G8-1 Area of Triangles
Workbook pages: 143–144
Goals: Students will develop and use the formula for the area of a triangle.
Prior Knowledge Required:
Can find the area of a rectangle using a formula
Can find the area of a right triangle
Understands that area is additive
Can identify congruent triangles
Can multiply decimals
Is proficient in measuring with a ruler
Can identify and draw right angles using a protractor or set square
Vocabulary:
area
base
height
Curriculum Expectations:
Ontario: 6m36, 6m37, 6m38, conceptual review; 8m2, 8m3, 8m7
WNCP: 7SS2, conceptual review; [C, R]
Review how to find the area of a rectangle and a right triangle. Emphasize the
connection between the two: a right triangle is half of a rectangle with the same
length and width as the sides of the triangle adjacent to the right angle.
Draw several acute and obtuse triangles on a grid so that the longest side is
horizontal.
This way:
Not this way:
Ask students to split the triangles into two right triangles. Then invite students to
draw a rectangle around each triangle and to use the rectangle to find the area of
both the original triangle and the two right triangles. ASK: What fraction of the
area of the big rectangle is the area of the big triangle? (half)
Introduce base and height. Point out that when we split a triangle into two
right triangles, we drop a perpendicular from one of the vertices to the
opposite side. The perpendicular is then called the height and the side we
draw the perpendicular to is called the base. Ask students to draw an acute
scalene triangle and to draw a perpendicular to each of its sides. Then
show the picture at right and have students explain how they could find the
area of the triangle. What fraction of the rectangle is the triangle? (half)
Ask students to identify the base and the height in the picture, then to write
the area of the rectangle in terms of base and height. What is the area of
the triangle in terms of base and height? (base × height ÷ 2)
4 cm
3 cm
2 cm
Draw a right triangle and choose one of the legs (legs are sides adjacent to the right angle) as a base.
Ask students to find the height. Since the triangle is a right triangle, the height is the second leg. Does the
formula “base × height ÷ 2” produce the same answer as “length × width ÷ 2”? (Yes: in the case of a right
triangle we can talk about length and width—as the length and width of the rectangle that was cut in half
to make the triangle—and they are the same as base and height if we regard one of the legs as the
base.)
Introduce the case of an obtuse triangle with one of the shorter sides chosen as the base. Ask
students to suggest how to draw the height to that base. If the solution does not arise, explain that when
we draw perpendiculars, they are perpendicular to the whole line containing the base, not just the line
segment itself. We can draw a perpendicular to the line that contains the base by first extending the
base. Would the formula “base × height ÷ 2” work in this case? How could we check?
One way to check is to find the area in two ways—using two different sides as bases—and compare the
answers. Ask students to do this with several obtuse triangles. Their answers are likely to differ slightly.
Why could this be? (because the measurements are imprecise) PA – Worksheets, Question 7 – [C, R],
7m3
A different way to check would be to look at the triangle as part of a
parallelogram (see picture). What is the formula for the area of a
parallelogram? Where in this picture is the base? Where is the height? Point
out, if necessary, that both dashed lines are heights of the parallelogram.
Are the dashed line segments the same length? (Yes: the top and bottom
sides of the parallelogram are parallel lines, so they are the same distance
apart.) Does it matter which one is used to find the area of the parallelogram? (no) What fraction of the
parallelogram is the triangle? (half) Does the formula “base × height ÷ 2” hold in this case? (yes) PS –
Revisiting conjectures that were true in one context
Turning triangles into rectangles with the same area. The next two questions could be done together
with Question 5 on the worksheet, as they present another method of finding the formula for the area of a
triangle by converting it to a rectangle with the same area as the triangle. In Question 5, the rectangle
has the same width as the base of the triangle but is half as high. In the first question below, the height of
the triangle and the rectangle are the same, but the rectangle is only half as wide as the triangle. The
second question compares the methods.
1. Ian wants to find the area of a triangle. He cuts the triangle and
rearranges the pieces as shown to form a rectangle.
a) What is the width of his rectangle? How is it related to the base of
the triangle?
b) What is the height of his rectangle? How is it related to the height
of the triangle?
c) Write a formula for the area of the triangle using the base and the
height of the triangle from Ian’s method.
ANSWERS:. The diagram shows that the width of the rectangle is half the base of Ian’s triangle and the
height of the rectangle is the same as the height of Ian’s triangle. The area of the rectangle, when
rewritten, produces the formula for the area of a triangle: (base ÷ 2) × height = base × height ÷ 2.
Jan and Ian start with the same triangle.
a) Draw how Jan and Ian cut and rearrange the triangle.
b) How are the bases and the heights of Jan’s and Ian’s triangles related?
c) Explain why they get the same area using the pictures and the formulas. PS – Reflecting on other
ways to solve a problem
ANSWERS:
Jan
Ian
base × (height ÷ 2) = base × height ÷ 2 = (base ÷ 2) × height
Jan’s rectangle has a base that is twice the base of Ian’s and a height that is half the base of Ian’s. So
the areas of the rectangles are the same.
Extensions
1. This extension goes with Question 7 on the worksheets, where students are likely to get three
different answers because the measurements are approximations. When you find the area of a
triangle by drawing heights and measuring the lengths of the sides and the heights, you perform two
imprecise operations—drawing a perpendicular and measuring with a ruler.
a) Suppose the triangle from Question 7 on the worksheets was much larger (for example, drawn on
the board) and the lengths were measured to the nearest centimetre, with area calculated to the
nearest centimetre squared (cm2). Are the three methods of calculating the area (using different
sides as bases) more likely to give the same answer in this case than in Question 7? Why or why
not?
