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Sagot :
To solve this problem, let's break it down step-by-step:
1. Height of the Pyramid:
The length of the base edge of the pyramid is given as [tex]\( x \)[/tex].
The height of the pyramid is 3 times this base edge length. Therefore, the height of the pyramid is:
[tex]\[ 3x \][/tex]
2. Area of an Equilateral Triangle:
The area of an equilateral triangle with side length [tex]\( x \)[/tex] is given by:
[tex]\[ \frac{x^2 \sqrt{3}}{4} \][/tex]
3. Area of the Hexagon Base:
A regular hexagon can be divided into 6 equilateral triangles. Therefore, the total area of the hexagon base is 6 times the area of one equilateral triangle:
[tex]\[ 6 \times \frac{x^2 \sqrt{3}}{4} = \frac{6x^2 \sqrt{3}}{4} = \frac{3x^2 \sqrt{3}}{2} \][/tex]
4. Volume of the Pyramid:
The volume of a pyramid is given by:
[tex]\[ \text{Volume} = \frac{1}{3} \times \text{Base Area} \times \text{Height} \][/tex]
Substituting the known values:
[tex]\[ \text{Volume} = \frac{1}{3} \times \left( \frac{3x^2 \sqrt{3}}{2} \right) \times 3x \][/tex]
Simplify the expression:
[tex]\[ \text{Volume} = \frac{1}{3} \times \frac{3x^2 \sqrt{3}}{2} \times 3x = \frac{3 \times 3 x^3 \sqrt{3}}{2 \times 3} = \frac{9x^3 \sqrt{3}}{6} = \frac{3x^3 \sqrt{3}}{2} \][/tex]
So, the detailed answers are:
- The height of the pyramid is [tex]\( 3x \)[/tex].
- The area of an equilateral triangle with side length [tex]\( x \)[/tex] is [tex]\( \frac{x^2 \sqrt{3}}{4} \)[/tex] square units.
- The area of the hexagon base is [tex]\( \frac{3x^2 \sqrt{3}}{2} \)[/tex] square units.
- The volume of the pyramid is [tex]\( \frac{3x^3 \sqrt{3}}{2} \)[/tex] cubic units.
1. Height of the Pyramid:
The length of the base edge of the pyramid is given as [tex]\( x \)[/tex].
The height of the pyramid is 3 times this base edge length. Therefore, the height of the pyramid is:
[tex]\[ 3x \][/tex]
2. Area of an Equilateral Triangle:
The area of an equilateral triangle with side length [tex]\( x \)[/tex] is given by:
[tex]\[ \frac{x^2 \sqrt{3}}{4} \][/tex]
3. Area of the Hexagon Base:
A regular hexagon can be divided into 6 equilateral triangles. Therefore, the total area of the hexagon base is 6 times the area of one equilateral triangle:
[tex]\[ 6 \times \frac{x^2 \sqrt{3}}{4} = \frac{6x^2 \sqrt{3}}{4} = \frac{3x^2 \sqrt{3}}{2} \][/tex]
4. Volume of the Pyramid:
The volume of a pyramid is given by:
[tex]\[ \text{Volume} = \frac{1}{3} \times \text{Base Area} \times \text{Height} \][/tex]
Substituting the known values:
[tex]\[ \text{Volume} = \frac{1}{3} \times \left( \frac{3x^2 \sqrt{3}}{2} \right) \times 3x \][/tex]
Simplify the expression:
[tex]\[ \text{Volume} = \frac{1}{3} \times \frac{3x^2 \sqrt{3}}{2} \times 3x = \frac{3 \times 3 x^3 \sqrt{3}}{2 \times 3} = \frac{9x^3 \sqrt{3}}{6} = \frac{3x^3 \sqrt{3}}{2} \][/tex]
So, the detailed answers are:
- The height of the pyramid is [tex]\( 3x \)[/tex].
- The area of an equilateral triangle with side length [tex]\( x \)[/tex] is [tex]\( \frac{x^2 \sqrt{3}}{4} \)[/tex] square units.
- The area of the hexagon base is [tex]\( \frac{3x^2 \sqrt{3}}{2} \)[/tex] square units.
- The volume of the pyramid is [tex]\( \frac{3x^3 \sqrt{3}}{2} \)[/tex] cubic units.
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