an rectangle has a perimeter of 42cm and an area of 108cm2. What is the length of a diagonal of the rectangle?
2007-03-11 15:55:06 · 2 answers · asked by Natasha 1 in Science & Mathematics ➔ Mathematics
Perimeter of a rectangle: l + l + w + w = 42 cm
Area of a rectangle : l x w = 108 cm2
l = 12
w = 9
now use pythagorean theorem (W^2 + L^2 = c^2)
12^2 + 9^2 = 225
sqrt(225) = 15 <---- the diagonal!
2007-03-11 16:23:25 · answer #1 · answered by nemlo23 2 · 0⤊ 1⤋
Here's a neat way to solve it:
P = 2L + 2W = 42, so L+W = 21.
A = LW = 108
Square both sides of the perimeter equation:(L + W)^2 = 441,
but (L+W)^2 = L^2 + 2LW + W^2
Double both sides of the area equation: 2LW = 216.
Substitute this into the perimeter equation:
L^2 + 216 + W^2 = 441.
Subtract 216 from both sides:
L^2 + W^2 = 225.
But the diagonal squared = L^2 + W^2.
So, diagonal = sqrt(225) = 15cm
2007-03-12 00:35:54 · answer #2 · answered by s_h_mc 4 · 1⤊ 0⤋