frac{A B + A C + B C}{2}$
$s = \frac{5 + 12 + 13}{2} = \frac{30}{2} = 15$ cm

Find the Inradius $r$

Inradius $r = \frac{Area}{s}$
$r = \frac{30}{15} = 2$ cm

Final Calculation: $2 R + r$

We need to find the value of $2 R + r$ .
Substitute $R = \frac{13}{2}$ and $r = 2$ :
$2 R + r = 2 (\frac{13}{2}) + 2$
$= 13 + 2 = 15$

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The Sigma Insight: Properties of Triangles

Solution Diagram

Analyzing the Setup

We are given a triangle

△ A B C

with side lengths

A C = 12

cm and

A B = 5

cm, and an area of

30

^{2}

. Our objective is to determine the value of

2 R + r

, where

R

is the circumradius and

r

is the inradius.

The Hidden Angle

To unlock the properties of this triangle, we utilize the sine formula for the area of a triangle:

Area = \frac{1}{2} \cdot A B \cdot A C \cdot sin A

Substituting the known values into this equation, we obtain:

30 = \frac{1}{2} \cdot 5 \cdot 12 \cdot sin A

Simplifying the expression yields

30 = 30 sin A

, which implies

sin A = 1

. Consequently, the angle

A

must be

9 0^{\circ}

, revealing that

△ A B C

is a right-angled triangle.

The Right Triangle Revelation

Since

△ A B C

is a right-angled triangle at

A

, we apply the Pythagorean theorem to calculate the length of the hypotenuse

B C

B C^{2} = A B^{2} + A C^{2}

B C^{2} = 5^{2} + 1 2^{2} = 25 + 144 = 169

Taking the square root, we find

B C = 13

cm. The sides of the triangle are therefore

5

12

, and

13

The Circles Within and Without

In a right-angled triangle, the circumcenter is located at the midpoint of the hypotenuse. Thus, the circumradius

R

is:

R = \frac{B C}{2} = \frac{13}{2} = 6.5 cm

To find the inradius

r

, we first calculate the semi-perimeter

s

s = \frac{5 + 12 + 13}{2} = \frac{30}{2} = 15 cm

Using the relationship

r = \frac{Area}{s}

, we find:

r = \frac{30}{15} = 2 cm

Final Calculation

We now compute the final expression

2 R + r

using our derived values:

2 R + r = 2 (6.5) + 2

2 R + r = 13 + 2 = 15

The final value is 15.

Common Traps & FAQs

Wait, why do we assume the angle is 90 degrees? Can't the area formula work for any angle?

That is a great question! The area formula

Area = \frac{1}{2} ab sin C

is indeed universal. However, when we plug in the given values, we find that

sin A = 1

. Since the maximum value of the sine function is 1, the only possible angle is

9 0^{\circ}

. It is not an assumption; it is a mathematical necessity derived from the given data!

Why is the circumradius

R

just half the hypotenuse?

In any right-angled triangle, the hypotenuse acts as the diameter of the circumcircle. This is a consequence of Thales's Theorem, which states that an angle inscribed in a semicircle is a right angle. Therefore, the midpoint of the hypotenuse is the center of the circumcircle, making the radius exactly half the length of the hypotenuse.

Can I use Heron's formula to find the area instead?

You absolutely could, but it would be like using a sledgehammer to crack a nut! Heron's formula is powerful for finding the area when you know all three sides. Here, we already know the area and two sides, so the sine formula is much faster and more direct. Always look for the most efficient path in JEE problems!

Enter Numerical Value:

Visualized Solution

Visualize the Triangle $A B C$

Apply the Area Formula

Substitute Known Values

Solve for $sin A$

Identify the Right Triangle

Calculate the Hypotenuse $B C$

Find the Circumradius $R$

Calculate Semi-perimeter $s$

Find the Inradius $r$

Final Calculation: $2 R + r$

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Hidden Angle

The Right Triangle Revelation

The Circles Within and Without

Final Calculation

Similar Questions

Let $a, b$ and $c$ be the length of sides of a triangle $A B C$ such that $\frac{a + b}{7} = \frac{b + c}{8} = \frac{c + a}{9}$ . If $r$ and $R$ are the radius of incircle and radius of circumcircle of the triangle $A B C$ , respectively, then the value of $\frac{R}{r}$ is equal to

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

In a triangle $A B C$ , let $∠ C = π /2$ . If $r$ is the inradius and $R$ is the circumradius of the triangle, then $2 (r + R)$ is equal to

If in a triangle $A B C$ , $A B = 5$ units, $∠ B = cos^{- 1} (\frac{3}{5})$ and radius of circumcircle of $Δ A B C$ is 5 units, then the area (in sq. units) of $Δ A B C$ is:

In a triangle PQR, let $∠ P QR = 3 0^{\circ}$ and the sides PQ and QR have lengths $103$ and 10, respectively. Then, which of the following statement(s) is (are) TRUE ?

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Enter Numerical Value:

Visualized Solution

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

Apply the Area Formula

Substitute Known Values

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

Identify the Right Triangle

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

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

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

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

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2R+r

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Hidden Angle

The Right Triangle Revelation

The Circles Within and Without

Final Calculation

Similar Questions

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Visualize the Triangle $A B C$

Solve for $sin A$

Calculate the Hypotenuse $B C$

Find the Circumradius $R$

Calculate Semi-perimeter $s$

Find the Inradius $r$

Final Calculation: $2 R + r$

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