Solve this problem of geometry?

Question

Let there be a rectangle ABCD
On each side of the rectangle, ANY point is taken. (The point must lie on the side. You can’t produce a side backwards, and then take a point on it.)
Let the points be P, Q, R, S on the sides AB, BC, CD, DA respectively.

Now prove:
2(AC)<=PQ+QR+RS+ST
[Hint : for a triangle ABC, AC
When you’ve solved this one, try proving it without using the theorem given in the hint.

Answer 1

Join BD
Now, in triangle ABC,P and Q are the mid-points of sides AB and BC.Therefore PQ=1/2 AC
Similarly we can prove that RS=1/2AC
or,PQ+RS=AC
We can also prove,that QR+SP=BD
But in a rectangle the diagonals are equal
Therefore,AC=BD
Therefore,2AC=PQ+QR+RS+SP proved

Answer 2

Given W = L +9 and section W*L = 36 ==> (L+9) * L = 36 ==> L^2 + 9L = 36 ==> L^2 + 9L - 36 = 0 ==> L^2 + 12L -3L - 36 = 0 ==> L(L + 12) - 3(L + 12) = 0 ==> (L -3) (L + 12) = 0 ==> L = 3 OR L = -12 ==> L = 3 (damaging cost not allowed) length = 3 cm and Width = 3 + 9 = 12 cm Dimensions are length 3 and width 12

Answer 3

Steps:

1. 2(AC)<=PQ+QR+RS+ST
2. this question = waste of time
3. 6 pack of Budweiser
4. Drink 1
5. 6-1 = 5

Answer 4

Take X the symmetric of S with respect to B and the Y the symmetric of S with respect to D. Then XY = 2BD =2AC, YQ
I would be surprised if you could do without the triangular inequality, at least implicitly.