ANSWER: Yes. The measurements would need to be less precise, so they would produce
answers that are closer to each other.
b) Here is a formula for area that uses only side lengths, which means you are only measuring
existing lines instead of a perpendicular that you drew. Find the area of the triangle in Question 7
using Heron’s formula (a, b, c are the sides):
Let s = (a + b + c) ÷ 2. The area of a triangle is
s ( s − a )( s − b)( s − c) .
2. a) Here are two ways to find the area of a trapezoid:
a
Method 1
length of rectangle =
h
a+b
2
height of rectangle = h
b
area of trapezoid = area of rectangle =
a
Method 2
a
h
h
+
h
a
b
area of trapezoid
a+b
×h
2
=
ah
b−a
+
(b − a) × h ÷ 2
i) Choose a value for h. h = _______
ii) Substitute the value you chose into both formulas and check that they produce the same answer.
iii) Do you think that the formulas will produce the same answer for any value of h?
ANSWERS: For h = 4:
a+b
) × 4 = (a + b ) ÷ 2 × 4 = (a + b ) × 2 = 2a + 2b
2
Method 1
(
Method 2
a × 4 + (b − a) × 4 ÷ 2 = 4a + (b − a) × 2
Applying the distributive law, we get 4a + 2b − 2a = 2a + 2b
The answer is the same for h = 4.
Explain that no matter what value of h you use, the two formulas will still give the same answer. So in
a+b
) × h = ah + (b − a ) × h ÷ 2
general, (
2
b) Substitute a = b into the formula for the area of a trapezoid. Explain the result geometrically.
ANSWER: The formula for the area of a trapezoid from Method 1 can be rewritten as (a + b) × h ÷ 2.
Substituting a = b will give 2a × h ÷ 2 = ah. This is the formula for the area of a parallelogram. Indeed, a
quadrilateral with equal parallel opposite sides is a parallelogram.
c) Substitute a = 0 into the formula for the area of a trapezoid. Explain the result geometrically.
ANSWER: Substituting a = 0 into the formula from b) will give b × h ÷ 2, which is the formula for the area
of a triangle. Indeed, the smaller the top base of a trapezoid, the closer the trapezoid comes to being a
triangle.
3. Do the triangles T1, T2, T3, and T4 all have the same area? Explain.
ANSWER: Yes, because they all have the same base and the same height.
T1 T2
T3
T4
4. Find the areas of the triangles below, then fill in the table.
a)
b)
c)
d)
Triangle
Number of dots on the
perimeter of the triangle
Half the number of dots on the
perimeter of the triangle
Number of dots
inside the triangle
Area
a)
8
4
3
6
What do you notice? Hint: How can you get the last column from the two columns previous columns?
Mathematicians have proven that the area of any polygon on dot paper can be found using the number of
dots inside and on the perimeter of the triangle.
Pick’s theorem: Area = # of dots inside the shape + # of dots on the perimeter the triangle ÷ 2 − 1.
G8-2 Squares on a Grid
Workbook pages: 145–147
Goals: Students will rotate line segments on a grid, identify right angles between slant line segments,
and create and identify squares on a grid.
Prior Knowledge Required:
Can identify and perform a 90° rotation clockwise or counter-clockwise
Can identify right angles
Can identify and draw congruent right triangles
Is familiar with angle and side properties of special quadrilaterals
Vocabulary:
clockwise
counter-clockwise
quarter turn
turn of 90°
horizontal, vertical, slant
Materials:
grid paper, protractors or set squares
BLM Clockwise and Counter-Clockwise (p 1)
Curriculum Expectations:
Ontario: 6m53, conceptual review; 8m2, 8m3, 8m5
WNCP: 6SS6, conceptual review; [CN, C]
Review relevant vocabulary (see the list above). A slant line is a line that is neither vertical nor
horizontal.
Rotating horizontal and vertical lines 90°. Using a square grid, students should be able to rotate an
arrow coinciding with the grid lines 90° clockwise or counter-clockwise. See Question 1 on the
worksheets. Struggling students might use BLM Clockwise and CounterClockwise.
Rotating slant lines 90°. Ask students to do some problems like those in
Questions 2 and 3 on the worksheets. Then ask students to measure the angle
between the slant line and its image using a protractor. What do they notice?
(The angle is 90°; by rotating the horizontal and the vertical lines 90°, the slant
line was automatically rotated 90°.)
O
?
Draw a slant line segment between two points on a gird and label one of the endpoints O. ASK: How
could we use right triangles to rotate this line segment 90° clockwise around the point O?
Invite volunteers to draw the right triangles that might help with the rotation (there are two possible 1 × 3
triangles, one above and the other below the slant line):
O
or
O
Invite more volunteers to perform the rotation. Does the answer depend on the triangle used? (no)
Have students practise rotating slant line segments as in Question 4.
Proving mathematically that slant lines are perpendicular. When students
are comfortable rotating slant line segments, present the following picture and
ask students to mark the angle equal to ∠AOB . What is the size of ∠BOB ' ? Of
∠AOA ' ? Ask students to explain how they know the size of ∠AOA ' . PS –
Justifying the solution
ANSWER:
B
O
A
A′
B′
∠AOA′ = ∠AOB′ + ∠B′OA′
∠BOB′ = ∠AOB′ + ∠AOB
Triangles AOB and A′OB′ are 90° rotations of each other, so they are congruent, so
∠AOB = ∠A′OB′. This means ∠AOA′ = ∠BOB′ = 90°.
