Question

The perimeter and area of a triangle are 16 cm and 12 cm' respectively. If one of its sides is 5 cm, find the
length of remaining two sides.
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Answer 1

Answer:

5,5,6

Step-by-step explanation:

S= perimeter /2 = 16/2 = 8

A^2 = S ( S-a) ( S-b) ( S-c)

12*12 = 8 * ( 8-5) (S-b) (S-c)

144= 24* (S-b) ( S-c)

( S-b) (S-c)= 6 = 3 * 2

if we put b and c value as 5 and 6 it will be resolved

Answer 2

Answer:

6cm,5cm

Step-by-step explanation:

According to the question

let a side of triangle be 5cm

Perimeter=16cm.......(i)

a+b+c=16

5+b+c=16

b+c=16-5=11

therefore, b+c=11.....(ii)

Area= 12cm^2...........

here, from eqn (i)

a+b+c=16cm.........(iii)

therefore, (putting the value of a+b+c) we get

S= a+b+c/2

=16/2

=8cm

Now,

A= √S (S-a)(s-b)(s-c)

12= √8 (8-5)(8-b)(8-c)

12= √8* (3)(8-b)(8-c)

12= √24*(8-b)(8-c)

12*2= 24*(8-b)(8-c)

144/24=(8-b)(8-c)

6=(8-b)(8-c)

6=8(8-c)-b(8-c)

6=64-8c-8b+bc

6=64-8(c+b)+bc

(putting the value of (c+b) from eqn (ii) we get)

6=64-8(11)+bc

6=64-88+bc

6=-24+bc

30=bc.....(iv)

30/b=c

c=30/b

here from eqn (ii)

c+b=11

30/b+b=11

30+b^2/b=11

30+b^2=11b

b^2-11b+30=0.......(v)

solving eqn(v) we get

(b-5)(b-6)=0

either b=5 or 6

so putting the value in end (ii)

b+c=11

5+c=11

c=6

(Thank you)