Question

15. The length of each edge of a regular tetrahedron,

whose faces are identical equilateral triangles, is
8 cm. Find its
(i) slant height
(ii) volume.
Hint: All sides are equal. The centre of an equilateral
triangle is of its height h.
​

Answer 1

Slant height of tetrahedron is=6.53cm

Volume of the tetrahedron is=60.35
`\mathrm{cm}^(3)`

Given:

Length of each edge a=8cm

To find:

Slant height and volume of the tetrahedron

Step by Step Explanation:

Solution;

Formula for calculating slant height is given as

Slant height=
`\sqrt{(2)/(3)} a`

Where a= length of each edge

Slant height=
`\sqrt{(2)/(3)} * 8`

=
`√(0.6667) * 8`

=
`0.8165 * 8`=6.53cm

Similarly formula used for calculating volume is given as

Volume of the tetrahedron=
`(a^(3))/(6 √(2))`

Substitute the value of a in above equation we get

Volume=
`(8^(5))/(6 √(2))`

=
`(512)/(6 √(2))`

=
`(512)/(6 * 1.414)`

Volume=
`512 / 8.484`=60.35
`\mathrm{cm}^(3)`

Result:

Thus the slant height and volume of tetrahedron are 6.53cm and 60.35
`\mathrm{cm}^(3)`