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given: circle k(o) with ot ⊥ xy, ou ⊥ wz , and ot≅ou, prove: △xoy ≅ △zow

ΔXOY ≅ ΔZOW ⇒ proved down

Step-by-step explanation:

* Lets study some facts on the circle

- If two chords equidistant from the center of the circles,

 then they are equal in length

* the meaning of equidistant is the perpendicular distances

 from the center of the circle to the chords are equal in length

* Lets check this fact in our problem

∵ XY and WZ are two chords in circle O

∵ OT ⊥ XY

- OT is the perpendicular distance from the center to the chord XY

∵ OU ⊥ WZ

- OU is the perpendicular distance from the center to the chord WZ

∵ OT ≅ OU

- The two chords equidistant from the center of the circle

∴ The two chords are equal in length

∴ XY ≅ WZ

* Now in the two triangles XOY and ZOW , to prove that

 they are congruent we must find one of these cases:

1- SSS ⇒ the 3 sides of the 1st triangle equal the corresponding

               sides in the 2nd triangle

2- SAS ⇒ the two sides and the including angle between them

                in the 1st triangle equal to the corresponding sides and

                including angle in the 2nd triangle

3- AAS ⇒ the two angles and one side in the 1st triangle equal the

                corresponding angles and side in the 2nd triangle

* Lets check we will use which case

- In the two triangles XOY and ZOW

∵ XY = ZW ⇒ proved

∵ OX = OZ ⇒ radii

∵ OY = OW ⇒ radii

* This is the first case SSS

∴ ΔXOY ≅ ΔZOW

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

or

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.

the answer is 5/3x+7/5y+7