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If the sides of a pentagon equals x what is the height?
1.78x
Volume for a pentagon?
A pentagon is a two-dimensional geometric figure and therefore has area but not volume. To have volume, a geometric figure must have height or thickness as well as its plane dimensions.
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?
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
What is the formula for measuring the area of a pentagon?
You didn't say it was a regular pentagon. For an arbitrary pentagon, you would calculate its area as you would for any polygon: divide it up into triangles, and add up the areas of the triangle. The area of a triangle is 1/2 times the base times the height, the height being the length of the perpendicular dropped to the base from the opposite vertex.
How do you find the surface area of pentagon-based pyramid with a slant height of 9?
There is not enough information to answer the question.
