The picture below shows Damarian standing at a corner of a field that is in the shape of a rhombus. His friend is standing in the opposite corner. The distance of each side of the field is 90 feet. The distance between the other two corners of the field is approximately 128 feet. What is the distance between Damarian and his friend, rounded to the nearest foot? (whole number)

The distance between Damarian and his friend is approximately 127 feet

Step-by-step explanation:

Look to the attached figure

∵ ADCF is a rhombus

∵ The distance of each side of the field is 90 feet

- The four sides of the rhombus are equal

∴ AD = DC = CF = FA = 90

∵ AC and DF are its diagonals

∴ AC ⊥ DF and bisects each other

∵ D is the position of Damarian

∵ F is the position of his friend

∵ The distance between the other two corners of the field is

   approximately 128 feet

∴ AC = 128 feet

∵ AC and DF bisects each other at point I

∴ AI = CI

- That means each part is one-half AC

∴ AI = CI = 128 ÷ 2 = 64 feet

Now let us use any right triangle of the four right triangles in the figure to find the length of one-half the other diagonal

In Δ FIC

∵ ∠FIC is a right angle

∵ FC = 90 ⇒ The hypotenuse of the Δ

∵ CI = 64 ⇒ one leg of the Δ

- By using Pythagoras Theorem

∴ (FC²) = (IF)² + (CI)²

- Substitute the values if CI and FC in the formula of Pythagoras

∵ (90)² = (IF)² + (64)²

∴ 8100 = (IF)² + 4096

- Subtract 4096 from both sides

∴ 4004 = (IF)²

- Take √  for both sides

∴ 63.277 = IF

∵ DI = IF = DF

- Multiply DI by 2 to find DF

∴ DF = 126.554

- Round it the the nearest whole number

∴ DF = 127 feet

DF represents the distance between Damarian and his friend

The distance between Damarian and his friend is approximately 127 feet

step-by-step explanation:

i found the koficent of the linear function and i foun c acordind to the strict formula y=kx+c and then i took 2 as an x and replacet it on the formula i found and then foun the y