the steps for finding **Area** if we are given the circumference The missing information for solving for the **area** **of** a **circle** is always radius, because Pi is always 3.14 Where do we get our radius? So, what if we start with the **Area**? If we have the **area**, and we need/find to solve for? We need the Diameter to find circumference, so to find the diameter if we are given **Area**, we have to work backwards from the **area**, and find the specific radius. This is harder than it seems. What might be a problem in solving this?/

Warm-Up Find the **area** and circumference **of** a **circle** with radius r = 4. Geometry 11-1 Lines **of** **Circles** Definitions Example Identify each line or segment that intersects P. chords: secant: tangent: diameter: radii: QR and ST ST PQ, PT, and PS UV ST Definition A common tangent is a line that is tangent to two **circles**. Theorem 1 If a line is tangent to a/

toString() in abstract superclass: public abstract class Shape implements Comparable { private String name; public Shape(String name) { this.name = name; } public abstract double **area**(); public int compareTo(Shape other) {... } public String toString() { return name + " **of** **area** " + **area**(); } } 22 New constructors public **Circle**(double radius) { super("cicle"); this.radius = radius; } public Rectangle(double length, double width) { super("rectangle"); this.length = length; this.width = width; } public Square/

center **circle** bigger? Keep staring at the black dot. After a while the gray haze around it will appear to shrink. Can you find the dog? Stare at the black light bulb for at least 30 seconds. Then immediately stare at a white **area** on the screen or at a sheet **of** /What do you see on the left? What do you see on the right? Stare at the blue **circles** and move your head back and forth from the screen. Do the outer **circles** move? Can you believe that this is a picture and not an animation? To check, just focus on/

= new Rectangle(5, 3); Rectangle r2 = new Rectangle(3, 3); System.out.println(“c and r1 have equal **area**: " + equalArea(c, r1)); System.out.println(“r1 and r2 have equal **area**: " + equalArea(r1, r2)); } // what should the definition **of** equalArea() look like? public static boolean equalArea( **Circle** o1, Rectangle o2){ return o1.getArea() == o2.getArea(); } public static boolean equalArea( Rectangle o1, Rectangle o2){ return/

) »Died from a burst bladder! –Stayed up EVERY night **of** his life »Mapping the sky »Keeping data –Volumes & VOLUMES **of** data Intro Stuff (That means “boring stuff..”) Kepler? Kepler?/ means “boring stuff..”) 3 LAWS? [Block 4] 3 LAWS? [Block 4] –Ellipses –Equal **Areas** –K = T 2 / R 3 Math or Science? Math or Science? Video Clip!!! YIPPEE!!!/Law (Block 8) –Planets travel in ELLIPTICAL orbits, not circular. –Definitions »**Circle** – ‘r’ »Ellipse – ‘a+b’ –Why? **Circle**, duh… R Ellipse, F1F1 F2F2 Earth a b Real Stuff 2 nd /

26.4 % Remarks:- This trial was held in exchange compound because it required permission to open manhole and mainly because **of** coastal **area** and monsoon season all manhole were fill up with water. Formed a 100m lengthy 110-100 mm conduit, do sub-/ - India Sub- ducting via 40-33mm in 110-100mm conduit Date: 4 & 5th June 2010 End User: BSNL Kerala **Circle**，Trivendrum Installer: Milliken Venue: StachueTelephone Exchange, Trivendrum Classification: Overlay Sub- ducting via 40-33mm in 110-100mm conduit 18mm cable /

Getting her Daughter ready for School Getting her Daughter ready for School All **of** the Above All **of** the Above What is a Lava Tube? Image Courtesy Hawaiian Volcano Observatory Ancient/**Circle** Wind Direction A A 1. **Circle** 2. Fly to Predicted Radar **Area** Wind Direction A A 1. **Circle** 2. Fly to Predicted Radar **Area** 3. Fly Predicted Radar **Area** Wind Direction A A 1. **Circle** 2. Fly to Predicted Radar **Area** 3. Fly Predicted Radar **Area** 4. Fly to Next Radar **Area** Wind Direction A A 1. **Circle** 2. Fly to Predicted Radar **Area**/

