Question

The perimeter of an isosceles triangle is 42cm, and its base is $1\dfrac{1}{2}$times its congruent sides. Find the length of the congruent sides.

Answer 1

Hint: Assume the length of the congruent sides of the isosceles triangle is x. Express the length of base in terms of x. Use the perimeter of a triangle as equal lengths of sides of the triangle to form a linear equation in x. Solve the equation to find the value of x. The value of x is the length of the congruent sides of the triangle.

Complete step-by-step answer:
Let the length of the congruent sides of the triangle be x.
Since the length of the base is $1\dfrac{1}{2}=1+0.5=1.5$times the length of the congruent sides, we have
Length of base = 1.5x
Now perimeter of triangle = sum of lengths of sides of triangle = 42
Hence, we have
x+x+1.5x=42
i.e.3.5x =42
Dividing both sides by 3.5, we get
$\begin{align}
  & \dfrac{3.5x}{3.5}=\dfrac{42}{3.5} \\
 & \Rightarrow x=\dfrac{420}{35} \\
 & \Rightarrow x=12 \\
\end{align}$
 Hence the length of congruent sides is equal to 12cm.

Note: Although we got the value of x, there is a possibility that the triangle may not exist.
A triangle with sides a,b,c exists if and only if a+b>c, b+c>a and c+a>b all hold.
Here the length of base = 1.5x= 18.
Hence a = 12, b =12 , c = 18
Now a+b =12+12 =24>18=c.
Hence a+b>c.
b+c = 12+18 =30>12=a.
Hence b+c>a.
a+c = 12+18 = 30>12=b.
Hence a+c>b.
Hence all of the identities a+b>c, b+c>a and a+c>b hold.
Hence a triangle with side a = 12, b= 12 and c= 18 is a valid triangle.