Rotating slant lines without using triangles. Ask students to rotate more slant line segments by only
imagining the triangles, not drawing them. How can they use the lengths of the short sides of the
triangles to draw the rotated line segment? (The triangle AOB, above, has horizontal side 3 and vertical
side 2. So its image under a 90° rotation will have horizontal side 2 and vertical side 3—the lengths of the
horizontal and the vertical lines are reversed.)
Have students copy the line segment at right onto grid paper and rotate it
90°. Emphasize how much harder this problem is to do by eye than the
simpler 2 × 3 slant line. Knowing the rule of interchanging the horizontal
and vertical distances really helps in this case. PE – Reflecting on what
makes a problem easy or hard
Then present two slant lines on the grid as shown, but do not
include the numbers. ASK: Do these lines look like they make
a 90° angle? How can we check for sure without using a
protractor? If the solution does not arise, explain that the top
line goes right 4 and down 13; if we go down 4 and left 13, we
should get a point on the second line. Invite a volunteer to
check if this is the case. (almost, but not on the line) What
does this mean? (the lines are not perpendicular) Invite
another volunteer to check whether the lines are perpendicular
using a protractor or a square corner.
4
13
5
16
Drawing squares on grid paper. Ask students to start with a slant line segment and to create a square
by rotating the line segments around the endpoints as in Questions 5 and 7 on the worksheets. PA –
Worksheets, Question 5 – [CN], 8m5
Discuss with the students how squares are different from other special quadrilaterals before you assign
Question 9 together with the Bonus.
Extension
These slant line segments have horizontal and vertical lengths reversed. Are they
perpendicular? (no) How can we more precisely describe the rule for drawing
perpendicular slant lines? PE – Communicating
ANSWER: If one perpendicular line is oriented from top right to bottom left, then the
other should be oriented from top left to bottom right, and the horizontal and the vertical
distances should be reversed.
G8-3 Length of Slant Line Segments
Workbook pages: 148–150
Goals: Students will find the area of squares with slant sides on grids and estimate the length of slant
line segments.
Prior Knowledge Required:
Can create squares by rotating slant line segments on grids
Knows what a square root is and is familiar with its notation
Knows the square roots of the first 10 perfect squares
Is familiar with a2 notation
Can estimate the square root of a whole number
Can measure with a ruler
Vocabulary:
right triangle
square root
Materials:
1 cm grid paper, dice of two colours, protractors or set squares
Curriculum Expectations:
Ontario: 8m41; 8m2
WNCP: 7SS2, 8N1; [CN, C, R, V]
Review. Review the area of right triangles. Review drawing squares by rotating slant line segments as
done in the previous lesson. Students should draw squares on 1 cm grid paper and divide them into four
right triangles and a square aligned with the grid, as in Question 2 on the worksheet (but don’t ask
students to find the area yet). Students who struggle with dividing squares might benefit from Activity 1.
When students are comfortable dividing squares with slant sides into triangles and squares with
horizontal and vertical sides, ask them to find the areas of the squares they divided. Students should
label the squares with the area.
Finding side length from area. Draw a square on the board (not on a grid), write the measure of one of
the sides (say, 3 cm) and ask students to find its area. Then draw two more squares and write the area in
each (say, 16 cm2 and 36 cm2). ASK: What is the length of the sides of this square? How do you know?
Which mathematical operation did you perform on 16 or 36 to get the answer? How do you find the side
length of a square from the area?
Draw several squares with slant sides like the ones in Question 4 on the worksheets (their side lengths
are whole numbers, so their areas are perfect squares). Ask students to copy the squares to grid paper
and find their areas. Students should again label the squares with the areas.
Sample slopes: 3 × 4, 8 × 6, 5 × 12, 12 × 9 (Corresponding areas: 25, 100, 169, 225)
Review square roots. Ensure that students know the square roots of the first 10 perfect squares. It
might be a good idea to keep the list of perfect squares and their square roots visible somewhere in the
classroom. Review the notation for a square root and the meaning of the square root of a whole
number—it is a number that, when multiplied by itself, gives the number under the square root sign
( 19 × 19 = 19 ). Remind students of the order of operations when dealing with square roots:
16 + 9 = 4 + 3 = 7 is not the same as 16 + 9 = 25 = 5 . What happens when other operations are
combined with square roots? Have students predict which of the following pairs are equal, then calculate
and compare the answers to check (students might use a calculator).
a)
64 + 36 to
64 + 36
b)
64 − 36 to
d)
64 ÷ 36 to
64 ÷ 36
e)
9 × 4 + 64 to
64 − 36
9 × 16 + 64
c)
64 × 36 to
f)
64 × 4 to
64 × 36
64 × 4
ANSWER: Only the expressions in c) and d) are equal.
Then have students practise finding square roots with questions such as
a)
20 × 3 + 4
b)
20 × 4 + 1
c)
4 ×10 − 15
d)
8×2 + 5
Finding the side length of squares. Ask students to look at the squares you provided (where the areas
are perfect squares) and to find the side lengths of these squares. Then have students look at the
squares they drew and separated into triangles and smaller squares at the beginning of the lesson, and
ask students to find the side lengths of these squares as square roots.
Estimating square roots as sides of squares. Review with students how to estimate square roots: first
say which two numbers the square root is between, then say which of the two numbers the square root is
closer to. Review estimating square roots to one decimal point by using mixed fractions. Then have
students estimate the square roots that they obtained as side lengths of the squares they drew earlier.
They can check their estimates using rulers. For additional practice, students can do Activity 2.
The Investigation on page 150 can be done at the end of this lesson or as a separate lesson leading into
the next lesson. Note that students have to make sure that their triangles and squares fit onto the grids
provided. (Students will discover which triangles do and do not fit by trial and error and can adjust the
sizes of their triangles accordingly.) Students can check larger triangles on grid paper.
