Download Practice Book, Practice 7-1 # 1 – 20 (even) - epawelka-math

Elizabeth Pawelka
Special Right Triangles
3/26/12
p.1
Geometry
Lesson Plans
Sections 7-3: Special Right Triangles
3/26/12
Warm-up (15 mins)
 Practice Book, Practice 7-1 # 1 – 20 (even)
 Practice Book, Practice 7-2 # 1 – 19 (odd)
Go over Warm-up (5 mins)
Go over Thm 7-6 and 7-7:
 If c2 > a2 + b2, the triangle is obtuse
 If c2 < a2 + b2, the triangle is acute
Homework Review (10 mins) – ask for any questions on homework.
Homework (H)
 p. 351 – 353: # 1-27 (odd), 36, 44-46
 p. 360 - 362: # 1-23 (odd), 24-26, 40, 41
Homework (R)
 p. 351 – 353: # 1-27 (odd), 36
 p. 360 - 362: # 1-23 (odd), 24-26
Statement of Objectives (5 mins)
The student will be able to use properties of

45∘-45∘-90∘ triangles and

30∘-60∘-90∘ triangles
Teacher Input (50 mins)
Ask the students for the angle measures of an isosceles right triangle (45∘-45∘-90∘)
If each leg has length x and the hypotenuse has length y, use the Pythagorean Theorem to solve for y in
terms of x:
x2 + x2 = y2
2x2 = y2
x 2=y
Elizabeth Pawelka
Special Right Triangles
45∘-45∘-90∘ Triangle Theorem: In a 45∘-45∘-90∘ triangle, hypotenuse =
3/26/12
p.2
2 * leg
Introduce the 30∘-60∘-90∘ Triangle Theorem with the proof using an equilateral triangle
ΔABC is equilateral, so segment AD is the perpendicular bisector and bisects ∠A, creating two 30∘-60∘90∘ triangles. AB is the hypotenuse of triangle ABD.
BD = ½ BC = ½ BA =>
AB = 2BD
hypotenuse = 2s (side BD)
BD2 + AD2 = AB2
s2 + AD2 = (2s)2
s2 + AD2 = 4s2
AD2 = 3s2
AD = s 3
30∘-60∘-90∘ Triangle Theorem: In a 30∘-60∘-90∘ triangle:
 hypotenuse = 2 * shortest leg
 longest leg = 3 * shortest leg
Elizabeth Pawelka
Special Right Triangles
3/26/12
p.3
Assume c = hypotenuse; a = shortest leg; b = longest leg
45∘-45∘-90∘ Triangle/Isosceles Right Triangle
30∘-60∘-90∘ Triangle
c=a 2
c = 2a
b=a 3
Elizabeth Pawelka
Special Right Triangles
3/26/12
p.4
Example 1: Find missing sides – leave answers in simplest radical form
Example 2: Find the area of this triangle – leave answer in simplest radical form.
Point out this is an isosceles triangle so the perpendicular segment is a perpendicular bisector, so ½ the
base = 10 in. Find the height using Pythagorean Thm: h2 + 102 = 112 . So, h = 21 . So, area = ½
20* 21 = 10 21
Now use the special right triangle theorems:
Elizabeth Pawelka
Special Right Triangles
3/26/12
p.5
Example 3: Find all sides of this triangle and then find the area – leave answer in simplest radical form.
First find all side lengths: Note that the perpendicular bisector is also the angle bisector in an isosceles
triangle.
Example 4: If a square has one side length of 5, what is the length of its diagonal? (5 2 )
Example 5: Find the angles and area of a rectangle with a diagonal of 6m and one side length of 3m.
Round to nearest tenth or leave in simplest radical form
or √27 = 3√3 so you can say that this is a 30-60-90 triangle.
Area = 3*3√3 = 9√3 = 15.6
Elizabeth Pawelka
Special Right Triangles
3/26/12
p.6
Softball Field Examples
1. On a high school softball field, the distance between bases is 60 ft. To the nearest foot, how far does
a catcher have to throw the ball from home to 2nd base? Round to nearest hundredth.
60 2 = 84.85 ft
2. What is the perimeter of the softball field? (4 x 60 = 240 ft)
3. What is the area of the softball field? (60 x 60 = 3600 ft2)
Closure (5 mins)
Today you learned to use properties of

45∘-45∘-90∘ triangles and
 30∘-60∘-90∘ triangles
Tomorrow you’ll learn to find areas of trapezoids, rhombuses, and kites.
Homework (H)
 p. 369 – 371 # 1-29, 35-39 (odd)
Homework (R)
 p. 369 – 371 # 1-10, 12-29, 33