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Sagot :
Let's solve the problem step by step.
1. Apothem Calculation:
For a regular hexagon with a side length of 6 units, the formula for the apothem [tex]\( a \)[/tex] is given by:
[tex]\[ a = \text{side\_length} \times \frac{\sqrt{3}}{2} \][/tex]
Substituting the given side length of 6 units:
[tex]\[ a = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3} \][/tex]
So, the apothem is [tex]\( 3\sqrt{3} \)[/tex] units long.
2. Slant Height Calculation:
The slant height [tex]\( s \)[/tex] is the hypotenuse of a right triangle formed with the apothem [tex]\( a \)[/tex] and the height [tex]\( h \)[/tex] of the pyramid. Using the Pythagorean theorem:
[tex]\[ s = \sqrt{a^2 + h^2} \][/tex]
Substituting [tex]\( a = 3\sqrt{3} \)[/tex] and [tex]\( h = 3 \)[/tex]:
[tex]\[ s = \sqrt{(3\sqrt{3})^2 + 3^2} = \sqrt{27 + 9} = \sqrt{36} = 6 \][/tex]
So, the slant height is 6 units long.
3. Lateral Area Calculation:
The lateral area [tex]\( A \)[/tex] of the pyramid is given by:
[tex]\[ A = \frac{1}{2} \times \text{perimeter of base} \times \text{slant height} \][/tex]
For a hexagon with side length 6, its perimeter [tex]\( P \)[/tex] is:
[tex]\[ P = 6 \times 6 = 36 \][/tex]
So, substituting [tex]\( P = 36 \)[/tex] and [tex]\( s = 6 \)[/tex]:
[tex]\[ A = \frac{1}{2} \times 36 \times 6 = 108 \][/tex]
Thus, the lateral area is 108 square units.
Therefore:
- The apothem is [tex]\( 3\sqrt{3} \)[/tex] units long.
- The slant height is 6 units.
- The lateral area is 108 square units.
1. Apothem Calculation:
For a regular hexagon with a side length of 6 units, the formula for the apothem [tex]\( a \)[/tex] is given by:
[tex]\[ a = \text{side\_length} \times \frac{\sqrt{3}}{2} \][/tex]
Substituting the given side length of 6 units:
[tex]\[ a = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3} \][/tex]
So, the apothem is [tex]\( 3\sqrt{3} \)[/tex] units long.
2. Slant Height Calculation:
The slant height [tex]\( s \)[/tex] is the hypotenuse of a right triangle formed with the apothem [tex]\( a \)[/tex] and the height [tex]\( h \)[/tex] of the pyramid. Using the Pythagorean theorem:
[tex]\[ s = \sqrt{a^2 + h^2} \][/tex]
Substituting [tex]\( a = 3\sqrt{3} \)[/tex] and [tex]\( h = 3 \)[/tex]:
[tex]\[ s = \sqrt{(3\sqrt{3})^2 + 3^2} = \sqrt{27 + 9} = \sqrt{36} = 6 \][/tex]
So, the slant height is 6 units long.
3. Lateral Area Calculation:
The lateral area [tex]\( A \)[/tex] of the pyramid is given by:
[tex]\[ A = \frac{1}{2} \times \text{perimeter of base} \times \text{slant height} \][/tex]
For a hexagon with side length 6, its perimeter [tex]\( P \)[/tex] is:
[tex]\[ P = 6 \times 6 = 36 \][/tex]
So, substituting [tex]\( P = 36 \)[/tex] and [tex]\( s = 6 \)[/tex]:
[tex]\[ A = \frac{1}{2} \times 36 \times 6 = 108 \][/tex]
Thus, the lateral area is 108 square units.
Therefore:
- The apothem is [tex]\( 3\sqrt{3} \)[/tex] units long.
- The slant height is 6 units.
- The lateral area is 108 square units.
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