Activities
1. Students can draw four identical right triangles on grid paper, cut them out, and put them together to
create a square with a square space in the middle. Students can trace the squares divided into triangles
and a smaller square on grid paper.
2. Students will need a pair of dice of different colours and 1 cm grid paper.
Each student draws a point on the grid and rolls the dice. Students move up or
down by the number rolled on the red die and left or right by the number rolled
on the blue die. (The line segment at left results from a 2 on the red die and a 5
on the blue die.) Students then draw a square by rotating the line segment
around the endpoint and find the area of the square and the length of the line
segment. They can also estimate the square root by measuring to a millimetre.
G8-4 The Pythagorean Theorem
Workbook pages: 151–152
Goals: Students will find missing sides of right triangles using the Pythagorean Theorem.
Prior Knowledge Required:
Is familiar with a2 and
Understands that
a notation
a × a = ( a )2 = a
Can estimate the square root of a whole number
Can measure with a ruler
Is familiar with variables
Understands that doing the same operation to both sides of an equation preserves the equality
Vocabulary:
right triangle
leg
hypotenuse
square root
Pythagorean Theorem
Materials:
a measuring tape
a stick at least 50 cm long
modelling clay to keep the stick in place while performing measurements
string (for Activity)
Curriculum Expectations:
Ontario: 8m50; 8m2
WNCP: 8SS1; [CN, PS, R, T, V]
Measure the stick and tell the students how long it is. Place one end of the stick 5 cm from the wall and
the other end against the wall. Secure the stick by putting some modelling clay in front of the end on the
floor. ASK: How high off the floor is the other end of the stick? Will it be the whole length of the stick or
slightly less? How much less? 5 cm less? Record the estimates on the board. Then invite a volunteer to
measure the height the other end reaches (to the nearest millimeter). Record the measurement in
centimetres.
Explain that you are going to slide the stick so that it is now 10 cm from the wall. Ask students to predict
how the height of the top end will change: will it be higher or lower than before? Will it be 5 cm lower?
10 cm lower? 2 mm lower? Again, record the predictions and invite a volunteer to make the
measurements. Continue sliding the stick 10 cm at a time and recording the guesses and the
measurements in the first three columns of the table below (do not show the next three columns yet).
Continue until the stick is as close to the floor as possible without being flat on the floor.
Distance from the wall (cm), b
Height (cm, to the nearest tenth), h
estimate
b2
h2
b2 + h2
measurement
5
10
20
…
ASK: What shape was the stick creating between the wall and the floor? (a right triangle) Ask students to
sketch the situation. Which sides of the triangle were known? Which side were they trying to predict?
Explain that problems that require finding a side of a right triangle first arose in ancient Egypt, where
people wanted their buildings (such as pyramids) to have certain geometric proportions. For example, if
they were building a pyramid with base x, they wanted to know what height would make each triangular
side equilateral. People noticed a relationship between the lengths of the sides in a right triangle, but this
relationship is not between the sides themselves, it’s between their squares.
Ask students to find the squares of the numbers in the table (assign different numbers to different
students for speed) and fill in the next two columns of the table. Then add the last column and ask
students to fill it in as well. What do they notice? Remind them of the length of the stick and ASK: What is
the square of the length of the stick? Explain that the relationship students have found was well known in
ancient Egypt, and was brought to Greece by the mathematician, philosopher, and traveller Pythagoras.
This relationship now bears his name.
State the Pythagorean Theorem: If a right triangle has sides a, b, and c, with c opposite the right angle,
then a2 + b2 = c2. Introduce the terms legs—sides adjacent to the right angle—and hypotenuse—the
side opposite the right angle. Draw several right triangles in a variety of orientations and ask students to
identify the legs and the hypotenuse. Ask students to state the theorem using the terms legs and
hypotenuse. (If a right triangle has legs a and b and hypotenuse c then a2 + b2 = c2.)
Review finding the length of a square when the area is known. ASK: How is solving the equation
produced by the Pythagorean Theorem related to solving an equation such as x2 = 25? Which operation
are you performing on both sides of the equation? (taking a square root)
Finding the hypotenuse. Draw several right triangles, mark the sides
with different letters, and ask students to write the Pythagorean
relationship for each of the triangles. Then replace the letters that mark
the legs (but not the hypotenuse) with whole numbers and have
students state the equality again, this time with numbers. After that,
have students solve the equation to find the hypotenuse. Use triangles
with different orientations and include examples where the answer will
be a whole number or a square root.
Example: w2 = 22 + 52
w
2
5
Review solving equations of the type 2 + x = 5 by subtracting the same number from both sides.
Finding legs. Remove the measurements and the variables from the sides of some triangles and write
new numbers, this time assigning numbers to one leg and the hypotenuse and a variable to the second
leg. Ask students to state the Pythagorean relationship for these triangles. Then solve the first equation
together as a class and have students solve the rest of the equations independently.
Review estimating square roots using number lines and mixed fractions and ask students to estimate
some of the square roots they obtained.
Finding sides—mixed problems. Present several triangles where either the hypotenuse or one of the
legs is a variable, and repeat the process of finding the sides. Point out that students will have to decide
what to do to solve the equation, depending on which side is unknown.
Finding sides when some sides are square roots. When students are
comfortable with finding sides when the known sides are whole numbers, review
the fact that the square of a square root returns the number itself. Then present
some triangles where the known sides are square roots and let students find the
unknown sides.
