Question

(a+b):(b+c):(c+a)=6:7:8 and a+b+c=14 then what is c

Answer 1

Answer

• Value of c is 6

Given

• (a+b) : (b+c) : (c+a) = 6 : 7 : 8

• a + b + c = 14

To Find

• Value of c

Solution

Let ,

a + b = 6k ... (1)

b + c = 7k ... (2)

c + a = 8k ... (3)

Add (1) , (2) & (3) .

⇒ ( a + b ) + ( b + c ) + ( c + a ) = 6k + 7k + 8k

⇒ 2a + 2b + 2c = 21k

⇒ 2 ( a + b + c ) = 21k

⇒ 2 (14) = 21k [ ∵ given ]

⇒ 28 = 21k

⇒ 21k = 28

⇒ k = 28/21

⇒ k = 4/3 ... (4)

Given ,

a + b + c = 14

⇒ c = 14 - ( a + b )

⇒ c = 14 - 6k   [ From (1) ]

⇒ c = 14 - 6(4/3)    [ From (4) ]

⇒ c = 14 - 8

⇒ c = 6

Answer 2

GIVEN :–

$\\ \: \: { \huge{.}} \: \: { \bold{(a+b):(b+c):(c+a) = 6:7:8 }}$

$\\ \: \: { \huge{.}} \: \: { \bold{a+b+c=14}}$

TO FIND :–

• Value of 'c' = ?

SOLUTION :–

$\\ \implies { \bold{(a+b):(b+c):(c+a) = 6:7:8 }}$

• We should write this as –

$\\ \dashrightarrow { \bold{(a+b) = 6x}}$

$\\ \dashrightarrow { \bold{(b+c) = 7x}}$

$\\ \dashrightarrow { \bold{(c+a) = 8x}}$

• Now second condition –

$\\ \implies { \bold{a+b+c=14}}$

• We should write this as –

$\\ \implies { \bold{2(a+b+c)=2 \times 14}}$

$\\ \implies { \bold{2a+2b+2c=28}}$

$\\ \implies { \bold{(a + b) + (b + c) + (c + a)=28}}$

• Put the values –

$\\ \implies { \bold{(6x) + (7x) + (8x)=28}}$

$\\ \implies { \bold{21x=28}}$

$\\ \implies { \bold{x= \cancel\dfrac{28}{21}}}$

$\\ \implies \large{ \boxed{ \bold{x= \dfrac{4}{3}}}}$

▪︎ So that –

$\\ \dashrightarrow { \bold{(a+b) = 8}}$

$\\ \dashrightarrow { \bold{(b+c) = \dfrac{28}{3}}}$

$\\ \dashrightarrow { \bold{(c+a) = \dfrac{32}{3}}}$

• Now Let's find 'c' –

$\\ \implies { \bold{a+b+c=14}}$

• Put the value of (a+b) –

$\\ \implies { \bold{8+c=14}}$

$\\ \implies { \bold{c=14-8}}$

$\\ \dashrightarrow \large { \boxed{ \bold{c = 6}}}$