The scale of a pitch is 1cm to 2.5m. If the length was drawn as 5cm, how long is the ground?
sketch a right-angled triangle representing your slope of 2.5/1
with a vertical of 5 and a horizontal of 2
then its hypotenuse is √29
In another similar triangle , let the "ground " be x
and its hypotenuse be 5
x /2 = √29/5
x = 2√29/5 = appr 2.154
in the first triangle the units were metres, in the second they were cm
In my equation, the units would cancel
Oooops .... My ratio should have been:
x/2 = 5/√29
x = 10/√29
x =1.8569
We could have used trig
from the slope of 2.5/1
tan P = 2.5/1
angle P = 68.1986°
now you want the base side
cos 68.1986 = x/5
x = 5cos68.1986 = appr 1.8569
nice trig, but what does it have to do with a scale drawing?
I think the "pitch" here is a sports field (although quite a small one)
1cm : 2.5m = 5cm : 12.5m
Was visualizing the pitch of a roof.
getting so tired of trying to make sense of poorly worded questions.
To determine the length of the ground, we need to establish the relationship between the length on the drawing (5cm) and the actual length (ground).
Given that the scale is 1cm to 2.5m, we can set up a proportion to solve for the ground length.
Let's define the variables:
- Length on the drawing: D (5cm)
- Length of the ground: G (unknown)
- Scale: 1cm = 2.5m
The proportion would be:
1cm/2.5m = 5cm/G
To solve for G, we can cross-multiply and then divide:
1cm * G = 5cm * 2.5m
G = (5cm * 2.5m) / 1cm
Now, let's calculate the value for G:
G = (5cm * 2.5m) / 1cm
G = 12.5m
Therefore, the length of the ground would be 12.5 meters.