Example:
94
p
85
Activity
Give students string divided into 12 equal pieces and tied into a circle, and challenge them to make a
right triangle. (To create this circle, tie 11 knots at equal intervals along a length of string, and then tie the
ends together creating the 12th knot.) What are the sides of the triangle? (ANSWER: 3, 4, and 5. You
might wish to point out that in Ancient Egypt such strings were used to check whether an angle is a right
angle by placing three parts of the string along one arm and four along the other arm of the angle in
question. If the rest of the string is taut, the angle is a right angle.) Repeat with a string divided into 30
pieces. (sides 5, 12, and 13)
Extensions
1.
Using 1 cm grid paper, I can draw a line of length exactly
Since 12 + 22 = 5, x =
5 cm as follows:
5 where x is the diagonal in this picture:
Find a pair of perfect squares that will add to the number under the square root and use 1 cm grid
paper to draw a line of length exactly:
a)
34
b)
74
c)
Hint: Use an organized list of perfect squares.
125
d)
85
SAMPLE SOLUTION for a):
a
2
34 − a2
e)
34
1
4
9
16
33
30
25
…
25
36
113
3
5
Look for the first perfect square in the second row; the numbers in that column
provide the answer.
ANSWERS:
a) 32 + 52 = 34
b) 52 + 72 = 74
c) 102 + 52 = 125
d) 22 + 92 = 85
e) 72 + 82 = 113
2. The sums of the squares of the three side lengths of a right triangle is 800. Find the hypotenuse.
SOLUTION: a2 + b2 = c2 and a2 + b2 + c2 = 800, so 2c2 = 800 and c2 = 400, so c = 20.
G8-5 Proving the Pythagorean Theorem
Workbook pages: 153–156
Goals: Students will prove the Pythagorean Theorem for specific cases and use it to check whether a
triangle is a right triangle.
Prior Knowledge Required:
Knows what a square root is and is familiar with its notation
Is familiar with variables
Is familiar with a2 and
Understands that
a notation
a × a = ( a )2 = a
Can estimate the square root of a whole number
Can measure with a ruler
Understands that doing the same operation to both sides of an equation preserves the equality
Vocabulary:
right triangle
leg
hypotenuse
square root
Pythagorean Theorem
Materials:
BLM Triangles (p 2)
Curriculum Expectations:
Ontario: 8m49; 8m2, 8m3
WNCP: 8SS1; [C, CN, R, V]
Do Investigation 1 together as a class. Activities 1 and 2 can be used to help students who have trouble
visualizing the transformations on the worksheet.
Review writing square roots in order, first using sets of square roots only, then using both square roots
and whole numbers:
a)
5, 3, 2
b)
64, 9, 5
f)
7,2,3
g) 11, 11, 5
c)
100, 49,12
h) 8, 12, 59
d)
12, 16,3
e) 7, 22, 81
i)
7, 10, 3
j)
8, 5, 19
You could do Questions 4 and 5 together as a class, or you could do Activity 3 in small groups and use
Questions 4 and 5 for assessment. PA – Worksheet, Question 5 – [R, C, CN, V], 8m2, 8m3
The longest side in a triangle is always opposite the largest angle. Students can use triangles from
BLM Triangles to compare the angles in triangles and the sides. They can fold the triangles to find the
longest side(s) and the largest angle. ASK: Where is the largest angle in a triangle relative to the longest
side? Where is the longest side relative to the largest angle? PE – Making and investigating conjectures
Exercise: Ask students to circle the side that is opposite the largest angle.
a) 2, 3, 4
ANSWERS: a) 4
b) 2,
b)
13
13 , 3
c)
c)
2, 3 , 1,
d)
2, 3 , 2
3 d) 2
Extra practice for Question 5:
Circle the angle that is opposite the longest side.
a) 60°, 50°, 70°,
b) 90°, 50°, 40°,
c) 160°, 10°, 10°,
d) 20°, 90°, 70°
ANSWERS: a) 70° b) 90° c) 160° d) 90°
ASK: In a right triangle, is the 90° angle always opposite the longest side? How do you know? (Yes, the
right angle is always the largest angle and therefore opposite the longest side; if there were an angle
larger than 90° in addition to the 90° angle, these two angles alone would add to more than 180°, which
is impossible in a triangle.)
Determining whether a triangle is a right triangle. Draw a triangle on the board and invite a volunteer
to write these dimensions on its sides: 24, 25, 7. Tell students that you want to check whether the triangle
is a right triangle. Ask students to find the squares of the sides. Which sides should be added together to
get the third side? (242 + 72) How do you know? (the longest side should be the hypotenuse, the side
opposite the right angle) Is the triangle a right triangle? (yes)
Now ask students to determine which triangles from the exercise above are right triangles. ANSWER: b)
and c)
Extra practice:
Determine which triangles from Question 6 on the worksheet are right triangles. ANSWER: c); d) is
almost a right triangle
Activities
1. Students can check the Pythagorean Theorem by direct comparison: Students draw right triangles on
grid paper, add squares on each side, and cut the squares out. They can check that the area of the
large square is equal to the sum of the areas of the small squares by cutting the small squares into
pieces and laying them on the large square. PE – Making and investigating conjectures
2. Students can compare the areas of the large squares in Figures 2 and 3 of the Investigation directly:
Students can use cut-out squares and sets of eight right triangles to check that the two smaller
squares and four triangles make the same square as the large square and four triangles.
