Perimeter of a rectangle is 22 m The perimeter of a triangle is 12 m. Find the dimensions of the rectangle. The triangles bases are l/2 and w with the hypotenuse of 5
The rectangle
Perimeter = 2 * length plus 2 * width
P = 2L + 2W
22 = 2L + 2W
L + W = 11 ------------------ (1)
The triangle
The triangle is a right-angled triangle; therefore the sun of the squares on the bases is equal to the square on the hypotenuse. i.e.
W^2 + (L/2)^2 = 5^2
W^2 + L^2/4 = 25
4W^2 + L^2 = 100 ------------- (2)
Perimeter of the triangle
p = W + L/2 + 5
12 = W + L/2 + 5
W + L/2 = 7
2W + L = 14 -------------------- (3)
Substituting for L = 11 – W from (1) into (3),
2W + (11 – W) = 14
2W – W = 14 – 11
W = 3
And L = 11 – W = 11 – 3 = 8
L = 8
There is another solution, using eqns (1) and (2).
Substituting for L = 11 – W from (1) into (2),
4W^2 + (11 – W)^2 = 100
4W^2 + 121 – 22W + W^2 = 100
5W^2 – 22W + 21 = 0
(5W – 7)(W – 3) = 0
W = 7/5, or W = 3
The 2nd solution is W = 7/5 = 1.4, from which L = 11 – W = 11 – 1.4 = 9.6
L = 9.6
The two solutions are: W = 3, L = 8, and W = 1.4, L = 9.6
