(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.
Complete the drawing shown in Fig. 9.14 to indicate where the free ends of the two wires should be joined to make the bulb glow