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Junior Executive (ATC) Official Paper 2: Held on Nov 2018 - Shift 2

Option 1 : \(\dfrac{1}{6}\) cubic units

Junior Executive (ATC) Official Paper 1: Held on Nov 2018 - Shift 1

20501

120 Questions
120 Marks
120 Mins

**Concept:**

**For triangular pyramid**

\(Volume = \frac 13 ~(area ~of ~triangular~ base)\times Height~ of ~pyramid \)

**Calculation:**

**Given:**

The bounded solid will be a equilateral triangular pyramid

side of equilateral triangular base = \(\sqrt {2}\)

Area of base of equilateral triangular pyramid = \(\frac {\sqrt {3}~a^2}{4}= \frac {\sqrt {3}~\times \sqrt {2}^2}{4}=\frac {\sqrt {3}}{2}\)

Height of pyramid = \(\frac 1{\sqrt {3}}\)(calculated by simple geometrical calculation)

\(Volume = \frac 13 ~(area ~of ~triangular~ base)\times Height~ of ~pyramid \)

\(Volume = \frac 13 ~\frac {\sqrt {3}}{2}\times \frac {1}{\sqrt {3}} = \frac 16\) cubic units