If perimeter of triangle ABC is 32m and sides AB= x+3, BC = x+ 5, AC = x, what is the value of x?

Answer 1

The perimeter is the sum of the sides.

So:
(x+3) + (x+5) + (x) = 32
3x + 8 = 32
3x = 24
x = 8

Check your work:
AB = x+3 = 8+3 = 11
BC = x+5 = 8+5 = 13
AC = x = 8
11 + 13 + 8 = 32
Looks good to me!

Answer 2

P=32=x+3+x+5+x=3x+8
3x=32-8=24
x=8

check
8+3+8+5+8=32
32=32

Answer 3

If the perimeter is 32 m, then if you add the length of all the sides together you will get 32

AB + BC + AC = 32

plug in the x values:
(x+3) + (x+5) + x = 32
x+3+x+5+x = 32

combine like terms: (there are 3 x's and 5+3=8)
3x + 8 = 32

now subtract 8 from both sides
3x + 8 - 8 = 32 - 8
3x=24

now divide both sides by 3 to isolate the x:
x = 8

Answer 4

The perimeter is AB+BC+AC=x+3+x+5+x=3x+8=32, so x=(32-8)/3=24/3=8.

Answer 5

The sum of the three sides equals the perimeter, so:

(x + 3) + (x + 5) + x = 32

Now simplify the left side:

3x + 8 = 32

Solve for x:

3x + 8 - 8 = 32 - 8

3x = 24

1/3*(3x) = 1/3*24

x = 8

Now, check the answer

x = 8
x + 3 = 11
x + 5 = 13

8 + 11 + 13 = 32

Answer 6

Perimeter of triangle ABC
=side AB+SIDE BC+SIDE AC.........(1)
=(x+3)+(x+5)+x
=x+x+x+3+5
=3x+8.................(2)

NOW FROM (1) AND (2), WE HAVE

32 m=3x+8
3x=32-8
3x=24
x=24/3
x=8 m

(Now compare the results in reverse:

Perimeter of triangle ABC
=side AB+SIDE BC+SIDE AC
=(x+3)+(x+5)+x
=3x+8
=3*8+8
=24+8
=32)

Answer 7

P = A + B + C AB = X + 3
A = X +3/B
BC = X +5
B = X +5/C
AC = X
C = X/A
therefor

32 = (x + 3)/B + (X+5)/C + X/A
= (x + 3)/B + A(X+5)/X , since C = X/A
= (x + 3)/B + (AX + 5X)/X
32 = (x + 3)/B +( A +5)
32 = X + 3 + (A +5)B
= X +3 + ((x + 3)/B + 5)(X+5)/C
32 = X + 13
X = 19

Answer 8

8

Answer 9

perimeter=AB+BC+AC
32 =X+3+x+5+x
24=3x
x=8

Answer 10

AB + BC + AC = 32m
(x+3) + (x+5) + x = 32
3x + 8 = 32
3x = 24
x = 8m = length side AC