Help please. The length of a rectangle is 2 times the width. If the perimeter is to be less than 108 meters. What are the possible values for the width.

If the only perimeters are that the length has to be twice the width and less than 108 meters then there are an infinite amount of possible values for width. For example: 1.0002 and 2.0004 or 4 and 8. You can use an infinite amount of decimal places and therefore have an infinite amount of values.

The largest the width can be is the limit approaching 18 and it has to be greater than 0 tho

solution:

we have to construct

∠m nt ≅ ∠p qr

the steps to draw this construction are

step 1: use a compass to draw an arc from point q which intersects the side pq at point a and the side qr at point b.

step 2: draw a segment nt and use the same width of the compass to draw an arc from point n which intersects the segment nt at a point x.

step 3:

step 4: join points n and y using a straightedge.

step 3:

option (d).maintaining the same width of the compass as ab, draw an arc from point x such that it intersects the arc drawn from n in a point y.

to draw   congruent angles , 1. width of compasses while drawing two angles should be same i.e from point at which angle has to be drawn, 2.the distance between the two points where the compass intersects the two sides should be same while drawing any two congruent angles.