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What is the area of a regular pentagon if the radius r of the circumcircle is 3.4 cm and half the side length a is 2 cm rounded to 1 decimal place?
Wiki User
∙ 7y ago
In a regular pentagon there are 5 vertices which connect to the circumcentre each a distance r = 3.4 cm away. This creates 5 triangles inside the pentagon; so the area of the pentagon is the sum of these five triangles; as the pentagon is regular, this is the same as 5 times the area of one of these triangles.
The length of each side of the pentagon is twice half the length of one side
→ side = 2 × a = 2 × 2 cm
Consider one side and the two radii connecting its ends to the circumcentre; this is an isosceles triangle, so dropping a perpendicular from the circumcentre to the side of the pentagon bisects the base of the triangle (side of the pentagon). From this the height of the triangle can be calculated using Pythagoras:
half_side² + height² = radius²
→ (a cm)² + height² = (r cm)²
→(2 cm)² + height² = (3.4 cm)²
→ height² = (3.4 cm)² - (2 cm)²
→ height = √(3.4² - 2²) cm
area_pentagon = 5 × area_triangle
= 5 × ½ × base × height
= 5 × ½ × (2 × 2 cm) × √(3.4² - 2²) cm
= 10 × √7.56 cm²
= 27.495... cm²
≈ 27.5 cm² to 1 dp
Wiki User
∙ 7y ago
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