Question

The three sides of a triangle are (3a-7b+8c) cm.(a-9+8b-10c)cm and (-9+8b-11c)cm.find the perimeter of the triangle.​

Answer 1

Answer:

Perimeter of the triangle (ABC) = 4a+9b-13c-18

Step-by-step explanation:

Let the triangle be as ABC.

The side AB = (3a-7b+8c)

side BC = (a-9+8b-10c)

side CA = (-9+8b-11c)

Now,

Perimeter of triangle (ABC) = AB + BC + CA

P (ABC) = (3a-7b+8c) + (a-9+8b-10c) + (-9+8b-11c)

P (ABC) = 3a-7b+8c +a-9+8b-10c -9+8b-11c

P (ABC) = 3a+a -7b+8b+8b +8c-10c-11c -9-9

P (ABC) = 4a+9b-13c-18

Answer 2

Answer:

8

Step-by-step explanation:

Let the three sides be x, y nd z respectively.

∴ Given that,

x=(3a−7b+8c)cm

y=(a−9b+10c)cm

z=(−9a+8b−11c)cm

Perimeter of triangle =−5a−mb+7c⟶(i)

As we know that,

perimeter of triangle = sum of all sides

∴ Perimeter of triangle =x+y+z=(3a−7b+8c)+(a−9b+10c)+(−9a+8b−11c)

⇒ Perimeter of triangle =−5a−8b+7c⟶(ii)

Comparing eq

n

(i)&(ii), we have

m=8

Hence, the correct answer is 8.