Answer:
The perimeter of the triangle ABC is 17 cm.
Step-by-step explanation:
Consider the Isosceles triangle ABC.
The sides CA and CB are equal with measures, 5 cm.
The base angles are assumed to be x° each. Hence, the angle ACB is 2x°.
The altitude CP divides the base AB into two equal halves and the angle ACB is also cut into halves.
Consider the right angled triangle ACP.
The sum of all the angles in a triangle is 180°.
Determine the value of x as follows:
x° + x° + 90° = 180°
2x° = 90°
x° = 45°
Compute the length of side AP as follows:
[tex]cos\ 45^{0}=\frac{AP}{CA}[/tex]
[tex]\frac{1}{\sqrt{2}}=\frac{AP}{5}[/tex]
[tex]AP =\frac{5}{\sqrt{2}}\\\\AP=3.5[/tex]
Then the length of side AB is:
AB = AP + PB
= 3.5 + 3.5
= 7 cm
The perimeter of triangle ABC is:
Perimeter = AB + CA + CB
= 7 + 5 + 5
= 17
Thus, the perimeter of the triangle ABC is 17 cm.
