Areas of Parallelograms and Triangles 7-1
Parallelogram A parallelogram is a quadrilateral where the opposite sides are congruent and parallel. A rectangle is a type of parallelogram, but we often see parallelograms that are not rectangles (parallelograms without right angles).
Area of a Parallelogram Any side of a parallelogram can be considered a base. The height of a parallelogram is the perpendicular distance between opposite bases. The area formula is A=bh A=bh A=5(3) A=15m 2
Area of a Triangle A triangle is a three sided polygon. Any side can be the base of the triangle. The height of the triangle is the perpendicular length from a vertex to the opposite base. A triangle (which can be formed by splitting a parallelogram in half) has a similar area formula: A = ½ bh.
Example A= ½ bh A= ½ (30)(10) A= ½ (300) A= 150 km 2
Complex Figures Use the appropriate formula to find the area of each piece. Add the areas together for the total area.
Example 24 cm 10 cm 27 cm Split the shape into a rectangle and triangle. The rectangle is 24cm long and 10 cm wide. The triangle has a base of 3 cm and a height of 10 cm.
Solution Rectangle Triangle A = lw A = ½ bh A = 24(10) A = 240 cm 2 A = ½ (3)(10) A = ½ (30) A = 15 cm 2 Total Figure A = A 1 + A 2 A = 240 + 15 = 255 cm 2
Practice! Pg. 353-355 # 1-14 all # 29, 36-43 all
7-2 The Pythagorean Theorem and Its Converse
Parts of a Right Triangle In a right triangle, the side opposite the right angle is called the hypotenuse. It is the longest side. The other two sides are called the legs.
The Pythagorean Theorem
Pythagorean Triples A Pythagorean triple is a set of nonzero whole numbers that satisfy the Pythagorean Theorem. Some common Pythagorean triples include: 3, 4, 5 5, 12, 13 8, 15, 17 7, 24, 25 If you multiply each number in the triple by the same whole number, the result is another Pythagorean triple!
Finding the Length of the Hypotenuse What is the length of the hypotenuse of ABC? Do the sides form a Pythagorean triple?
The legs of a right triangle have lengths 10 and 24. What is the length of the hypotenuse? Do the sides form a Pythagorean triple?
Finding the Length of a Leg What is the value of x? Express your answer in simplest radical form.
The hypotenuse of a right triangle has length 12. One leg has length 6. What is the length of the other leg? Express your answer in simplest radical form.
Triangle Classifications Converse of the Pythagorean Theorem: If the square of the length of the longest side of a triangle is equal to the sum of the squares of the lengths of the other two sides, then the triangle is a right triangle. If c 2 = a 2 + b 2, than ABC is a right triangle. Theorem 8-3: If the square of the length of the longest side of a triangle is great than the sum of the squares of the lengths of the other two sides, then the triangle is obtuse. If c 2 > a 2 + b 2, than ABC is obtuse. Theorem 8-4: If the square of the length of the longest side of a triangle is less than the sum of the squares of the lengths of the other two sides, then the triangle is acute. If c 2 < a 2 + b 2, than ABC is acute.
Classifying a Triangle Classify the following triangles as acute, obtuse, or right. 85, 84, 13 6, 11, 14 7, 8, 9
Practice!! Pg. 360-363 #1-35 odd #36-38 all, 45 # 66 5 extra credit points!
Two Special Right Triangles 45-45 - 90 30-60 - 90 7-3
45-45 - 90 1 The 45-45-90 1 1 triangle is based on the square 1 with sides of 1 unit.
45-45 - 90 1 1 45 45 1 45 45 1 If we draw the diagonals we form two 45-45-90 triangles.
45-45 - 90 1 1 45 45 45 1 Using the Pythagorean Theorem we can 1 45 find the length of the diagonal.
