in the given figure ABCD is a trapezium in which ab parallel to dc. prove that: area of ∆acd =area of ∆bcd and area of ∆aod = ar
1 answer:
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Answer:
The kilometer pedestrian bridge B can be found = 14.1km
Question:
The following graph shows a highway that runs from north to south, on which there is a pedestrian bridge A. Another bridge B is built, such that bridge A is 600m from bridge B. At which kilometer is the pedestrian bridge B found? With explanation.
The complete information and diagram can be found in:
https://brainly.lat/tarea/17900202
Explanation:
The question requires us to find the distance of Pedestrian B from the north.
Bridge A is 600m from bridge B = distance between pedestrian bridge A and pedestrian bridge B
First we need to find the distance between them in km
1000m = 1km
600m = (600m × 1km)/1000m
= 0.6km
distance between pedestrian bridge A and pedestrian bridge B = 0.6km
To determine the kilometer pedestrian bridge B can be found, we would sum the distance of Pedestrian A from the north and the distance between pedestrian bridge A and pedestrian bridge B
The distance of Pedestrian A from the north = 13.5km
The kilometer pedestrian bridge B can be found = 13.5km + 0.6km
The kilometer pedestrian bridge B can be found = 14.1km
