Show an equilateral triangle with the lengths of its side given in terms of a and b : 4a cm and 3b cm. The perimeter of the triangle is (a+b+3)cm.
Let's represent the equilateral triangle as ΔABC, where A, B, and C are the vertices, and AB = BC = AC.
Since AB = 4a cm and BC = 3b cm, we can write the side lengths as follows:
AB = 4a cm
BC = 3b cm
AC = length of the third side
As the triangle is equilateral, all sides are equal. Therefore, we have:
4a = 3b = AC
Since the perimeter of the triangle is (a + b + 3) cm, we can write this as:
AB + BC + AC = (a + b + 3) cm
Substituting the values we have:
4a + 3b + AC = a + b + 3
Now, let's solve for AC:
AC = (a + b + 3) - (4a + 3b)
AC = a + b + 3 - 4a - 3b
AC = (a - 4a) + (b - 3b) + 3
AC = -3a - 2b + 3
So, the length of the third side, AC, is -3a - 2b + 3 cm.
To summarize, the equilateral triangle with side lengths of 4a cm and 3b cm has a third side length of -3a - 2b + 3 cm.