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secondary 2 | Maths
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How to do (b)?
Date Posted: 4 years ago
Views: 231
Use Pythagoras theorem again.
A regular hexagon can be divided into 6 equilateral triangles.
For each of these triangles, you can divide further into 2 right angled triangles.
The hypotenuse is one side of the right angled triangle, 6cm long. The adjacent is half the base of the triangle, 3cm long.
What you want to find is the
opposite, which is used as the height of the triangle to calculate the area later.
opp² + (3cm)² = (6cm)²
opp² + 9cm² = 36cm²
opp² = 27cm²
opp = √27 cm = √(3x9)cm = 3√3cm
Area of hexagon base
= Area of all 6 equilateral triangles
= 6 x ½ x base x height
= 3 x 6cm x 3√3 cm
= 54√3 cm²
Volume of any pyramid
= ⅓ x base area x vertical height
= ⅓ x 54√3 cm² x 8cm
= 144√3 cm³
≈ 249cm³ (3.sf)
A regular hexagon can be divided into 6 equilateral triangles.
For each of these triangles, you can divide further into 2 right angled triangles.
The hypotenuse is one side of the right angled triangle, 6cm long. The adjacent is half the base of the triangle, 3cm long.
What you want to find is the
opposite, which is used as the height of the triangle to calculate the area later.
opp² + (3cm)² = (6cm)²
opp² + 9cm² = 36cm²
opp² = 27cm²
opp = √27 cm = √(3x9)cm = 3√3cm
Area of hexagon base
= Area of all 6 equilateral triangles
= 6 x ½ x base x height
= 3 x 6cm x 3√3 cm
= 54√3 cm²
Volume of any pyramid
= ⅓ x base area x vertical height
= ⅓ x 54√3 cm² x 8cm
= 144√3 cm³
≈ 249cm³ (3.sf)
See 1 Answer
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Remember to memorize your formulas, they will come in handy.
Date Posted:
4 years ago
Area of hexagon formula not covered in secondary 2
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