(a) \(\text{Perimeter of}\; \triangle\; ABC\) = \(AB + BC + CA\)
= \(3\) \(cm\) + \(5\) \(cm\) + \(4\) \(cm\)
= \(12\) \(cm\)
(b) \(\text{Perimeter of equilateral}\; \triangle \;ABC\) = \(3 \times side\)
= \(3 \times 9\) \(cm\)
= \(27\) \(cm\)
(c) \(\text{Perimeter of}\; \triangle\; ABC\) = \(AB + BC + CA \)
= \(8\) \(cm\) + \(6\) \(cm\) + 8 \(cm\)
= \(22\) \(cm\)
Let's find the perimeter of each shape:
1. Triangle of sides 3 cm, 4 cm, and 5 cm:
The perimeter P of a triangle is the sum of the lengths of its three sides.
\(P = 3 \text{ cm} + 4 \text{ cm} + 5 \text{ cm} = 12 \text{ cm}\)
2. An equilateral triangle of side 9 cm:
In an equilateral triangle, all sides are equal. So, the perimeter is:
\(P = 9 \text{ cm} + 9 \text{ cm} + 9 \text{ cm} = 27 \text{ cm}\)
3. Isosceles triangle with equal sides 8 cm each and the third side 6 cm:
In an isosceles triangle, two sides are equal, and the third side is different. So, the perimeter is:
\(P = 8 \text{ cm} + 8 \text{ cm} + 6 \text{ cm} = 22 \text{ cm}\)
So, the answer is 12cm, 27cm and 22cm.