3. Give each student three triangles: one acute isosceles, one right, and one obtuse isosceles. (A good
set would be i) 5 cm, 5 cm, 3 cm; ii) 4 cm, 5 cm, and 6 cm; iii) 6 cm, 6 cm, and 5 cm. Different
students can have different triangles; you can use the triangles on BLM Triangles as samples). Ask
students to measure the sides of their triangles and find the squares of the side lengths. Then ask
students to find the largest angle on each triangle (if there are two angles of the same size larger than
the third angle, they can choose one of the two). Ask students to check what is larger—the sum of the
squares of the sides adjacent to the largest angle or the square of the side opposite to it. Then ask
students to colour the largest angle of the triangle according to the result:
•
green, if the sum is the same as the square of the opposite side;
•
red, if the square of the opposite side is larger than the sum of the squares of the adjacent sides;
•
blue, if the square of the opposite side is smaller than the sum of the squares of the adjacent
sides.
Ask students to share their findings with students who have different triangles (they can work in groups of
4). Is there a correlation between the colours and the size of the angles? (yes: blue are acute, red are
obtuse, green are right) Ask students to make a conjecture about the size of the angle and the
relationship between a2 + b2 and c2. ASK: Do you think the conjecture will be true for triangles that are not
isosceles? Students can test their conjecture in groups—two students draw acute scalene triangles and
two students draw obtuse scalene triangles, and each checks the conjecture for the largest angle in their
triangles. PS – Making and investigating conjectures, Revisiting conjectures that were true in one context.
Extension
1. a) The conjecture in Question 4 on the worksheet can be rewritten as follows:
In a triangle with sides a, b, and c, with c opposite angle C:
When ∠C < 90° , then c < a 2 + b 2
When ∠C = 90° , then c = a 2 + b 2
When ∠C > 90° , then c > a 2 + b 2
The reason for that can be seen from the following sketches:
The arc centred at C has radius 4. CB = 3. So all triangles
based on CB with a vertex on the arc have sides 3 and 4. The
third side grows larger as the angle C grows larger. From the
Pythagorean theorem we know that when ∠C = 90° , then the
third side is
C
32 + 4 2 = 5 .
B
When ∠C < 90° , the side opposite C is less than 5, and when ∠C > 90° , the side is more than 5.
C
C
B
B
b) Determine whether the triangle is a right, acute, or obtuse triangle without drawing it. Use the
conjecture above.
i) 6, 7, 8
ii) 16, 17, 18
iii) 16, 17, 28
ANSWERS: i) acute ii) acute iii) obtuse iv) right v) right
iv) 7, 24, 25
v) 12, 35, 37
2. A different way to verify the Pythagorean Theorem.
b
Draw the triangle and squares as shown at right for different values
of a and b. Cut out the triangle and the squares, and then cut the
larger square as shown. Use the shapes to prove the Pythagorean
Theorem for
a) a = 7, b = 5
b) a = 8, b = 4
a
c) a = 8, b = 3
(a − b) ÷ 2
ANSWER: Rearrange the small square and the four pieces of the
large square as shown to get the square with the side length equal
to the hypotenuse.
EXPLANATION:
The length of AD is b + (a − b) ÷ 2 = (a + b) ÷ 2
A
The length of CE is a − (a − b) ÷ 2 = (a + b) ÷ 2.
b
AD || CE, so ADEC is a parallelogram. This means DE = AC.
The triangles BHD, HCE, EFI and IGD are all right triangles, and
because their corresponding legs are equal, they are all congruent (by
SAS). This means that DH = HE = EI = ID, and they are perpendicular,
because the horizontal and the vertical distances are reversed for each
adjacent pair of sides. So the angles of the shape DHEI are right
angles, and DHEI is a square.
The diagonals of the square DHEI, IH and DE, are equal, perpendicular,
and bisect each other. They cut the square DHEI into four congruent
right isosceles triangles.
This means that the line segments IH and DE cut the square BCFD
into four congruent quadrilaterals. Let’s summarize what we know
about these four quadrilaterals, taking BHJD as an example:
•
The angles ∠B and ∠J are right angles.
•
DJ = JH = AC ÷ 2
•
BD = (a − b) ÷ 2 (the shortest side)
•
BH = a − (a − b) ÷ 2 = (a + b) ÷ 2 (the longest side)
•
Moreover, from the sum of the angles in a quadrilateral, ∠D
+ ∠H = 180°.
Rearrange the four quadrilaterals as shown. Angles B, F, C, and G
are 90°. All sides of BGCF are equal to the difference between the
longest side and the shortest side of the quadrilaterals congruent to
BHJD. This means BGCF is a square.
BG = (a + b) ÷ 2 − (a − b) ÷ 2 = b, so the smaller square fits exactly into
the whole, proving the Pythagorean Theorem again.
a H
B
C
D
J
G
E
(a − b) ÷ 2
F
I
B
H
D
J
F
B
C
G
D
J
H
G8-6 Problem-Solving — Using a Formula
Workbook pages: 157–160
Goals: Students will develop strategies for solving problems using formulas.
Prior Knowledge Required:
Understands square roots
Knows the area formulas for a triangle, parallelogram, and rectangle
Can find the perimeter of polygons
Can solve equations of the form ax ÷ b = c
Can find the sides of a right triangle using the Pythagorean Theorem
Can convert units of length and area
Vocabulary:
right triangle
leg
hypotenuse
square root
Pythagorean Theorem
base
height
Curriculum Expectations:
Ontario: 8m50; 8m3, 8m4, 8m5, 8m6, 8m7
WNCP: 8SS1; [C, CN, R, V]
Discuss with students subjects and situations in which they might need to use a formula. Examples: in
physics, chemistry and other sciences; to find area and volume.
Have students complete the worksheets page by page. You can use the questions below as extra
practice for students who work more quickly than others (to keep the class working through the questions
at approximately the same pace) or for students who are struggling with a particular concept or step.
