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In a DeltaABC if r(1):r(2):r(3)=2:4:6, t...

In a `DeltaABC` if `r_(1):r_(2):r_(3)=2:4:6,` then `a:b:c `=

A

`3:5:7`

B

`1:2:3`

C

`5:8:7`

D

none of these

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The correct Answer is:
To solve the problem, we need to find the ratio \( a:b:c \) given the ratio of the inradii \( r_1:r_2:r_3 = 2:4:6 \). ### Step-by-step Solution: 1. **Understanding the Relationship**: The inradii \( r_1, r_2, r_3 \) correspond to the sides opposite to vertices \( A, B, C \) respectively. The relationship can be expressed as: \[ \frac{r_1}{\Delta} = \frac{1}{s-a}, \quad \frac{r_2}{\Delta} = \frac{1}{s-b}, \quad \frac{r_3}{\Delta} = \frac{1}{s-c} \] where \( \Delta \) is the area of triangle \( ABC \) and \( s \) is the semi-perimeter. 2. **Setting Up the Ratios**: Given \( r_1:r_2:r_3 = 2:4:6 \), we can write: \[ \frac{1}{s-a} : \frac{1}{s-b} : \frac{1}{s-c} = 2 : 4 : 6 \] 3. **Introducing a Constant**: Let \( k \) be a constant such that: \[ \frac{1}{s-a} = 2k, \quad \frac{1}{s-b} = 4k, \quad \frac{1}{s-c} = 6k \] 4. **Finding Expressions for \( s-a, s-b, s-c \)**: From the above, we can express: \[ s-a = \frac{1}{2k}, \quad s-b = \frac{1}{4k}, \quad s-c = \frac{1}{6k} \] 5. **Using the Semi-Perimeter**: The semi-perimeter \( s \) is given by: \[ s = a + b + c \] We can express \( a, b, c \) in terms of \( s \): \[ a = s - (s-a) = s - \frac{1}{2k}, \quad b = s - (s-b) = s - \frac{1}{4k}, \quad c = s - (s-c) = s - \frac{1}{6k} \] 6. **Substituting for \( s \)**: Adding the expressions for \( s-a, s-b, s-c \): \[ (s-a) + (s-b) + (s-c) = \frac{1}{2k} + \frac{1}{4k} + \frac{1}{6k} \] To simplify, find a common denominator (which is \( 12k \)): \[ \frac{6}{12k} + \frac{3}{12k} + \frac{2}{12k} = \frac{11}{12k} \] Thus: \[ 3s - (a + b + c) = \frac{11}{12k} \] Since \( a + b + c = 2s \), we have: \[ 3s - 2s = \frac{11}{12k} \implies s = \frac{11}{12k} \] 7. **Finding \( a, b, c \)**: Substitute \( s \) back into the expressions for \( a, b, c \): \[ a = s - \frac{1}{2k} = \frac{11}{12k} - \frac{6}{12k} = \frac{5}{12k} \] \[ b = s - \frac{1}{4k} = \frac{11}{12k} - \frac{3}{12k} = \frac{8}{12k} \] \[ c = s - \frac{1}{6k} = \frac{11}{12k} - \frac{2}{12k} = \frac{9}{12k} \] 8. **Finding the Ratio**: The ratio \( a:b:c \) is: \[ a:b:c = \frac{5}{12k} : \frac{8}{12k} : \frac{9}{12k} = 5:8:9 \] ### Final Result: Thus, the ratio \( a:b:c = 5:8:9 \).

To solve the problem, we need to find the ratio \( a:b:c \) given the ratio of the inradii \( r_1:r_2:r_3 = 2:4:6 \). ### Step-by-step Solution: 1. **Understanding the Relationship**: The inradii \( r_1, r_2, r_3 \) correspond to the sides opposite to vertices \( A, B, C \) respectively. The relationship can be expressed as: \[ \frac{r_1}{\Delta} = \frac{1}{s-a}, \quad \frac{r_2}{\Delta} = \frac{1}{s-b}, \quad \frac{r_3}{\Delta} = \frac{1}{s-c} ...
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OBJECTIVE RD SHARMA ENGLISH-PROPERTIES OF TRIANGLES AND CIRCLES CONNECTED WITH THEM-Exercise
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  10. If in a !ABC,angleA=pi//3 and AD is a median , then

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  11. In a triangle ABC cos^(2)A/2+cos^(2)B/2+cos^(2)C/2=

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  12. The base of a triangle is 80cm and one of the base angles is 60^(@).If...

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  13. In a DeltaABC if r(1)=16,r(2)=48 and r(3)=24, then its in-radius ,is

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