A right pyramid has an equilateral triangular base of side 4 units. If

1.	A right pyramid has an equilateral triangular base of side 4 units. If the number of square units of its whole surface area be three times the number of cubic units of its volume, find its height.(a) 6 units (b) 10 units (c) 8 units (d) 4 units
Answer» Answer: (c) = 8 units Let a be the length of each side of the base, h be the height and l be the slant height of the pyramid. Here, a = 4 cm. ∴ Slant height (l) = \(\sqrt{h^2 +\frac{a^2}{12}}\) = \(\sqrt{h^2 +\frac{16}{12}}\) = \(\sqrt{h^2 +\frac{4}{3}}\) According to the given question, Lateral surface area + Area of the base = 3 (Volume) ⇒ \(\frac{1}{2}\times 12 \times \sqrt{h^2+\frac{4}{3}} +\frac{\sqrt{3}}{4} \times (4)^2\) = \(3 \times \frac{1}{3}\times \big(\frac{\sqrt{3}}{4} \times 4^2 \times h\big)\) ⇒ \(6\sqrt{h^2+\frac{4}{3}} +4\sqrt{3}\) = \(4\sqrt{3}\) h ⇒ \(6\sqrt{h^2+\frac{4}{3}} \) = \(4\sqrt{3}(h-1)\) ⇒ \(36\big(h^2+\frac{4}{3}\big) =48(h-1)^2\) ⇒ \(3\big(h^2+\frac{4}{3}\big) = 4(h-1)^2\) ⇒ 3h² + 4 = 4h² – 8h + 4 ⇒ h² – 8h = 0 ⇒ h(h – 8) = 0 ⇒ h = 8 as h ≠ 0.

A right pyramid has an equilateral triangular base of side 4 units. If the number of square units of its whole surface area be three times the number of cubic units of its volume, find its height.(a) 6 units (b) 10 units (c) 8 units (d) 4 units

Answer»

Answer: (c) = 8 units

Let a be the length of each side of the base, h be the height and l be the slant height of the pyramid.

Here, a = 4 cm.

∴ Slant height (l) = \(\sqrt{h^2 +\frac{a^2}{12}}\) = \(\sqrt{h^2 +\frac{16}{12}}\) = \(\sqrt{h^2 +\frac{4}{3}}\)

According to the given question,

Lateral surface area + Area of the base = 3 (Volume)

⇒ \(\frac{1}{2}\times 12 \times \sqrt{h^2+\frac{4}{3}} +\frac{\sqrt{3}}{4} \times (4)^2\)

= \(3 \times \frac{1}{3}\times \big(\frac{\sqrt{3}}{4} \times 4^2 \times h\big)\)

⇒ \(6\sqrt{h^2+\frac{4}{3}} +4\sqrt{3}\) = \(4\sqrt{3}\) h

⇒ \(6\sqrt{h^2+\frac{4}{3}} \) = \(4\sqrt{3}(h-1)\)

⇒ \(36\big(h^2+\frac{4}{3}\big) =48(h-1)^2\)

⇒ \(3\big(h^2+\frac{4}{3}\big) = 4(h-1)^2\)

⇒ 3h² + 4 = 4h² – 8h + 4

⇒ h² – 8h = 0 ⇒ h(h – 8) = 0

⇒ h = 8 as h ≠ 0.