If 6 is the side of a regular pentagon, the area is 61.937
21.4
Area = 110.11 cm2
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
The 5 sided regular pentagon will consist of 5 congruent isosceles angles with equal sides of 6.379881063 cm by using the sine rule and 2 equal base angles of 54 degrees with an apex angle of 72 degrees opposite side 7.5 cm Area of the pentagon: 0.5*6.379881063^2 *sin(72)*5 = 96.77685379 square cm
17 cmImproved Answer:-Area = 1/2*6*9 = 27 square cm
61.90(:
21.4
45 cm
27.50 (:
Area = 110.11 cm2
It's got 5 sides, so multiply 6 cm by 5...
2.5 cm. If the area of a pentagon is increased by 12.5 cm and each side of the pentagon is the same length, then each side would increase by 2.5 cm. (12.5 cm/ 5 sides = 2.5 cm/ side.
124
27.50
we know that regular pentagon has 5 equilateral triangle . now ,perimeter of pentagon = 5 Xside . now,perimeter of a regular pentagon with a side length of 7.2 cm . perimeter = 7.2 X 5 =36 cm .
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
A pentagon has 5 sides A regular pentagon has all sides equal → perimeter of regular pentagon with side 17 cm = 5 × 17 cm = 85 cm