Class **Circle** public class **Circle** { public double x, y; // centre **of** the **circle** public double r; // radius **of** **circle** //Methods to return circumference and **area** public double circumference() { return 2*3.14*r; } public double **area**() { return 3.14 * r * r; } Method Body 9 Data Abstraction Declare the **Circle** class, have created a new data type – Data Abstraction Can define variables (objects) **of** that type: **Circle** aCircle; **Circle** bCircle; 10 Class **of** **Circle** cont/

, we diagram the first premise Next, we diagram the first premise. The premise states that all snakes are reptiles. We represent this information by shading the **area** **of** the Snakes **circle** that does not overlap with the Reptiles **circle**. Next, we diagram the second premise Next, we diagram the second premise. The second premise states that all reptiles are cold-blooded animals. We represent/

bits. Click the mouse Click to see Click the mouse 3 The **area** **of** a **circle** Tandi Clausen-May Click the mouse 4 We saw in **Circle** Formulae 1 that… Circumference = × Diameter Now, what about the **area**? 5 Imagine a **circle** made out **of** strands **of** beads. Open it out. Click the mouse Click to see the **circle** open 6 circumference radius (half the diameter) Let’s watch that/

234 567 8 ? ? ? ? ?? ? ? There is a much more accurate way! Mathematical Genius! www.mathsrevision.com **Area** **of** a **circle** A = r² **Area** = x radius There is a special formula for the **area** **of** a **circle**. Remember: r² means r x r Example 1 What is the **area** **of** this **circle**? A = r²A = x 4 x 4 Press Then x 4 x4 = A = 50.3m² (1 d.p.) 4m www/

calls it: Shape *sPtr = new **Circle**(); double a = sPtr->**area**(); So **Circle**::**area** is called, and passed an implicit this pointer: **Circle**::**area**(sPtr); // sPtr is implicitly passed // as “this” vtable ptr ptr to **Circle**::draw sPtr xCoord ptr to Shape::rotate yCoord ptr to **Circle**::**area** color ptr to **Circle**::circ... radius **Circle**::**area** can then do its thing, making use **of** **Circle**-specific data like radius: double **Circle**::**area**() { return PI * radius * radius; } Remember/

to write a sort procedure —many already exist. Avoid duplication **of** effort! **Circle**@x … **area**() … **Circle** Shape …Object b Rect@y … **area**() … Rect Shape …Object Trian@z … **area**() … Trian Shape …Object **area**() 0 1 2 3 4 … … Trian@z … **area**() … Trian Shape …Object **area**() Sort array **of** Shapes Solution: Write a function compareTo that tells whether one shape has bigger **area** than another. Tell sort procedure to use it. 16 Look/

Apothem The shortest distance between center point and chord Example: OA A Segment **Area** which bordered by arc and chord Shaded **area** is minor segment Plain **area** is major segment O Sector **Area** which bordered by two radii and an arc Shaded **area** is minor sector Plain **area** is major sector O Tangents **of** the **circle** Requirements:- Compass Pencils Eraser Scale Set Square Tangent Chord Secent If line/

? h : 0; } 137 138 double Cylinder::getHeight() { return height; } 139 140 double Cylinder::**area**() const 141 { 142 // surface **area** **of** Cylinder 143 return 2 * **Circle**::**area**() + 144 2 * 3.14159 * getRadius() * height; 145 } // end function **area** 146 147 double Cylinder::volume() const 148 { return **Circle**::**area**() * height; } 149 150 void Cylinder::print() const 151 { 152 **Circle**::print(); 153 cout << "; Height = " << height; 154 } // end function print 155 // Fig. 20.1/