45-45 - 90 1 45 1 2 + 1 2 = c 2 1 45 2 45 1 1 + 1 = c 2 1 45 2 = c 2 2 = c
45-45 - 90 Conclusion: the ratio of the 1 45 2 sides in a 45-1 45 45-90 triangle is 1-1- 2
45-45 - 90 Practice 4 45 4 2 45 SAME 4 leg* 2
45-45 - 90 Practice 9 45 9 2 45 SAME 9 leg* 2
45-45 - 90 Practice 2 45 2 2 45 SAME 2 leg* 2
45-45 - 90 Practice 7 45 14 45 SAME 7 leg* 2
45-45 - 90 Practice
45-45 - 90 Practice 45 3 2 45 hypotenuse 2
45-45 - 90 Practice 3 2 2 = 3
45-45 - 90 Practice 3 45 3 2 45 SAME 3 hypotenuse 2
45-45 - 90 Practice 45 6 2 45 hypotenuse 2
45-45 - 90 Practice 6 2 2 = 6
45-45 - 90 Practice 6 45 6 2 45 SAME 6 hypotenuse 2
45-45 - 90 Practice 45 11 2 45 hypotenuse 2
45-45 - 90 Practice 11 2 2 = 11
45-45 - 90 Practice 11 45 11 2 45 SAME 11 hypotenuse 2
45-45 - 90 Practice 45 8 45 hypotenuse 2
45-45 - 90 Practice 8 2 2 * = 8 2 2 2 = 4 2
45-45 - 90 Practice 4 2 45 8 45 SAME 4 2 hypotenuse 2
45-45 - 90 Practice 45 4 45 hypotenuse 2
45-45 - 90 Practice 4 2 2 * = 4 2 2 2 = 2 2
45-45 - 90 Practice 2 2 45 4 45 SAME 2 2 hypotenuse 2
45-45 - 90 Practice 45 6 45 Hypotenuse 2
45-45 - 90 Practice 6 2 2 * = 6 2 2 2 = 3 2
45-45 - 90 Practice 3 2 45 6 45 SAME 3 2 hypotenuse 2
30-60 - 90 The 30-60-90 2 2 60 60 2 triangle is based on an equilateral triangle with sides of 2 units.
30-60 - 90 The altitude (also the angle bisector 2 30 30 60 60 2 and median) cuts the triangle into two 1 2 1 congruent triangles.
Long Leg 30-60 - 90 30 60 Short Leg This creates the 30-60-90 triangle with a hypotenuse a short leg and a long leg.
30-60 - 90 Practice 30 We saw that the hypotenuse is twice the short leg. 1 2 60 We can use the Pythagorean Theorem to find the long leg.
30-60 - 90 Practice A 2 + B 2 = C 2 30 A 2 + 1 2 = 2 2 3 2 A 2 + 1 = 4 1 60 A 2 = 3 A = 3
30-60 - 90 3 30 2 Conclusion: the ratio of the sides in a 30-60-90 triangle 1 60 is 1-2 - 3
30-60 - 90 Practice 4 3 30 8 The key is to find the length of the short side. 4 Long Leg = short leg * 3 60 Hypotenuse = short leg * 2
30-60 - 90 Practice 30 5 3 10 Hypotenuse = short leg * 2 60 Long Leg = short leg * 3 5
30-60 - 90 Practice 30 7 3 14 Hypotenuse = short leg * 2 60 Long Leg = short leg * 3 7
30-60 - 90 Practice 30 3 2 3 Hypotenuse = short leg * 2 60 Long Leg = short leg * 3 3
30-60 - 90 Practice 30 30 Long Leg = short leg * 3 10 2 10 60 Hypotenuse = short leg * 2
30-60 - 90 Practice
30-60 - 90 Practice 30 11 3 22 Short Leg = Hypotenuse 2 Long Leg = short leg * 3 11 60
30-60 - 90 Practice 30 2 3 4 Short Leg = Hypotenuse 2 60 Long Leg = short leg * 3 2
30-60 - 90 Practice 30 9 3 18 Short Leg = Hypotenuse 2 60 Long Leg = short leg * 3 9
30-60 - 90 Practice 30 15 3 30 Short Leg = Hypotenuse 2 Long Leg = short leg * 3 15 60
30-60 - 90 Practice 30 23 3 46 Hypotenuse = Short Leg * 2 60 Short Leg = Long leg 3 23
30-60 - 90 Practice 30 14 3 28 Hypotenuse = Short Leg * 2 60 Short Leg = Long leg 3 14
30-60 - 90 Practice 30 16 3 32 Hypotenuse = Short Leg * 2 60 Short Leg = Long leg 3 16
30-60 - 90 Practice 9 30 6 3 Hypotenuse = Short Leg * 2 Short Leg = Long leg 3 3 3 60
30-60 - 90 Practice 12 30 8 3 Hypotenuse = Short Leg * 2 Short Leg = Long leg 3 4 3 60
30-60 - 90 Practice 30 27 18 3 Hypotenuse = Short Leg * 2 60 Short Leg = Long leg 3 9 3
30-60 - 90 Practice 30 21 14 3 Hypotenuse = Short Leg * 2 60 Short Leg = Long leg 3 7 3
30-60 - 90 Practice 30 33 22 3 Hypotenuse = Short Leg * 2 60 Short Leg = Long leg 3 11 3
Practice!! Pg. 369-370 # 1-30 all #32
Areas of Trapezoids, Rhombuses, and Kites 7-4
