Download Solutions to Exam III

Math 124 - Exam 3 Solutions - Spring 2009 - Jaimos F Skriletz
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1. Angles
(a) A line intersects two parallel lines as shown. Find the value of x.
By straight line angles it follows that
(7x + 3) + y = 180◦
By the parallel axiom and vertical angles, y = 10x + 7. Thus the value of x satisfies the equation:
(7x + 3) + (10x + 7) = 180
17x + 10 = 180
17x = 170
x = 10
(b) Find all the unknown angles in the below figure.
The vertical angels are shown on the diagram so it is sufficient to find the measure of the angels A, B, C and D
in the above diagram. Angle C and 130◦ form a straight line angle so
C + 130◦ = 180◦
C = 50◦
Angle A, C and 20◦ complete a triangle so
A + C + 20◦ = 180◦
A = 160◦ − C = 110◦
Angle A and B form a straight line angle so
A + B = 180◦
B = 180◦ − A = 70◦
Last angle D and 20◦ form a straight line angle so
D + 20◦ = 180◦
D = 160◦
Thus the measures of the angles are A = 110◦ , B = 70◦ , C = 50◦ and D = 160◦ . These angles along with their
vertical angle will gives every angle in the above figure.
Math 124 - Exam 3 Solutions - Spring 2009 - Jaimos F Skriletz
2. Find the area and perimeter of the following objects.
(a) Find the area by splitting the figure into two pieces as shown.
The area of the rectangle is
A1 = (4)(6) = 24
The area of the semicircle (half circle) with radius r = 2 is
A2 =
1
1
(πr2 ) = (π(4)) = 2π
2
2
Thus the area of the whole figure is
A = A1 + A2 = 24 + 2π square feet
The perimeter around the semicircle is 1/2(2πr) = 2π, Thus the total perimeter of the figure is
P = 2π + 6 + 4 + 6 = 16 + 2π feet
(b) Find the area by splitting the figure into two pieces as shown.
The area of the rectangle is
A1 = (4)(8) = 32
The area of the triangle with height h = 4 and base b = 3 is
A2 =
1
1
bh = (3)(4) = 6
2
2
Thus the total area of the figure is
A = A1 + A2 = 32 + 6 = 38 square feet
The perimeter of the figure is
P = 8 + 5 + 11 + 4 = 28 feet
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Math 124 - Exam 3 Solutions - Spring 2009 - Jaimos F Skriletz
3. Similar Triangles:
(a) Describe similar triangles and give the sufficient condition for which two triangles are similar.
Two triangles are similar if
i. Their three angles are equivalent (the same).
ii. Ratios of the corresponding sides are constant. If A, B and C are the lengths of the legs of one triangle and
a, b and c are the corresponding leg lengths on the other triangle, then
a
A
=
B
b
A
a
=
C
c
B
b
=
C
c
If two of the three angles of two triangles are equivalent (the same), then the two triangles are similar.
(b) If a 6 foot man standing 15 feet away from a street light casts a 5 foot shadow as shown, what is the height of the
street light?
Let h be the height of the street light as shown. Since the large right triangle (formed by the street light and the
tip of the shadow) and the inner right triangle (formed by the man and the tip of his shadow) both have a right
angle and share a common angle they are similar (have two equivalent angles). Thus the ratio of corresponding
legs is constant.
6
h
=
20
5
6
h = (20) = 24
5
Thus the street light is 24 feet high.
Math 124 - Exam 3 Solutions - Spring 2009 - Jaimos F Skriletz
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4. A 4 foot tall chest is constructed such that its length is three times its width (x) as shown.
(a) Find the volume and surface area of this box in terms of the variable x.
The volume of the chest is
V = x(3x)(4) = 12x2
The area of the front/back sides are
A1 = x(4) = 4x
The area of the left/right sides are
A2 = (3x)(4) = 12x
The area of the top/bottom sides are
A3 = x(3x) = 3x2
Thus the surface area of the whole chest is
A = 2A1 + 2A2 + 2A3 = 2(4x) + 2(12x) + 2(3x2 ) = 8x + 24x + 6x2 = 32x + 6x2
(b) If the volume of the chest is known to be 108 cubic feet, what is the value of x?
From part (a), V = 12x2 . Thus if the volume of the chest is 108, then
V = 108
12x2 = 108
x2 = 9
x=3
Thus if the volume is 108 cubic feet, the width of the box is x = 3 feet.
5. For any integer n bigger than 1, the values a = 2n, b = n2 −1 and c = n2 +1 form a Pythagorean triple (i.e. a2 +b2 = c2 ).
(a) What is the Pythagorean triple generated by n = 3?
If n = 3 then a = 2(3) = 6, b = 32 − 1 = 8 and c = 32 + 1 = 10.
(b) Show that the numbers found in part (a) form a Pythagorean triple.
If a = 6, b = 8 and c = 10 then
a2 + b2 = 62 + 82 = 100 = 102 = c2
Thus the values for a, b and c from part (a) form a Pythagorean triple.
(c) Show that the given values of a, b and c satisfy the relation a2 + b2 = c2 for all values of n.
First notice that
a2 + b2 = (2n)2 + (n2 − 1)2
= 4n2 + (n4 − 2n2 + 1)
= n4 + 2n2 + 1
On the other hand
c2 = (n2 + 1)2 = n4 + 2n2 + 1
Since the expressions for a2 + b2 and c2 are the same, they must be equal for any chosen value of n. Thus
a2 + b2 = c2 for all possible choices of n.