// with a member initializer then initializes radius. **Circle**::**Circle**( double r, int a, int b ) : Point( a, b ) // call base-class constructor { setRadius( r ); } // Set radius **of** **Circle** void **Circle**::setRadius( double r ) { radius = ( r >= 0 ? r : 0 ); } // Get radius **of** **Circle** double **Circle**::getRadius() const { return radius; } Esempi Ereditarietà6 // Calculate **area** **of** **Circle** double **Circle**::**area**() const { return 3.14159 * radius * radius; } // Output a **Circle** in the form: // Center = [x, y]; Radius/

own designs. Click and drag an eBlock off **of** the “Available eBlocks” panel to add it to your design. To connect two blocks, click and drag from an output port (colored **circle**) to an input port (gray **circle**). A connection can be destroyed by clicking on/Frank Vahid, UC Riverside22/29 Graphical Simulator Welcome to the eBlocks Simulator! In this **area**, you’ll find helpful hints on creating your own designs. Click and drag an eBlock off **of** the “Available eBlocks” panel to add it to your design. To connect two /

method, this is called “method overriding” –Method Overriding is not Variable Shadowing –We actually will often want to override methods (eg, a subclass computes **area**() differently than the **Circle** **area**() method) –Cannot use casting to access overridden methods Example **of** Method Overriding Consider this example: public class A { public f( ) {….} } public class B extends A{ public f( ) {….} } B b = new B( ); A a = b/

own designs. Click and drag an eBlock off **of** the “Available eBlocks” panel to add it to your design. To connect two blocks, click and drag from an output port (colored **circle**) to an input port (gray **circle**). A connection can be destroyed by clicking on/Frank Vahid, UC Riverside21/24 Graphical Simulator Welcome to the eBlocks Simulator! In this **area**, you’ll find helpful hints on creating your own designs. Click and drag an eBlock off **of** the “Available eBlocks” panel to add it to your design. To connect two /

blurry with color distortion How big should the “Rotating Segmented **Circles**” be? Dyslexic **Areas** “Rotating Segmented **Circles**” Allan’s view with Single Vision Lenses Copyright©2015 – Dyop® Vision Associates – All Rights Reserved Inherent Dyslexic Vision Loss Astigmatic **Area** Dyslexic **Area** Dyslexic **Areas** caused blurry and distorted vision and almost four years **of** functional blindness Astigmatic **Area** Dyslexic **Area** Allan’s Problem Copyright©2015 – Dyop® Vision Associates – All Rights Reserved/

= 14 ft. C = 3.14 14 =43.96 ft Answer: 87.92 ft Your Turn Problem #1 Find the circumference **of** a **circle** if the diameter is 28 ft. (Use 3.14 for ) 3.4 **Area** and Circumference 4 **Area** Example 2. Find the **Area** **of** a **circle** if the radius is 7 ft. (Use 3.14 for ) The formula is A = r 2 or A = r/

Height? Formula chart ½Base Height (½ bh) **Area** **of** a **Circle**?1 **Area** **of** a **Circle**?2 **Area** **of** a **Circle**?3 **Area** **of** a **Circle**?4 **Area** **of** a **Circle**?5 RED: Half the circumference = Length ORANGE: Radius = Width **Area** **of** a **Circle**? RED: Half the circumference = Length ORANGE: Radius = Width Formula chart How Much Land is Wasted? Wasted Land Calculations **Area** **of** FarmSum **of** the **Circle**’s **Areas** Land NOT Used Percent **of** Farm NOT Used Farm **Area** 2.70 km 2 Farm NOT/

out the circumference **of** each **circle**. C = d The Circumference **Area** **of** a **circle** To find the **area** we could try counting the squares inside the **circle**… 1 234 567 8 ? ? ? ? ?? ? ? There is a much more accurate way! Mathematical Genius! **Area** **of** a **circle** A = r² **Area** = x radius There is a special formula for the **area** **of** a **circle**. Remember: r² means r x r Example 1 What is the **area** **of** this **circle**? A = r/

