Question

the perimeter of an isosceles triangle is 27 cm and the length of the congruent sides is greater than twice the length of the base by 1 cm find the length of each side of the triangle

Answer 1

let the length of congruent sides be x and length of its base be y
then x=2y+1
and perimeter = x+x+y = 2(2y+1) +y = 5y+2
27 = 5y+2
y = 5cm
x = 2y+1 = 2*5+1 = 11cm
hence the sides of the triangle are 11,5,11cm respectively

Answer 2

Let the triangle be ABC, where AB and AC are equal sides and BC is the rest.
Then,
AB = AC = x (say)
BC = y (say)

Therefore,
AB + AC + BC = 27cm
I.e. x + x + y = 27cm

2x + y = 27cm

Acc. to the question
x = 2y + 1

Therefore,
2(2y + 1) + y = 27cm

Now,
4y + 2 + y = 27
5y + 2 = 27
5y = 25
y = 5

Putting the value of y in eq. x = 2y + 1
We get x = 11

Hence, AB = 11cm, AC = 11cm and BC = 5cm.