State whether each of the following statements is true or false:
$(1)$ If the side of an equilateral triangle is $10 \, cm,$ then its semi-perimeter is $30 \, cm.$
$(2)$ In an isosceles triangle,the length of two equal sides is $20 \, cm,$ then its semi-perimeter is $20 \, cm.$
$(3)$ In $\Delta ABC,$ $\angle B = 90^{\circ},$ $AB = 6 \, cm$ and $BC = 8 \, cm,$ then its semi-perimeter is $12 \, cm.$
$(1)$ If the side of an equilateral triangle is $10 \, cm,$ then its semi-perimeter is $30 \, cm.$
$(2)$ In an isosceles triangle,the length of two equal sides is $20 \, cm,$ then its semi-perimeter is $20 \, cm.$
$(3)$ In $\Delta ABC,$ $\angle B = 90^{\circ},$ $AB = 6 \, cm$ and $BC = 8 \, cm,$ then its semi-perimeter is $12 \, cm.$
(A) $(1)$ False. The perimeter of an equilateral triangle with side $a = 10 \, cm$ is $3a = 30 \, cm.$ The semi-perimeter $s$ is $\frac{30}{2} = 15 \, cm,$ not $30 \, cm.$
$(2)$ False. An isosceles triangle has two equal sides of $20 \, cm.$ Without the third side,the semi-perimeter cannot be determined as $20 \, cm.$ Even if the third side were $0$ (which is impossible),the semi-perimeter would be $\frac{20+20+0}{2} = 20 \, cm.$ Since the third side must be greater than $0,$ the semi-perimeter must be greater than $20 \, cm.$
$(3)$ False. In a right-angled triangle with sides $AB = 6 \, cm$ and $BC = 8 \, cm,$ the hypotenuse $AC = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, cm.$ The perimeter is $6 + 8 + 10 = 24 \, cm.$ The semi-perimeter $s = \frac{24}{2} = 12 \, cm.$ Wait,the statement is True.
$(2)$ False. An isosceles triangle has two equal sides of $20 \, cm.$ Without the third side,the semi-perimeter cannot be determined as $20 \, cm.$ Even if the third side were $0$ (which is impossible),the semi-perimeter would be $\frac{20+20+0}{2} = 20 \, cm.$ Since the third side must be greater than $0,$ the semi-perimeter must be greater than $20 \, cm.$
$(3)$ False. In a right-angled triangle with sides $AB = 6 \, cm$ and $BC = 8 \, cm,$ the hypotenuse $AC = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, cm.$ The perimeter is $6 + 8 + 10 = 24 \, cm.$ The semi-perimeter $s = \frac{24}{2} = 12 \, cm.$ Wait,the statement is True.
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