Transcript practice - Edmonds

TODAY IN GEOMETRY…
 Practice: Solving missing sides using the
Pythagorean Theorem
 Learning Target 1: Use the Converse of the
Pythagorean Theorem determine if a
triangle is a right triangle
 Independent Practice
 Learning Target 2: Use the properties of
Special Right Triangles to find missing
sides.
REVIEW:
USING THE PYTHAGOREAN
THEOREM TO FIND MISSING
SIDES
PRACTICE: Identify the unknown side as a leg or hypotenuse. Then,
find the unknown side length of the right triangle. Write your
answer in the simplest radical form.
The unknown side is attached to the
right angle so it is a leg of the triangle.
26
58
𝑥
Use the Pythagorean theorem to find
the missing leg:
(𝑙𝑒𝑔)2 +(𝑙𝑒𝑔)2 = (ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒)2
(26)2 +𝑥 2 = (58)2
676 + 𝑥 2 = 3364
−676
− 676
𝑥 2 = 2688
𝑥 2 = 2688
𝒙 = 𝟖 𝟒𝟐 ≈ 𝟓𝟏. 𝟖
PRACTICE: Identify the unknown side as a leg or hypotenuse. Then,
find the unknown side length of the right triangle. Write your
answer in the simplest radical form.
𝑥
6
𝑎 = 10
12
The unknown side is not attached to the
right angle so it is a hypotenuse of the
triangle.
(There are two right triangles. You must
find a before you can find x.)
Use the Pythagorean theorem to find a:
(𝑙𝑒𝑔)2 +(𝑙𝑒𝑔)2 = (ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒)2
8
(6)2 +(8)2 = 𝑎2
Use the Pythagorean theorem again to
36 + 64 = 𝑎2
find x:
2
100
=
𝑎
(𝑙𝑒𝑔)2 +(𝑙𝑒𝑔)2 = (ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒)2
2
100
=
𝑎
2
2
2
(10) +(12) = 𝑥
𝟏𝟎 = 𝒂
100 + 144 = 𝑥 2
244 = 𝑥 2
244 = 𝑥 2
2 61 = 𝒙 ≈ 𝟏𝟓. 𝟔
CONVERSE TO THE
PYTHAGOREAN THEOREM:
3 PARTS!!!
CONVERSE TO THE
PYTHAGOREAN
THEOREM:
𝐵
𝑐
𝑎
𝐶
𝐴
𝑏
PART 1:
PART 2:
PART 3:
RIGHT
TRIANGLE if…
ACUTE
TRIANGLE if…
OBTUSE
TRIANGLE if…
2
2
𝑎 +𝑏 =𝑐
2
2
2
𝑎 +𝑏 >𝑐
2
2
2
𝑎 +𝑏 <𝑐
2
PRACTICE: Tell whether a triangle with the given side lengths is a
right triangle.
Assign the given lengths to the sides of
the triangle. The largest side is always
4, 43, 8
assigned the hypotenuse!
Use the Pythagorean theorem to check
if it is a right triangle.
43
4
8
4
8
43
(𝑙𝑒𝑔)2 +(𝑙𝑒𝑔)2 = (ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒)2
(4)2 +(8)2 = 432 ?
16 + 64 = 1849 ?
80 = 1849 ?
𝟖𝟎 ≠ 𝟏𝟖𝟒𝟗
NOT A RIGHT ANGLE!
80 < 1849
TRIANGLE IS OBTUSE!
PRACTICE: Tell whether a triangle with the given side lengths is a
right triangle.
Assign the given lengths to the sides of
the triangle. The largest side is always
10, 11, 14
assigned the hypotenuse!
Use the Pythagorean theorem to check
if it is a right triangle.
14
10
11
11
10
14
(𝑙𝑒𝑔)2 +(𝑙𝑒𝑔)2 = (ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒)2
(10)2 +11)2 = 142 ?
100 + 121 = 196 ?
221 = 196 ?
𝟐𝟐𝟏 ≠ 𝟏𝟗𝟔
NOT A RIGHT ANGLE!
221 > 196
TRIANGLE IS ACUTE!