Trapezoids: b 1 = base 1 leg h = height leg Height distance between the 2 bases. * Must be b 2 = base 2 Area of trapezoid base base A = ½ h(b 1 + b 2 ) Height
Find the area of the following trapezoid. 30in 20in 18in This is the height!! 36in A = ½ h(b 1 + b 2 ) = ½ (18in)(36in + 20in) = ½ (18in)(56in) = 504in 2
h = 3.5cm Find the area of following trapezoid. This is a 30-60-90 Δ 5cm A = ½ h(b 1 + b 2 ) 60 h 7cm = ½ (3.5cm)(5cm + 7cm) = ½ (3.5cm)(12cm) = 20.8cm 2 Need to find h first! Short side = 2cm h = 2 3
Area of a Rhombus or a Kite Rhombus 4 equal sides. Diagonals bisect each other. Diagonals are. Kite Adjacent sides are. No sides //. Diagonals are. Area of Kites or Rhombi A = ½ d 1 d 2 Diagonal One Diagonal Two
Find the Area of the following Kite. 4m A = ½ d 1 d 2 3m 3m = ½ (6m)(9m) 5m =27m 2
Find the area of the following Rhombus 15m 12m b d 1 = 24m d 2 = 18m 12m a 2 + b 2 = c 2 12 2 + b 2 = 15 2 144 + b 2 = 225 b 2 = 81 b = 9 15m A = ½ d 1 d 2 = ½ (24m)(18m) = 216m 2
What have I learned?? Area of Trapezoid A = ½ h(b 1 + b 2 ) Area of Rhombus or Kite A = ½ d 1 d 2
Practice!!! Pg. 376-377 #1-35 all
7-5 Area of Regular Polygon Apothem
Find the Area of an Equilateral Triangle Area is ½ ab a is Altitude b is Base A 1 4 3 8 8 a 4 3 8 A 2 4 4 3 b 8 A 16 3
Find the Area of an Equilateral Triangle (there is an easier way) Theorem Area is 1 4 3 s 2 s is Side of triangle 8 a 4 3 b 8 8 A A A 1 2 4 1 4 3 64 3 16 3 8
Find s of an equilateral triangle with area of 25 3 Area is 1 4 3 s 2 s is Side of triangle
Find s of an equilateral triangle with area of 25 3 Area is 1 4 3 s 2 s is Side of triangle 25 3 1 25 4 100 s s 2 1 4 2 3 s 2 s 100 s 10
Finding the Area of a Regular Hexagon inscribed in a circle. Parts of the inscribed hexagon Center Central Angle 360 n Apothem from center to side
Central Finding the Area of a Regular Hexagon inscribed in a circle. Parts of the inscribed hexagon 360 Angle 6 60 30 60 60 60 Apothem from center to side
Central Finding the Area of a Regular Hexagon inscribed in a circle. Parts of the inscribed hexagon 360 Angle 6 60 1 Area ap 2 a is apothem p is Perimeter 30 60 60 60 Apothem from center to side
Finding the Area of a Regular Hexagon inscribed in a circle. 10 10 1 Area ap 2 a is apothem p is Perimeter 5 3 10
Finding the Area of a Regular Hexagon inscribed in a circle. apothem Perimeter A 1 2 5 5 A 30 5 A 150 3 6 10 3 60 3 3 60 10 5 3 10 10 1 Area ap 2 a is apothem p is Perimeter
Finding the Area of a Regular Octagon inscribed in a circle. Sides of 4, what the Central Angle 4 Central 4 Angle 360 8 45 4 4 4 4 1 Area ap 2 a is apothem 4 4 p is Perimeter
How do you find the Apothem Sides of 4 a 22.5 22.5 a 2 Tan 22.5 a 2 a Tan 22.5 2 4.83 67.5 67.5 2 2
Finding the Area of a Regular Octagon inscribed in a circle. Sides of 4, what the Central Central Angle 4 4 4 Angle 360 8 45 4 4 apothem 4.83 Perimeter 8 4 32 4 4 2 2 1 Area ap 2 a is apothem p is Perimeter
Finding the Area of a Regular Octagon inscribed in a circle. Sides of 4 A A 1 2 77.28 4.83 32 4 Central Angle 4 4 4 360 8 45 apothem 4.83 Perimeter 8 4 32 4 4 2 2 1 Area ap 2 a is apothem p is Perimeter
Find the Area of a 12-gon Sides of 1.2; Radius of 2.3 Apothem a a 1.2 2 0.6 2.3 a 2 2 a a 2.22 2 0.6 2.3 2 2.3 0.6 2 2.3 0.6 2 2 2 Perimeter 1.2 12 14.4
Find the Area of a 12-gon Sides of 1.2; Radius of 2.3 Apothem a 2.22 p 14.4 Area 1 2 4.83 14.4 34.776
Practice!! Pg. 382-383 # 1-32 all
and Objective: Find the measures of central angles and arcs.