Lesson 6-9 Pages 275-277 Geometry: **Circles** and Circumference (and **Area**) Lesson Check 6-8 What you will learn! How to find the circumference **of** **circles**. CircleCenter Diameter Radius Circumference What you really need to know! A **circle** is the set **of** all points in a plane that are the same distance from a given / = 24.5 cm C = 2 r C = 2 3.14 24.5 C = 153.86 cm 49 cm Example 2: Find the **area** **of** a **circle** with a diameter **of** 49 centimeters. d = 49 cm r = 24.5 cm A = rr A = 3.14 24.5 24.5 A = 1884.785 cm 2 /

Objective: TLW a) Apply the given formula to find the **area** **of** a **circle**, the circumference **of** a **circle**, or the volume **of** a rectangular solid. Performance Indicator 9 Radius & Diameter Radius (r): the line HALFWAY through the **circle** Diameter (d): the line to the END **of** the **circle** radius diameter Radius & Diameter 5 ft. r =_____ d= ____ 3.2 m r =_____ d= ____ 6 ¼ cm r/

circumference, given its radius. The program should repeatedly continue calculating and displaying the **area** and circumference until the user enter 0 for radius. Use pseudocode and flow chart. (**Area**= r 2, Circumference=2 r) Answer Problem Analysis Input: radius Output: **area** and circumference **of** a **circle** Formula: **Area**= r 2 Circumference=2 r Constraint: none Answer Pseudocode Begin Set pi = 3.14 Read radius, r Calculate/

7-6 **Circles** & Arcs 7-7 **Area** **of** **Circles** and Sectors Vocabulary Central Angle – angle whose vertex is the center **of** the **circle**

**Circle** public class **Circle** { public double x, y; // centre **of** the **circle** public double r; // radius **of** **circle** //Methods to return circumference and **area** public double circumference() { return 2*3.14*r; } public double **area**() { return 3.14 * r * r; } Method Body / Method Definition 8 Data Abstraction Declare the **Circle** class, have created a new data type – Data Abstraction Can define variables (objects) **of** that type: **Circle** aCircle; **Circle** bCircle; 9 Class **of** **Circle**/

Class **Circle** public class **Circle** { public double x, y; // centre **of** the **circle** public double r; // radius **of** **circle** //Methods to return circumference and **area** public double circumference() { return 2*3.14*r; } public double **area**() { return 3.14 * r * r; } Method Body 7 Data Abstraction Declare the **Circle** class, have created a new data type – Data Abstraction Can define variables (objects) **of** that type: **Circle** aCircle; **Circle** bCircle; 8 Class **of** **Circle** cont/

Class **Circle** public class **Circle** { public double x, y; // centre **of** the **circle** public double r; // radius **of** **circle** //Methods to return circumference and **area** public double circumference() { return 2*3.14*r; } public double **area**() { return 3.14 * r * r; } Method Body 8 Data Abstraction Declare the **Circle** class, have created a new data type – Data Abstraction Can define variables (objects) **of** that type: **Circle** aCircle; **Circle** bCircle; 9 Class **of** **Circle** cont/

SPI 6.4.4 I CAN identify parts **of** a **circle**. I CAN find the circumference and **area** **of** a **circle**. **Circle** – named from the center point. Radius – one endpoint at the center; the other endpoint on the **circle** Diameter – a segment that passes through the center and has both endpoints on the **circle** Chords – endpoints on the **circle** Central angle – an angle with its vertex at the center. Diameter/

visualize the rectangles. PerimeterLengthWidthArea 20 units7 units21 square units 20 units1 unit 20 units20 square units 20 units6 units 20 units8 units Quick Quiz Here are some rectangles with an **area** **of** 24 square units. **Circle** the rectangles with a perimeter greater than 24 units. Which rectangle has a larger perimeter? How much larger is that perimeter? Rectangle #1 Rectangle #2

Training 2015-16 School Year **CIRCLE** Progress Monitoring Reporting o Completion Report: tracks completion **of** required assessments o Summary Report: allows districts and communities view children’s performance across all subject **areas** o Growth Report: allows districts, communities, and teachers to view children’s gains over time o Group Report: groups children with scores below age-related benchmarks and recommends activities for /