PRACTICE: What kind of triangle (acute, obtuse or right) can be
formed with segment lengths 4.3, 5.2, 6.1
Assign the given lengths to the sides of
the triangle. The largest side is always
assigned the hypotenuse!
4.3
6.1
Use the Pythagorean theorem to check
if it is a right triangle.
5.2
4.3
5.2
6.1
(𝑙𝑒𝑔)2 +(𝑙𝑒𝑔)2 = (ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒)2
(4.3)2 +(5.2)2 = (6.1)2 ?
18.49 + 27.04 = 37.21 ?
45.53 = 37.21 ?
𝟒𝟓. 𝟓𝟑 ≠ 𝟑𝟕. 𝟐𝟏
NOT A RIGHT ANGLE!
45.53 > 37.21
TRIANGLE IS ACUTE!
PRACTICE: Graph points A, B and C. Connect the points to form
△ 𝐴𝐵𝐶. Decide whether the triangle is acute, right or obtuse.
𝐴 −2, 4 𝐵 6, 0 𝐶(−5, −2)
Use the distance formula to find the segment
lengths: 𝑑 = (𝑥2 − 𝑥1 )2 +(𝑦2 − 𝑦1 )2
𝑨
𝐴𝐵 = (6 − −2 )2 +(0 − 4)2 = 80
𝐵𝐶 = (−5 − 6)2 +(−2 − 0)2 = 125
𝐶𝐴 = (−5 − −2 )2 +( −2 − 4)2 = 45
𝑩
𝑪
Use the Pythagorean theorem to check if it is a
right triangle.
(𝑙𝑒𝑔)2 +(𝑙𝑒𝑔)2 = (ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒)2
( 45)2 +( 80)2 = ( 125)2 ?
45 + 80 = 125 ?
125 = 125 ?
𝟏𝟐𝟓 = 𝟏𝟐𝟓
IT IS A RIGHT TRIANGLE!
HOMEWORK #2:
Pg. 444:1-6, 8, 10, 12
If finished, work on other assignments:
HW #1: Pg. 436: 3-29 odd
SPECIAL RIGHT TRIANGLES:
𝟒𝟓° − 𝟒𝟓° − 𝟗𝟎°
𝟑𝟎° − 𝟔𝟎° − 𝟗𝟎°
𝟒𝟓° − 𝟒𝟓° − 𝟗𝟎° 𝐑𝐈𝐆𝐇𝐓 𝐓𝐑𝐈𝐀𝐍𝐆𝐋𝐄𝐒:
45°
𝟐𝒙
𝒙
𝒙
45°
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆 = 𝒍𝒆𝒈 · 𝟐
Example:
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 𝑥 · 2
= 2𝑥
PRACTICE: Find the value of x. Write your answer in simplest radical
form.
8
8 2=𝑥
45° − 45° − 90° 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒
Substitute known values
8
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆 = 𝒍𝒆𝒈 · 𝟐
𝑥 = 8∙ 2
𝒙=𝟖 𝟐
PRACTICE: Find the value of x. Write your answer in the simplest
radical form.
3 2
𝑥= 6
3 2
45° − 45° − 90° 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒
Substitute known values
Multiply square roots
Simplify
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆 = 𝒍𝒆𝒈 · 𝟐
𝑥 =3 2∙ 2
𝑥=3 4
𝑥 = 3(2)
𝒙=𝟔
PRACTICE: Find the value of x. Write your answer in the simplest
radical form.
9=𝑥
9 2
𝑥= 9
45° − 45° − 90° 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒
Substitute known values
Divide by 2
Simplify
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆 = 𝒍𝒆𝒈 · 𝟐
9 2 = 𝑥∙ 2
2
𝟗=𝒙
2
PRACTICE: Find the value of x. Write your answer in simplest radical
𝑥= 5 2
form.