PA – Worksheets:
Questions 2, 3 – [C, V], 8m6
Question 6 – [R], 8m4
Question 11 – [C, R], 8m3
Question 14 – [C, CN], 8m5, 8m7
Extra practice for page 157:
For each problem, write out the measurements you are given. What do you need to find? Write the
formula you will use. Make a sketch for the problem and mark the information on the sketch. (Note:
Students do not need to solve the problems.)
a) A book cover is 30 cm long and 20 cm wide. What is the area of the cover?
b) A parallelogram-shaped window has base 2 m and height 75 cm. How much glass is needed for the
window?
c) A parallelogram-shaped window has base 1 m and slant side 85 cm. A rubber band is placed around
the perimeter of the window for insulation. What is the length of the band needed for the window?
d) A parallelogram ABCD has base AB = 1 m and height DE = 75 cm. E is the midpoint of AB. What is
the length of AD?
ANSWERS: c) Formula: perimeter = 2(a + b)
d) Formula: c 2 = a 2 + b 2
D
C
85 cm
1m
75 cm
A
E
B
1m
Extra practice for page 158:
For each problem, write the area formula you will use. Make a sketch for the problem and mark the
measurements you are given on the sketch. Underline the value that you are given directly in the problem
in the formula. Find the missing value that you need to substitute in the formula, then use the formula to
find the area.
a) A parallelogram is long and thin. The height to a long side is 5 cm. The long sides of the
parallelogram are three times as long as the height. What is the area of the parallelogram?
b) In a rectangle, one of the sides is 3 m long. The perimeter of the rectangle is 10 m. What is the area
of the rectangle?
BONUS William cuts out a paper right trapezoid that is 10 cm high. When he folds his trapezoid, he sees
that it makes a square and a right isosceles triangle. What is the area of the trapezoid?
ANSWERS:
5 cm
a) Formula: area = base × height; base = 15 cm, area 75 cm2
3m
3 × 5 cm
b) Formula: area = length × width; width = 2 m, area 6 m2
3m
BONUS Area of square = a2, area of triangle = base × height ÷ 2, base of triangle = 10 cm.
Area of square = 100 cm2, area of triangle = 50 cm2, area of trapezoid = 150 cm2.
Extra practice for page 159:
10 cm
Before they do the next problems, remind students that a rhombus is a parallelogram with equal sides.
Draw a rhombus, mark its sides as 50 cm, and say that the area of the rhombus is 1500 cm2. Ask
students to draw the heights to two adjacent sides, then have them find the height of the rhombus to each
side from the area. ASK: Do you think that the heights will be equal in any rhombus? Ask students to
explain their thinking. (A very good explanation would be to say that since area—which doesn’t change—
is base × height and the base is the same length regardless of the chosen side, the height must also be
the same.)
1. A rhombus has all sides 2.4 m long. The height of the rhombus is three times shorter. What is the
area of the rhombus?
ANSWER: 1.92 m2
2. A rhombus has sides that are 15 cm long. Its area is 120 cm2. What is the height of the rhombus?
a) Underline what you need to find. Circle the information you know.
b) There is no formula for the height. Which formula that you know involves height?
c) Here is a sketch for the problem. Add the information you know. Mark the
value you need to find with an x.
d) Which part of the formula is x?
e) Solve the equation for x.
ANSWERS:
b) area of parallelogram = base × height
c)
d) height
x
e) x = 8 cm
15 cm
2
Area = 120 cm
Extra practice for page 160:
5. A park lawn has the shape of an isosceles triangle. The area of the
lawn is 400 m2. The shortest side of the lawn is 16 m long. How far
is the opposite corner of the lawn from the shortest side of the lawn?
ANSWER: 25 m
6. A parallelogram is made from two right triangles joined along the
longer leg. The legs of the triangles are 50 cm and 1.2 m. What is
the perimeter of the parallelogram?
ANSWER: The hypotenuse is 1.3 m. The perimeter of the parallelogram is 3.6 m.
Extension
Estimate the square roots to decide if the triangles are drawn to scale. Then make a better sketch.
a)
85
b)
c)
10
81
ANSWERS:
116
a)
85
81
17
4
b)
116
10
17
c)
4
G8-7 Solving Problems
Workbook pages: 161–162
Goals: Students will solve problems that involve applying the Pythagorean Theorem.
Prior Knowledge Required:
Understands square roots
Knows the area formulas for a triangle, parallelogram, and rectangle
Can find the perimeter of polygons
Can solve equations of the form ax ÷ b = c
Can find the sides of a right triangle using the Pythagorean Theorem
Can convert units of length and area
Can solve a problem by converting a formula to an equation
Vocabulary:
right triangle
leg
hypotenuse
square root
Pythagorean Theorem
base
height
Curriculum Expectations:
Ontario: 8m50; 8m1, 8m5, 8m6, 8m7
WNCP: 8SS1; [C, CN, PS, R, V]
Work through the following problem as a class, then have students practise solving problems
independently using the worksheets and the extra questions below.
Problem: Find the area of the equilateral triangle in the sketch.
10 cm
Prompts:
•
What information are you given? What do you have to find?
•
What formula are you going to use?
•
What information in the formula are you not given in the problem? Show it on the sketch and mark it
with an x.
10 cm
ASK: Are there right triangles in the sketch? (yes) What formula could you use to find the height? (the
formula given by the Pythagorean Theorem) Is the height a leg or a hypotenuse of the right triangle? (a
leg) After they find the height (in terms of a square root), ask students to estimate the height in
centimetres and millimetres, and then have students find the area of the triangle.
