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Chapter 3: Problem 14

The sides of a triangle are in the ratio \(3: 4: 5\), the relation between \(r\) and \(R\) for the triangle is (a) \(r=\frac{R}{2}\) (b) \(r=\frac{2 R}{5}\) (c) \(r=\frac{R}{5}\) (d) None of these

Short Answer

Expert verified
The relation between the inradius 'r' and the circumradius 'R' of a triangle with sides in the ratio \(3:4:5\) can be calculated using standard formulas and the area of the triangle. The relation between 'r' and 'R' is found to be \(r=\frac{2R}{5}\), which corresponds to option (b).

Step by step solution

01

Assign variables to the triangle sides

Let's assume the sides of the triangle are 3x, 4x, and 5x, where x is a positive constant.
02

Calculate semi-perimeter 's'

The semi-perimeter 's' is half the sum of the sides of the triangle. Therefore, \[ s = \frac{3x + 4x + 5x}{2} = \frac{12x}{2} = 6x \]
03

Calculate the area '√{'s'(s-a)(s-b)(s-c)}$'

Use Heron's formula to find the area of the triangle: \[ A = \sqrt{s(s-a)(s-b)(s-c)} \] \[ A = \sqrt{6x(6x-3x)(6x-4x)(6x-5x)} \] \[ A = \sqrt{6x(3x)(2x)(x)} \] \[ A = 6x^2 \]
04

Find the inradius 'r'

The inradius 'r' of the triangle can be found using the formula: \[ r = \frac{A}{s} \] \[ r = \frac{6x^2}{6x} = x \]
05

Use the circumradius formula for right triangles

As our given triangle sides are in the ratio \(3:4:5\), it is a right-angle triangle (Pythagorean triple rule). The circumradius 'R' of the right-angle triangle can be found using the formula: \[ R = \frac{Hypotenuse}{2} \] \[ R = \frac{5x}{2} \]
06

Find the relation between 'r' and 'R'

Now, we have r = x and R = \(\frac{5x}{2}\). We can find the relation between 'r' and 'R' : \[ R = \frac{5}{2}r \] Hence, the relation between 'r' and 'R' is: \[ r=\frac{2R}{5} \] This corresponds to option (b).

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Most popular questions from this chapter

The sides of a \(\Delta A B C\) satisfy the equation, \(2 a^{2}+4 b^{2}+c^{2}=4 a b+2 a c\), then find \(\angle A\) and \(\angle B\).

\begin{aligned} &\text { In a triangle } A B C ; \cos A+\cos B+\cos C=\frac{7}{4}, \text { then }\\\ &\text { evaluate } \frac{R}{r} \end{aligned}

If \(a, b, c\) are the sides of \(\Delta A B C\) and \(\sin \theta\) and \(\cos \theta\) are the roots of equation \(a x^{2}-b x+c=0\), then \(\cos B\) equals (a) \(\frac{c}{a}-1\) (b) \(\frac{c}{2 a}-1\) (c) \(1-\frac{c}{a}\) (d) \(1+\frac{c}{a}\)

In a triangle \(A B C, 2 a c \sin \frac{A-B+C}{2}\) is equal to (a) \(a^{2}+b^{2}-c^{2}\) (b) \(c^{2}+a^{2}-b^{2}\) (c) \(b^{2}-c^{2}-a^{2}\) (d) \(c^{2}-a^{2}-b^{2}\)

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