Tetrahedron MCQ Quiz in मल्याळम - Objective Question with Answer for Tetrahedron - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Given:

A box in the form of a tetrahedron having side 2√2m

Concept:

A tetrahedron is a pyramid having base as equilateral triangle and all face also as equilateral triangle.

Formula Used:

Slant height of tetrahedron = ½ × √3 × side

Calculation:

Slant height of tetrahedron = ½ × √3 × side = ½ × √3 × 2√2

= √3 × √2

= √6 m

Volume of a tetrahedron = \(V = \;\frac{{{a^3}}}{{6\sqrt 2 }}\)

Where, a = length of each side

V = (24 × 24 × 24)/6√2

∴ V = (4 × 576)/√2

Given:

Regular tetrahedron with side length (a) = 10m.

Formula used:

Volume (V) of a regular tetrahedron = (a³√2) / 12

Calculation:

V = (10³√2) / 12

V = (1000√2) / 12

V = (250√2) / 3 m³

∴ The correct answer is option 4.

Given:

The side of the regular tetrahedron = 12 cm

Formula Used:

The lateral surface area of the regular tetrahedron = 3√3 × a²/4

Where,

a = side of the regular tetrahedron

Calculation:

The lateral surface area of the regular tetrahedron = 3√3 × a²/4

⇒ 3√3 × (12)²/4

⇒ 3√3 × 144/4

⇒ 3√3 × 36

⇒ 108√3 cm²

Given, side of tetrahedron = 8 cm

As we know, surface area of tetrahedron = √3 × a² and volume of tetrahedron = √2/12 × a³

Given:

Side length of a regular tetrahedron (a) = \(\sqrt{3}\)

Formula Used:

Height (h) of a regular tetrahedron with side length 'a' is given by the formula: \(h = a \sqrt{\frac{2}{3}}\)

Calculation:

Substitute the value of the side length 'a' into the formula:

h = \(\sqrt{3} \times \sqrt{\frac{2}{3}}\)

h = \(\sqrt{3 \times \frac{2}{3}}\)

h = \(\sqrt{2}\)

∴ The height of the regular tetrahedron is \(\sqrt{2}\).

Length of each edge of tetrahedron = 6 cm

∴ Volume of the tetrahedron,

\(\Rightarrow \frac{{\sqrt 2 }}{{12}} \times {6^{3\;}}\) 

⇒ 18√2 cm³

Area of regular hexagon = 3√3a²/2, where a is the side of regular hexagon.

⇒ 72√3 = 3√3a²/2

⇒ a² = 144/3

⇒ a = 4√3

Given:

The side of the regular tetrahedron = 1 cm

Formula Used:

The total surface area of regular tetrahedron = a2 × √3

Where,

a = side of the regular tetrahedron

Calculation:

The total surface area of regular tetrahedron = a2 × √3

⇒ (1)2 × √3

⇒ 1 × √3

⇒ √3 cm2

Let the upper length of frustum be x

Side of top square base of the frustum:

16/(16 – 4) = 12/x

x = 9 cm

Lateral surface area of the pyramid with square base of side 12 cm = (Circumference of base × Lateral Height)/2

= (12 × 4) × 16/2

= 384 cm²

Lateral surface area of the pyramid with square base of side 9 cm = (Circumference of base × Lateral Height)/2

= (9 × 4) × 12/2

= 216 cm²

Lateral surface area of the frustum = 384 – 216 = 168 cm²

Total surface area of the frustum = Lateral surface area + Flat surface area

⇒ 168 + (9)² + (12)²

⇒ 393 cm²