ANSWER: All sides of the triangle are equal (10 cm), so AD = 5 cm. By the Pythagorean Theorem, the
height of the triangle is
102 − 52 = 100 − 25 = 75 ≈ 8.7 cm. The area is about 43.5 cm2.
Extra practice:
ANSWER: (3 × 4 ÷ 2) + (5 × 7 ÷ 2) = 6 + 17.5 = 23.5 cm2
7 cm
D
1. What is the area of ABCD? (Careful—ABCD is not a
trapezoid!)
A
4 cm
L
2. A quadrilateral KLMN has KL = KN = 5 cm, LM = 2 cm and
two right angles, LKN and LMN. What is the length of MN?
K
Estimate your answer to one decimal point.
M
C
3 cm
B
N
ANSWER: MN = LN 2 − LM 2 = (25 + 25) − 4 = 46 cm, or about 6.8 cm
window
3. Rick is 1 m 60 cm tall. He needs to wash a 1 m × 1 m square window
that is 5.7 m above the ground. He has a 5 m ladder.
a) If he places the ladder against the wall so that the bottom of the ladder
is 2 m from the wall, can he reach the top of the window?
b) For safety reasons, the distance from the wall to the foot of the ladder
should be at least one quarter of the distance from the top of the ladder to
the ground. How close to the wall should Rick place the ladder so that he
can wash the window and be safe?
ladder
5.7 m
PE – [PS] Note: In b), students will need to decide how to check whether Rick can reach the top of the
window. They might first estimate how high they can reach with their hands. They must also keep in mind
that a person can’t actually stand at the very top of a ladder—the last step is below that! They can
proceed by trial and error to see whether different distances from the wall to the foot of the ladder give
Rick a high enough foothold and a safe ratio.
Sample solution: a) The top of the ladder will be 4.58 cm from the ground. It is over 2.2 m from the top of
the ladder to the top of the window. Rick will have trouble washing it without a mop on a long handle!
b) To reach the very top of the window without additional tools, Rick’s feet need to be about 4.6 m from
the ground. If he places the ladder 1.25 m from the wall, the ladder will reach 4.84 m (1.25 is more than a
quarter of 4.84, so the ladder is safe) so he can comfortably stand below the top of the ladder and wash
the window.
PA – Worksheets, Question 13 – [C, CN, V], 8m7, 8m5
Extensions
1. Find the length of the diagonal of a cube with sides of length 1:
a) Find the length of the diagonal of the top face, say BG.
C
B
E
D
b) What special quadrilateral is ABGH? (Hint: BG is horizontal, and so is AH)
c) What is the length of BH? How do you know?
G
A
H
F
ANSWER: ABGH is a rectangle, and BG = 2 . Then BGH is a right triangle with legs 1 and 2 and
hypotenuse BH. By the Pythagorean Theorem, BH = 3 .
2. A rectangular box has base 50 cm by 80 cm and height 40 cm. Will a metre stick fit flat in the bottom
of the box from one corner to the opposite corner? Will it fit in the box if placed from a bottom corner
to the opposite top corner?
ANSWER: The diagonal of the bottom is 8900 cm long. 8900 <100, so the metre stick will not fit
on the bottom. The diagonal of the box is
will fit into the box diagonally.
10500 > 102 cm. Even if the metre stick is 102 cm long, it
3. In triangle ABC, AX is perpendicular to BC, BX = 32,
AX = 24, and XC = 10. Find the perimeter of triangle ABC.
ANSWER: AC = 26, AB = 40.
So the perimeter is 32 + 10 + 26 + 40 = 108 cm.
A
B
4.
C
X
The distance between the bases on a baseball diamond is about
27.4 m. A first base player throws a ball from first to third base. How
far did she throw the ball? If the ball travels 50 km/h, about how long
does it take for the ball to reach third base?
ANSWER: She threw the ball about 34.9 m.
50 km/hr = 50 000 m/3600 sec ≈13.9 m/sec.
It takes the ball about 34.9 m ÷ 13.9 m/sec = about 2.5 seconds to get
there.
5. a) On grid paper, draw 3 non–congruent triangles with base 4 and height 5.
5
b) Is it possible to draw an isosceles triangle with base 4 and height 5? (yes) If yes,
draw it.
4
c) Can you draw another isosceles triangle with one side of length
4 and height 5 (perpendicular to this side) that is not congruent
to the first one? Explain.
ANSWER: No. In order to have a triangle that is not congruent,
4 must be one of the equal sides. If the height is 5, both sides
adjacent to the vertex from which the height is drawn have to
be longer than 5 because they are hypotenuses of right
triangles with height being a leg. Since 4 < 5 , the height
cannot be 5 and a non-congruent triangle is not possible.
4 or
4
5
5
4
d) Can you draw a right triangle with base 4 and height 5? (yes) Can
you draw another triangle like that, but not congruent to the first
one?
ANSWER: No. There are 2 options: 4 is either a leg or a
hypotenuse. If 4 is a leg (base), then 5 is another leg (height), and
by SAS the triangle is fixed. That was the first triangle we drew.
1st triangle
5
4
If 4 is a hypotenuse, both legs are less than 4 because they are
shorter than the hypotenuse. When the height to the hypotenuse is
drawn, it divides the triangle into two right triangles; the height of the
big triangle is a leg in each of the smaller triangles, and the legs of
the big triangle are hypotenuses of the smaller triangles. So the
height of the big triangle is smaller than both its legs, which in turn
are smaller than the hypotenuse. This means that the height is
smaller than 4. So 4 cannot be the hypotenuse of a right triangle
with height 5, and a non-congruent triangle is not possible.
4