A CIRCLE is the set of all points equidistant from a given point called the center. This is circle P for Pacman. Circle P P
A CENTRAL ANGLE of a circle is an angle with its vertex at the center of the circle. Central angle
An arc is a part of a circle. In this case it is the part Pacman would eat. Arc
One type of arc, a semicircle, is half of a circle. C Semicircle ABC m ABC = 180 P A B
A minor arc is smaller than a semicircle. A major arc is greater than a semicircle.
RS is a minor arc. mrs = m RPS. LMN is a major arc. mlmn = 360 mln R N S P M O L
Identify the following in circle O: 1) the minor arcs A C O E D
Identify the following in circle O: 2) the semicircles A C O E D
Identify the following in circle O: 3) the major arcs containing point A A C O E D
The measure of a central angle is equal to its intercepted arc. R 53 o O 53 o P
Find the measure of each arc. 1. BC = 32 2. BD = 90 3. ABC = 180 4. AB = 148
Here is a circle graph that shows how people really spend their time. Find the measure of each central angle in degrees. 1. Sleep 2. Food 3. Work 4. Must Do 5. Entertainment 6. Other
Practice!!! Pg. 389-392 #1-14 all # 27-41 odd and #59 *59 may be turned in for 3 extra credit points!!
7-7 Areas of Circles and Sectors
Quick Review What is the circumference of a circle? What is the area of a circle? The interior angle sum of a circle is? 2 r r 2 360 o What is the arc length formula? ma B ) 360 2 r ma B ) 360 C
Sector of a Circle Formula is very similar to arc length Notation is slightly different! - The center pt is used when describing a sector. - The is not used for sectors.
How can we find area, based on what we already know?
Area of a sector = ma ) B 360 r2
1. Find the area of the shaded sector 2. Find the arc length of the shaded sector.
Segment any ideas of how to find the shaded area?
Finding the Area of a Segment of a Circle = Area of Sector minus Area of Triangle equals Area of Segment
Find the area of segment RST to the nearest hundredth. Step 1 Find the area of sector RST. = 4π m 2 Use formula for area of sector. Substitute 4 for r and 90 for m.
Continued Find the area of segment RST to the nearest hundredth. Step 2 Find the area of RST. ST = 4 m, and RS = 4m. = 8 m 2 Simplify.
Continued Find the area of segment RST to the nearest hundredth. Step 3 area of segment = area of sector RST area of RST = 4π 8 4.57 m 2
A segment of a circle
Finding the Area of a Segment Find the area of segment LNM to the nearest hundredth. Step 1 Find the area of sector LNM. = 27π cm 2 Use formula for area of sector. Substitute 9 for r and 120 for m.
Continued Find the area of segment LNM to the nearest hundredth. Step 2 Find the area of LNM. Draw altitude NO.
Continued Find the area of segment LNM to the nearest hundredth. Step 3 area of segment = area of sector LNM area of LNM = 49.75 cm 2
Find the area of the shaded portion
Find the area of the red portion Of the Tube Sign.
Find the shaded area with r = 2, 4, & 10
What is the Area of the Black part? 1cm 3cm 5cm
If you have 2 circles A and B that intersect at 2 points and the distance between the centers is 10. What is the area of the intersecting region? A B 10
Practice!! Page 397-398 7-27 odd 31-33 36-38
7-8 Geometric Probability
Finding a Geometric Probability A probability is a number from 0 to 1 that represents the chance an event will occur. Assuming that all outcomes are equally likely, an event with a probability of 0 CANNOT occur. An event with a probability of 1 is just as likely to occur as not.
Finding Geometric probability continued... In an earlier course, you may have evaluated probabilities by counting the number of favorable outcomes and dividing that number by the total number of possible outcomes. In this lesson, you will use a related process in which the division involves geometric measures such as length or area. This process is called GEOMETRIC PROBABILITY.
Geometric Probability probability and length Let AB be a segment that contains the segment CD. If a point K on AB is chosen at random, then the probability that that it is on CD is as follows: P(Point K is on CD) = Length of CD Length of AB A C D B
Geometric Probability probability and AREA Let J be a region that contains region M. If a point K in J is chosen at random, then the probability that it is in region M is as follows: Area of M P(Point K is in region M) Area = of J J M
Ex. 1: Finding a Geometric Probability Find the geometric probability that a point chosen at random on RS is on TU.
Practice!!! Pg 404-405 #1-31 ODD