College Math – Geometry Chapter Practice Set 5, “Introduction to Area Formulas”.

College Math – Geometry Chapter Practice Set 5, “Introduction to Area Formulas”.


Practice Set 5; solve each area problem below. In problem number 5
the figure you’re given is a circle. In the geometric world, the definition of a circle is a two-dimensional figure whose points are
all the same distance from a particular point. That particular
point referred to as the center. In everyone of these points along this
continuous curve here are the same distance forming are circle. To find the area, the second question of
problem 5 we’ll use the area of a circle formula.
Which is pi times the diameter squared divided by 4. We’ll use the approximate value 3.14 for pi. The diameter of a circle is a line segment whose end points are on the circle and the
line segment contains the center. In this figure we have the value of
67. Order of operations, we will take care of exponents first, 67 to the power of 2 means 67 times 67 giving us a resulting value of 4489. Next we’ll simplify the muliplication
in the numerator giving us a value of 14095.46 And last we’ll do the division which gives us a value of
3523.865 The instructions tells to round to
the nearest tenth. We look in the hundreds, 5 for
greater is going to increase our tenths value by 1.
So our final answer is 3523.9. And as far as units, the only place that
units were located was in the diameter. That diameter was squared, centimeters times centimeters gives us
centimeters squared. Pi and 4 aren’t altering our units so our
final answer is square centimeters. And we can indicate that with the centimeter to the second power. In problem number
7, as far as naming that object what we have is a four sided
figure with only one pair of parallel sides opposite one another and that’s called a
trapezoid. In a trapezoid, we have special names for the sides that
are parallel, we call those bases. And I’ll refer to these as base 1 and base 2. The distance between the two
bases is called the height, and I’ll use H to represent that value. To find the area we use the formula b1 + b2, in other words, the sum of the two bases
times the height all over 2. Essentially what this formula is doing,
anytime you add up 2 things and divide by 2 your
averaging. And what we’re doing is we’re averaging
these 2 sides that are opposite one
another to an average length and then multiplying it by the height. Plugging in the values that are
given we have a 23 for one base 18 for the other, the distance between the
two is 7.5 all over 2. Order of operations, the parenthesis would have the priority
so we’ll add those two values giving us 41. Next, we’ll times that by 7.5 which gives us 307.5. And last we’ll divide that by 2 to give us 153.75, rounding this to the nearest tenth 5 or greater. Well we have a 5 so our
(.7) will turn into (.8) so we have 153.8. As far as units, if we look at
the formula we would have been adding yards together
here. Which just gives us yards times the
height which would have been in yards, yards times yards gives us square yards. Problem number 9, they’re asking us to identify this shape and what we have is a
four-sided figure where both pairs of opposite sides are parallel we refer
to this as a parallelogram. To find the area of a parallelogram we take the base times the height. And the height is the distance between 2 parallel sides. Well the only distance
given here between two parallel sides is this value indicated. We’ll call that the height which means
these are going to act as our base here. In this problem, we have additional
information we’re not going to use that .8″ on either side that would
only be necessary if we were interested in the perimeter of this figure. So to calculate this value we’ll take
the base again either one of these would have worked, 1.25 inches times it by the height, the distance between the parallel sides, which is .65 of an inch. Only operation is multiplication, when you multiply we end up with
8.8125. Inches times inches is square inches. And in this problem also rounding it to the nearest tenth when we
look at the hundreds position, 5 or greater
is not the case, so we’ll leave it as (.8) and it is inches squared.

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