= 36π 3 units 2 Diameter = Radius = What is the **area** **of** the shaded region? 20cm 10cm A(square) = A(**circle**) = A (shaded) 20cm 20 * 20 = 400cm 2 πr 2 = π*(10) 2 = 100π = 314cm 2 = 400 - 314 = 86cm 2 What is the **area** **of** a **circle** that has a circumference **of** 94.2cm C = πd 94.2 = πd 94.2 = /d π d = 30cm r = 15cm A = πr 2 A = π(15) 2 A = 225π A = 706.5cm 2 What is the circumference **of** a **circle** that has an **area** **of** 452.16cm 2 C = πd C = π*24 C = 75.36cm d = 24cm r = 12cm A = πr 2 452.16 = πr 2 r 2 = 144cm /

life. 3 in 2 in 1.5 in What is the scale for this blueprint What is the ACTUAL **area** **of** the bedroom Bedroom Bathroom Hint: What type **of** polygon is the bathroom? **Area** & Circumference Objective: TLW a) Apply the given formula to find the **area** **of** a **circle**, the circumference **of** a **circle**, or the volume **of** a rectangular solid. Radius & Diameter Radius (r): the line HALFWAY through the/

157 cm. What is the Radius? **AREA**: A = Πr 2 32cm **AREA**: A = Πr 2 4.2 ft **AREA**: A = Πr 2 R = ? The **area** **of** the following **circle** is 28.26 ft. What is the radius? **AREA**: A = Πr 2 R = ? The **area** **of** the following **circle** is 490.625 in. What is the radius? **AREA**: A = Πr 2 D = ? The **area** **of** the following **circle** is 78.5 cm. What is the/

**Area** **of** **Circles** Tiana Coley and Brianna Alexander Basic Rule **Area** **of** a **circle** is: pi*r 2, where ‘r’ is the radius And Pi= 3.14 r Problems with **area** In **circle** A, the length **of** line AT is 10cm, and the length **of** AB is 5cm. What is the **area** between the two **circles**? The question is asking us to identify the difference between the two **circles**; therefore, we need to subtract/

value is 3.14 or as an improper fraction is 22/7. Pi is an approximation. Circumference The circumference **of** a **circle** is the distance around a **circle**. Formula: C = πd C = 2πr 4 in. **Area** **of** a **Circle** The **area** **of** a **circle** is the number **of** square units needed to cover the inside **of** a **circle**. (Think **of** a blanket) Formula: A = πr 2 7 cm. Find the circumference and/

) = 1 + tan²(π/8) = Objectives Find volumes **of** non-rotated solids with known cross-sectional **areas** Vocabulary Cross-section – a slice **of** a volume – an **area** obtained by cutting the solid with a plane Typically we see: –Squares: **Area** = s² –Rectangles: **Area** = l w –Semi-**circles**: **Area** = ½ π r² –Triangles: **Area** = ½ b h Volume **of** a Known Cross-Sectional **Area** Volume = ∑ **Area** thickness (∆variable) V = ∫ **Area** dx or V = ∫ **Area** dy Integration endpoints are based on/

.3mi C= 3.14(9.7) C ≈ 30.5in C= 2(3.14)(14) C ≈ 87.9cm C= 2(3.14)(7.6) C ≈ 47.7m r d **Area** **of** a **circle** - what’s inside the **circle** Formula : A = πr 2 10 in A = πr 2 A = π(5) 2 A = π(25) A = 3.14(25) A = 78.5 in 2 d= 40mi r= 14cm/ 530.66 square feet. Find the radius. A = πr 2 530.66 = πr 2 π π 169 = r 2 13 ft = r Find the radius **of** a **circle** that has an **area** **of** 12,070 square feet A = πr 2 12070 = πr 2 π π 3843.95 = r 2 62 ft = r C = 56.52 in C = 2πr 56.52 = 2πr 56.52 = /

