Answer:
Maximum slope for hand-propelled wheelchair ramps should be 1" of rise to every 12" of length (4.8 degree angle; 8.3% grade).
Maximum slope for power chairs should be 1.5" rise to 12" length (7.1 degree angle; 12.5% grade).
Minimum width should be 36" (inside rails) - (48" is ideal).
The "deck" or surface of the ramp should be set down between a side-rail assembly such that there is about a 2" curb or lip along the edges of the ramp surface. Decking could consist of 1" X 6" pressure treated pine, (or 3/4" pressure treated plywood applied to a frame).
If possible, the end of the deck (where it meets the lower ground surface) should be beveled to provide a smooth transition from the ramp to level ground. Alternatively, a sheet of 10 Ga. steel at least 10" long and sized to fit the width of the ramp could be used to span the space between the deck surface and the walk or driveway surface at the end of the ramp. This piece should overlap the ramp deck by 2" and be fastened securely with 4 large countersunk flat-head wood screws.
A level platform of at least 5' X 5' should be at the top of ramp to allow for wheelchair maneuvering. If the entrance way opens outward, there should be 1' of surface area extending from the side of the door opening to allow motion to the side without backing the chair during door opening. This landing should not be considered part of the overall "run"/length of the ramp. Any turning point along the ramp needs a level landing. If the turn is a right angle (90 degrees), the landing should be a minimum of 5' by 4'. If a "switchback" of 180 degrees is constructed, the level landing should measure at least 5' X 8'. Ramps longer than 30' should provide a platform every 30' for purposes of safety and to create opportunity for rest
Step-by-step explanation:
Answer:
ΔRCQ = ΔAPS
QR = PS (CPCTC)
ΔSRD ≅ ΔQPB
PQ = RS (CPCTC)
Step-by-step explanation:
Given that ABCD is a parallelogram, we have;
AD = BC, and DC = AB (Definition of a parallelogram)
AD = AS + SD Given S is midpoint of AD
Similarly,
DC = DR + RC
BC = BQ + QC
AB = AP + PB
RC = DR = AP = PB
BQ = QC = QC = BQ segments on either sides of midpoint
∠BCA = ∠DAB (opposite interior angles of a parallelogram)
ΔRQC ≅ ΔSPA
ΔRCQ = ΔAPS (Side Angle Side SAS rule of congruency)
Therefore, segment QR = segment PS (QR = PS) (Corresponding Parts of Congruent Triangles are Congruent CPCTC)
Similarly we have;
∠CDA = ∠CBA (opposite interior angles of a parallelogram)
ΔSRD ≅ ΔQPB (opposite interior angles of a parallelogram)
Therefore, segment PQ = segment RS (PQ = RS) (Corresponding Parts of Congruent Triangles are Congruent CPCTC).
I'm pretty sure the answer is c. hope this helps
Answer:
the answer missing length is 7.6 and yes i rounded it to the nearest tenth.
Step-by-step explanation:
