SOLUTION: Given a rectangle with a width of 43 cm and a diagonal 228 cm long, what is its length? Answer: The length is exactly cm. What is the length to the nearest t

Algebra ->  Proportions  -> Lessons -> SOLUTION: Given a rectangle with a width of 43 cm and a diagonal 228 cm long, what is its length? Answer: The length is exactly cm. What is the length to the nearest t      Log On


   

Question 1202039: Given a rectangle with a width of
43
cm and a diagonal
228
cm long, what is its length?
Answer: The length is exactly
cm.
What is the length to the nearest tenth of a cm?
Answer: The length is approximately
cm.
Found 2 solutions by mananth, Theo:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Apply Pytagoras theorem
l^2+w^2 = diagonal^2
L^2= diagonal^2-w^2
L= sqrt(diagonal^2-w^2)
l= sqrt(228^2-43^2)
Calculate and round off
.

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
the width, length, and diagonal of the rectangle form a right triangle.
the width of the rectnangle forms the vertical side of the trians which is equal t o 43 centimeter and the diagonal of the rectangle forms the hypotenuse of the triangle which is equal to 228 centimeters.

by pythagorus, the length squared equals the hypotenuse squared minus the width squared = 228^2 - 43^2 = 50135.
the exact length of the rectangle is sqrt(50135).
the approximate length of the rectamg;e os sqrt(50135) = 223.9084634.
round to the nearest tenth of a centimeter = 223.9.