10
45° − 45° − 90° 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒
Substitute known values
Divide by 2
Rationalize the denominator
𝑥= 5 2
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆 = 𝒍𝒆𝒈 · 𝟐
10 = 𝑥 ∙ 2
2
2 10
∙ 2=𝑥
2
10 2
=𝑥
4
10 2
=𝑥
2
2
=𝟓 𝟐
𝟑𝟎° − 𝟔𝟎° − 𝟗𝟎° 𝐑𝐈𝐆𝐇𝐓 𝐓𝐑𝐈𝐀𝐍𝐆𝐋𝐄𝐒:
60°
𝒙
𝟐𝒙
30°
𝟑𝒙
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆 = 𝟐 · 𝒔𝒉𝒐𝒓𝒕𝒆𝒓 𝒍𝒆𝒈
EX: ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 2 ∙ 𝑥 = 2𝑥
𝒍𝒐𝒏𝒈𝒆𝒓 𝒍𝒆𝒈 = 𝒔𝒉𝒐𝒓𝒕𝒆𝒓 𝒍𝒆𝒈 · 𝟑
EX: 𝑙𝑜𝑛𝑔𝑒𝑟 𝑙𝑒𝑔 = 𝑥 ∙ 3 = 3𝑥
PRACTICE: Find the value of x and y. Write your answer in simplest
radical form.
𝑦= 4 3
*We’re given the shorter
leg first, we can start with
any of the two equations
for 30-60-90 triangles!
30°
8 =𝑥
4
60°
𝟑𝟎° − 𝟔𝟎° − 𝟗𝟎° 𝑻𝒓𝒊𝒂𝒏𝒈𝒍𝒆 𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆 = 𝟐 · 𝒔𝒉𝒐𝒓𝒕𝒆𝒓 𝒍𝒆𝒈
𝑥 = 2∙4
Substitute known values
Multiply
𝒙=𝟖
𝟑𝟎° − 𝟔𝟎° − 𝟗𝟎° 𝑻𝒓𝒊𝒂𝒏𝒈𝒍𝒆
Substitute known values
Simplify
𝒍𝒐𝒏𝒈𝒆𝒓 𝒍𝒆𝒈 = 𝒔𝒉𝒐𝒓𝒕𝒆𝒓 𝒍𝒆𝒈 · 𝟑
𝑦 =4 ∙ 3
𝒚=𝟒 𝟑
PRACTICE: Find the value of x and y. Write your answer in simplest
radical form.
*We’re given the hypotenuse first, we can
start with the first equation for 30-60-90
triangles!
20
60°
30°
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆 = 𝟐 · 𝒔𝒉𝒐𝒓𝒕𝒆𝒓 𝒍𝒆𝒈
20 = 2 ∙ 𝑥
Substitute values
Divide
2
2
Simplify
𝟏𝟎 = 𝒙
𝑦
= 10 3
𝑥 = 10
𝒍𝒐𝒏𝒈𝒆𝒓 𝒍𝒆𝒈 = 𝒔𝒉𝒐𝒓𝒕𝒆𝒓 𝒍𝒆𝒈 · 𝟑
Substitute values
𝑦 = 10 ∙ 3
𝒚 = 𝟏𝟎 𝟑
PRACTICE: Find the value of x and y. Write your answer in simplest
radical form.
*We’re given the shorter
leg first, we can start with
any of the two equations
for 30-60-90 triangles!
3
60°
𝑥=2 3
30°
𝑦= 3
𝟑𝟎° − 𝟔𝟎° − 𝟗𝟎° 𝑻𝒓𝒊𝒂𝒏𝒈𝒍𝒆 𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆 = 𝟐 · 𝒔𝒉𝒐𝒓𝒕𝒆𝒓 𝒍𝒆𝒈
Substitute known values
𝑥 =2∙ 3
Multiply
𝒙=𝟐 𝟑
𝟑𝟎° − 𝟔𝟎° − 𝟗𝟎° 𝑻𝒓𝒊𝒂𝒏𝒈𝒍𝒆
Substitute known values
Multiply
Simplify
𝒍𝒐𝒏𝒈𝒆𝒓 𝒍𝒆𝒈 = 𝒔𝒉𝒐𝒓𝒕𝒆𝒓 𝒍𝒆𝒈 · 𝟑
𝑦 = 3∙ 3
𝑦= 9
𝒚=𝟑
HOMEWORK #3:
Pg. 461: 3-18
If finished, work on other assignments:
HW #1: Pg. 436: 3-29 odd
HW #2: Pg. 444; 1-6, 8, 10, 12