(c) 10 cm, 132 cm2
For the first prism, a = 5 cm, b = 12 cm, c = 13 cm
⇒ s = \(\frac{a+b+c}{2}=\frac{5+12+13}{2}\) = 15 cm
Area of the base = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{15\times10\times3\times2}\) cm2 = 30 cm2
Let V1 be the volume of the prism. Then,
V1 = Area of the base x height
⇒ V1 = (30 × 18) cm3 = 540 cm3
Let S1 be the total surface area of the prism. Then
S1 = Lateral surface area + 2 (Area of base)
= (Perimeter of the base × height) + 2 (Area of base)
= (30 × 18 + 2 × 30) cm2 = 600 cm2
Let h be the height of new prism. Then, for the new prism
a = 9 cm, b = 12 cm, c = 15 cm ⇒ s = \(\frac{9+12+15}{2}=18\) cm
Area of the base = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{18\times9\times6\times3}\) cm2 = 54 cm2
∴ V2 = Volume of new prism = (54 × h) cm3
V1 = V2 ⇒ 54 × h = 540 ⇒ h = 10 cm.
Let S2 be the total surface area of the new prism. Then,
S2 = (36 × 10 + 2 × 54) cm2 = 468 cm2
Change in the whole surface area = S1 – S2
= (600 – 468) cm2 = 132 cm2.