Download Math 350 Section 2.1 Answers to Classwork

Math 350
Section 2.1
Answers to Classwork
I Definitions: Any side of a rectangle or parallelogram or triangle may be called the base. Once the
base is designated, an altitude is a segment that is perpendicular to the base and has one of its
end points on the base and the other endpoint on the other side parallel to the base. The length
of the altitude is called the height.
II Conclusion: If b is the length of the base of a rectangle and h is the height, then the Area A
of the rectangle is given by the formula A = bh.
III Classwork1 Below each of the diagrams below, a side is named. Sketch an altitude if that side is
considered to be the base of the figure.
E
N
M
A
J
C
B
CB
Base:
D
F
EF
G
H
GH
K
L
KN
IV Cut out the parallelogram with base b and height h on the Models page at the end of this
Section. Then it cut along the altitude. Rearrange the two pieces to form a rectangle. How is the area
of this rectangle related to the area of the original parallelogram? By the Fundamental Property of
Area the original parallelogram and the new rectangle have the same area.
Conclusion: If b is the length of the base of a parallelogram and h is the height, then the Area A
of the parallelogram is given by the formula A = bh.
V Cut out the two triangles with base b and height h shown on the Models page at the end of this
Section. Arrange them to form a new figure whose area we know how to determine from the
conclusions above.
What kind of figure is formed? A parallelogram What is the area of this figure in terms of b
and h? A = bh. How is the area of the original triangle related to the area of this figure? The area of
the original triangle is half the area of this parallelogram.
Conclusion: If b is the length of the base of a triangle and h is the height, then the Area A
1
of the triangle is given by the formula A = bh .
2
1
Classwork 2: A student says the area of the triangle shown below is (9 × 8) = 36 square units.
2
Is this correct? What do you say to the child?
7
8
9
Answer: The segment whose length is 8 is not an altitude for the base whose length is 9. The
base associated with the shown altitude has length 7, so A = 0.5(8)(7) = 28.
VI Definitions: The bases of a trapezoid are the two parallel sides. An altitude of the trapezoid is a
segment that is perpendicular to the bases and has its end points on the bases.. The height of the
trapezoid is the length of an altitude.
Cut out the two trapezoids shown on the Models page at the end of this Section. Each has bases
of lengths b1 and b2 and height h. Arrange the two trapezoids to form a new figure whose area we
already know how to determine from the area formulas above. What kind of figure is formed? A
parallelogram. What is the area of this figure in terms of b1, b2 and h ?
Each base has length (b1 + b2), so the area is A = (b1 + b2)h. How is the area of the original
trapezoid related to the area of this figure? The area of the original trapezoid is half the area of
this parallelogram. Conclusion: If b1 and b2 are the lengths of the bases of a trapezoid and h is the
1
height, then the Area A of the trapezoid is given by the formula A = ( b1 + b2 ) h .
2
Classwork 3: Find the area of this trapezoid two ways, first using the formula in VI, and then
again using the formulas in II and V. Do the answers agree?
8 cm
1
2
The formula in VI gives A = (8 + 11)4 = 38 cm
2
4 cm
The dotted altitude shown divides the trapezoid into a
11 cm
rectangle and a triangle. Their areas are 8x4 = 32 cm2 and
1
3(4) = 6 cm2 , respectively, so the total area is 32 + 6 = 38 cm2. The two methods agree.
2
Classwork 4: Proving the Pythagorean Theorem with Algebra
Four copies of a right triangle with sides a and b and hypotenuse c are arranged at right so that the
outer sides line up to form a square. The region inside the arrangement is a rhombus since all its
sides are of length c
1. Recall that the sum of the angles in a triangle is always 180 degrees. Why does this guarantee that
the angles of the inner quadrilateral are right angles? Why does this mean the inner quadrilateral is a
square? At the point where the corner of the inner quadrilateral meets the side of the outer
quadrilateral three angles are formed. Two of those angles are the angles of a right triangle, so
the sum of their measures is 90 degrees. The other angle is the interior angle of the inner
quadrilateral. so the interior angle of the inner quadrilateral must measure 180 – 90 = 90
degrees. All the sides of the inner quadrilateral have length c, so the inner quadrilateral is a
square.
2. Write the area of the big square as the sum of the areas of the 4 triangles plus the area of the inner
square. Then use algebra to reduce this equation to the Pythagorean Theorem!
Area of outer square = area of inner square + total area of the 4 triangles.
⎛ 1⎞
(a + b)2 = c 2 + 4 ⎜ ⎟ ab
⎝ 2⎠
a 2 + 2ab + b 2 = c 2 + 2ab
a 2 + b 2 = c 2
Classwork 5: Find the area and perimeter of the figure at right. (All
measurements are in inches.) Use the Pythagorean Theorem as needed.
The figure consists of a 10 x 10 square, a right triangle with sides 4
and 10 inches, and a semicircle of diameter 10 inches. The area of
the figure is the sum of the areas of these pieces:
1
1
A = (10)(10) + (4)(10) + π (5)2
2
2
= 100 + 20 + 12.5π = 120 + 12.5π ≈ 158.875 in 2
The perimeter is the sum of the lengths of the 14 inch bottom, the
10 inch right side, the semicircle, and the hypotenuse of the
triangle.
1
P = 14 + 10 + (10)π + 4 2 + 10 2 inches
2
= 24 + 5π + 116 inches, about 50.47 inches.
10
10
14