Find the probability of drawing a pawn a knight and another pawn from the bag of white pieces
1 answer:
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Answer:
The distance between Damarian and his friend is approximately 127 feet
Step-by-step explanation:
- The four sides of the rhombus are equal
- The diagonals of the rhombus bisects each other and perpendicular to each other
<em>Look to the attached figure</em>
∵ 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
<em>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</em>
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