on group risk Golden **circle** “We see the Ruhr region as a polycentric urban **area** and therefore you need real city centres” Our key thought Future Fundamental thoughts Current Let’s apply this to Marl Depopulation Placemaking Multimodal axis between core **areas** Green belt How does this/ -Media Marketing -Activities for City centre -Awareness Cost StructureRevenue Streams Business model Short summary **of** BeUrban “We see the Ruhr region as a polycentric urban **area** and therefore you need real city centres”

it belongs to a certain class) 9 type MethodName (parameter-list) { Method-body; } ADDING METHODS TO CLASS **CIRCLE** 10 public class **Circle** { double radius; // radius **of** **circle** double x, y; // center **of** the **circle** // …Constructors go here //Methods to return circumference and **area** public double circumference() { return (2*Math.PI*radius); } public double **area**() { return (Math.PI * Math.pow(radius,2)); } Method Body ALL TOGETHER 11 public class/

Lafferty 17 But the **area** inside this rectangle is also the **area** **of** the **circle** www.mathsrevision.com **Area** **of** a **Circle** **Area** = πr 2 Level 4+ 1-Jul-16Created by Mr. Lafferty Maths Dept. **Area** **of** a **circle** Q.Find the **area** **of** the **circle** ?Solution 4cm www.mathsrevision.com Level 4+ 1-Jul-16Created by Mr. Lafferty Maths Dept. **Area** **of** a **circle** Q. The diameter **of** the **circle** is 60cm. Find **area** **of** the **circle**?Solution www.mathsrevision.com/

a calculator to try to solve the following problems on **area** **of** a **circle**. **Area** **of** a **Circle** Discovery 6 in. Find the **area** **of** the **circle**. **Area** **of** a **Circle** Discovery 11 m. Find the **area** **of** the **circle**. Find the **area** **of** the **circle**. **Area** **of** a **Circle** Discovery What is the radius if the diameter is 10 ft? 10 ft. Find the **area** **of** the **circle**. Take out your study guide! **Area** **of** a **Circle** Discovery Thats all Folks! The End! Take out your/

carry out an investigation and see if you can determine the ‘largest possible’ **area** in each case. Record your results in the table below. Is there another shape which might give a bigger **area** with the 320m **of** fencing available? Design Shape Maximum possible **area** Square Rectangle Pentagon Hexagon Octagon **Circle** Semi **Circle** Parallelogram Other shape ? Solution Square - 6400 Rectangle – 6400 Pentagon – 7040 Hexagon – 7380.8/

What about the **AREA** **of** a **circle**? 2r Now consider a **circle** inside the square The **area** **of** the **circle** must be less than the are **of** the square A < 4r 2 r **Area** = ? xr 2 Finding a formulae for the **area** **of** a **circle** C= πd or C=2πr Semi-**circle**=πr πrπr r **Area** **of** Rectangle= Base x Height **Area** = πr x r **Area** =πr 2 The **Area** and Perimeter **of** a **Circle** A **circle** is defined/

in. Circumference Find the circumference **of** a **circle** with radius 2. Find the circumference **of** a **circle** with diameter 8. **Area** **of** a **Circle** AreaArea **of** a **circle** is the amount **of** space the **circle** covers. **Area** **of** a **Circle** Find the **area** **of** a **circle** with radius 9. 9 feet Find the **area** **of** the **circles** below. 1)2) 6 in. 16 m **Area** **of** a **Circle** Find the **area** **of** a **circle** with radius 5. **Area** **of** a **Circle** Find the **area** **of** a **circle** with diameter 24. Instructional Conversation/

+ 5 + 7 + 11 + 13 + 17 + 19 = 75 **Area** **of** **Circles** Lesson 11-6 Review **of** **Circles** r d The radius **of** a **circle** is half **of** the diameter. Divide the diameter by 2 to get the radius. **Area** **of** a **Circle** **Area** = r² The **area** **of** a **circle** equals the product **of** pi () and the square **of** its radius. r **Area** = r² Find the **Area** **of** the **Circle** Round the answer to the nearest tenth. **Area** = r² A = (4²) A = (16) A/

Ads by Google