Question

In a triangle ABC; let r1 =1, r2=2, r3=3, find the length of a, b, c.

Answer 1

Answer:

true

Step-by-step explanation:

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Answer 2

Answer:

r

1

,r

2

,r

3

in H.P. ⇒ a,b,c, are in A.P. as in (b)

or 2b=a+c

2s=a+b+c=24

∴ 3b = 24 ⇒ b = 8

Hence a+c=16, by (1) and (2)

Again we are given that S=24

or 12(s−a)(s−b)(s−c)=24

2

or 12(12−a)(12−8)(12−c)=576

or (12−a)(12−c)=12

or 144−12(a+c)+ac=12

Put a+c=16 ∴ac=60

Thus a,c are the roots of t

2

- 16t + 60 = 0$$

t=10,6∴a=10,a=6,b=8,a=6,c=10,b=8

Hence sides are 6,8,10 or 10,8,6.

Step-by-step explanation:

r

1

,r

2

,r

3

in H.P. ⇒ a,b,c, are in A.P. as in (b)

or 2b=a+c

2s=a+b+c=24

∴ 3b = 24 ⇒ b = 8

Hence a+c=16, by (1) and (2)

Again we are given that S=24

or 12(s−a)(s−b)(s−c)=24

2

or 12(12−a)(12−8)(12−c)=576

or (12−a)(12−c)=12

or 144−12(a+c)+ac=12

Put a+c=16 ∴ac=60

Thus a,c are the roots of t

2

- 16t + 60 = 0$$

t=10,6∴a=10,a=6,b=8,a=6,c=10,b=8

Hence sides are 6,8,10 or 10,8,6.