Square Roots Worksheet
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4Automotive Technology (47.0604) T-Chart
Apply properties of rational and irrational numbers to
Diagnose mechanical & electrical system problems
=
solve real-world or mathematical problems
Program Task: Diagnose electrical circuit, HVAC cooling and
PA Core Standard: CC.2.1.HS.F.2
engine drivability problems.
Description: Apply properties of rational and irrational
numbers to solve real-world or mathematical problems.
Program Associated Vocabulary:
Math Associated Vocabulary:
SQUARE ROOT, RESISTANCE, COFFICIENT OF DRAG
SQUARE ROOT
Program Formulas and Procedures:
Formulas and Procedures:
It is important for automotive technicians to properly diagnose fluid
and gaseous (i.e. air conditioning, emissions) system problems.
Find Square Roots:
Many devices change size in diameter, altering the amount of
3 in.
pressure or volume of fluid/gas in the system. They must be
diagnosed to determine the root cause of a malfunction. Here we’ll
learn how to diagnose an air conditioning system.
If a particular liquid enters the 1” diameter inlet with 10 lbs. of force
Area = 3 in x 3 in
3 in.
2
and exits the outlet with 2.5 lbs. of force, what is the radius of the
Area = 9 in
outlet?
9 = 3
Inlet
Outlet
Nearest Estimation Method to find Square Root:
Example: Estimate the square root of 7.
ILLUSTRATION IS NOT TO SCALE
1.
Pick two perfect squares closest to the number you want
to find the square root of; choose one perfect square
Find the cross-sectional area of the Inlet (A
).
inlet
greater than the number you want to find the square root
r = d/2 = 0.5”
of and one perfect square less than the number you want
to find the square root of. Two perfect squares below
2
2
2
= πr = π(0.5) = .784 in.
A
and above 7 are 4 and 9.
inlet
F
F
2.
Since 7 is closer to 9 than it is to 4, then
7 must be
inlet
outlet
=
between
4 = 2 and
9 = 3 but closer to
9 = 3.
A
A
inlet
outlet
3. An estimate around 2.6 to 2.7 would be fine.
10 lbs.
2.5lbs.
=
Cross-multiply
.784 sq.in.
A outlet
(2.5lbs.)(.784 sq.in.)
.
A
=
= .196 sq.in
outlet
10 lbs.
To find the radius of the outlet:
2
A=πr
πr
2
.196 sq.in. =
.196sq.in.
.
r =
= .0624 = .25 in
π
Note: While the radius of the outlet is half the radius of the inlet, the
force and area decreases by a factor of 4.
Originated June 2011
CC.2.1.HS.F.2
Reviewed June 2015
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