Question

Find the side of triangle a if s-a=4,s-b=8,s-c=12

Answer 1

Answer:

The sides of triangles are 20, 16, 12

Step-by-step explanation:

Given that-

$s - a = 4$           ------------------ 1

$s-b = 8$            ------------------- 2

$s-c = 12$           ------------------- 3

Where s is semi-perimeter and a, b and c are sides of a triangle.  

Adding equation 1, 2 and 3

We have-

$(s-a) + (s-b) + (s-c) = 4+8+12$

3s - (a + b + c) = 23

We know that-

The semi-perimeter of a triangle-

$s = \dfrac{a+b+c}{2}$

So, $\dfrac{3(a+b+c)}{2}- (a+b+c) = 24$

$\dfrac{3(a+b+c)-2(a+b+c)}{2} = 24$

$\dfrac{a+b+c}{2} = 24$

So, s = 24

Semi-perimeter of a triangle is 24.

Now, from equations 1, 2 and 3

s - a = 4

s = 24

So, 24 - a = 4

-a = 4 - 24

-a = -20

a = 20

s - b = 8

24 - b = 8

-b = 8 - 24

-b = -16

b = 16

And s - c = 12

24 - c = 12

-c = 12 - 24

-c = -12

c = 12

Hence, the sides of triangles are 20, 16, 12.