Aregular octagon has side length 10.9 in. the perimeter of the octagon is 87.2 in and the area is 392.4 in2. a second octagon has side lengths equal to 16.35 in. find the area of the second octagon.

To solve this problem, let us first calculate for the Perimeter of the other octagon. The formula for Perimeter is:

Perimeter = n * l

Where n is the number of sides (8) and l is the length of one side. Let us say that first octagon is 1 and the second octagon is 2 so that:

Perimeter 2 = 8 * 16.35 in = 130.8 inch

We know that Area is directly proportional to the square of Perimeter for a regular polygon:

Area = k * Perimeter^2

Where k is the constant of proportionality. Therefore we can equate 1 and 2 since k is constant:

Area 1 / Perimeter 1^2 = Area 2 / Perimeter 2^2

Substituting the known values:

392.4 inches^2 / (87.2 inch)^2 = Area 2 / (130.8 inch)^2

Area 2 = 882.9 inches^2

Therefore the area of the larger octagon is about 882.9 square inches.

11.2

step-by-step explanation:

when doing tangent the tangent of the angle is equal to the opposite side from the angle over the adjacent side to the angle.

tan43 = x / 12

move 12 over by multiplying each side by 12

12tan43 = x

x = about 11.2

answer:

total cost for tiles and paints is $924.  

step-by-step explanation:

we have been given that a community hall is in the shape of a cuboid. the hall is 40m long 15m high and 3m wide.

the paint will be required for 4 walls and ceiling.

let us find area of walls and ceiling.

therefore, the area of walls and ceiling is 1410 square meters.

given: cost for 10 litre of paint is $10 and 10 litre paint covers 25 square meter. therefore,

therefore, the total painting cost is $564.

tiles will be required for floor. let us find the area of floor.

given: 1m squared floor tiles costs $3. so,

therefore, the total tiles cost is $360.

now let us find combined total cost of tiles and paint.

therefore, the combined total cost of tiles and paint is $924.