Mathematics, 23.07.2020 09:01 bollipard1183

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Answer from: aleilyg2005

s = 25.33m

θ = 60.65°

12.37m

A = 160m^2

Step-by-step explanation:

The pyramid has a side base of 35m and a height of 22m.

side base = b = 35m

height of the pyramid = h = 22m

To calculate the slant edge of the pyramid, you first calculate the diagonal of the squared base of the pyramid.

You use the Pythagoras theorem:

$d=\sqrt{(\frac{35}{2})^2+(\frac{35}{2})^2}=24.74$

With the half of the diagonal and the height, and by using again the Pythagoras theorem you can calculate the slat edge:

$s=\sqrt{(\frac{24.74}{2})^2+(22)^2}=25.23$

The slant edge of the pyramid is 25.33m

The angle of the base is given by:

$\theta=sin^{-1}(\frac{h}{s})=sin^{-1}(\frac{22}{25.23})=60.65\°$

The angle of the base is 60.65°

The distance between the corner of the pyramid and its center of its base is half of the diagonal, which is 24.74/2 = 12.37m

The area of one side of the pyramid is given by the following formula:

$A=\frac{(b/2)l}{2}$ (1)

l: height of the side of pyramid

then, you first calculate l by using the information about the side base and the slant.

$l=\sqrt{s^2-(\frac{b}{2}^2)}=\sqrt{(25.33)^2-(\frac{35}{2})^2}\\\\l=18.31m$

Next, you replace the values of l and b in the equation (1):

$A=\frac{(35/2)(18.31)}{2}=160m^2$

The area of one aside of the pyramid is 160m^2

Answer from: Quest

perimeter of the first rectangular board is 9 feet.

step-by-step explanation:

let p = perimeter of first board.

l = length of first board

w = width of first board

length of first board = 2 * width of first board - 3

l = 2w - 3

perimeter of first board = 2l + 2w

since l = 2w-3, then

p = 2*(2w-3) + 2w which equals 4w-6 + 2w which equals 6w-6

length and width of the second board are reciprocals of corresponding length and width of the first board.

length of second board = 1/(2w-3)

width of second board = 1/w

perimeter of second board is 1/5th perimeter of the first board.

perimeter of second board = p/5

perimeter of the second board = (2/(2w-3) + 2/w)

since the perimeter of the first board is 5 times the perimeter of the second board, this means that:

6w-6 = 5*(2/(2w-3) + 2/w)

which is the same as:

6w-6 = (10/(2w-3) + 10/w)

if we multiply both sides of this equation by (2w-3), we get:

(6w-6)*(2w-3) = 10 + 10*(2w-3)/w

if we multiply both sides of this equation by w, we get:

(6w-6)*(2w-3)*w = 10*w + 10*(2w-3)

we can simplify this to become:

(6w-6)*(2w-3)*w = 10w + 20w - 30

which becomes:

(6w-6)*(2w-3)*w = 30w-30

if we divide both sides of this equation by (6w-6), we get:

(2w-3)*w = (30w-30)/(6w-6)

which becomes:

(2w-3)*w = 5

this can be simplified to:

2w^2 - 3w = 5

subtract 5 from both sides of this equation to get:

2w^2 - 3w - 5 = 0

which can be factored into:

(2w-5)*(w+1) = 0

which makes:

w = 5/2

w = -1.

w can't be negative so the only possible answer is w = 5/2.

if w = 5/2, then

p = 2w-3 = 5-3 = 2

we have:

p = 2

w = 5/2

p = 2l + 2w gets p = 4 + 5 = 9

the perimeter of the first board is 9 feet.

since the perimeter of the second board is 1/5 the perimeter of the first board, the perimeter of the second board is 9/5.

to see if that's correct, we substitute known values for l and w into the second board.

the length of the second board is 1/l.

this becomes 1/2.

the width of the second board is 1/w

this bgecomes 1/(5/2) = (2/5)

let l2 = length of second board.

let w2 = width of second board.

l2 = (1/2)

w2 = (2/5)

let p2= perimeter of second board.

then p2 = 9/5

since p2 = 2*l2 + 2*w2, we should get p2 = 9/5 using the dimensions of l2 and w2.

2 * l2 = 2 * (1/2) = 1

2 * w2 = 2 * (2/5) = 4/5

1 is the same as 5/5

5/5 + 4/5 = 9/5

p2 checks out good.

the dimensions as given are good.

answer to the problem is:

perimeter of the first rectangular board is 9 feet.

Answer from: Quest

-22

times both sides by 2, times () by 8, add alile terms, move terms, add alike and caculate divide both sides by 3

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