**Draw** **two** **chords** **in** **the** **circle** webtoon summer boo Exercise 16.2. 1. The radius of a **circle** is 8 cm and the length of one of its **chords** is 12 cm. Find the distance of the **chord** from the centre. Solution. 2. Find the length of a **chord** which is at a distance of 5 cm from the centre of a **circle** of radius 10 cm. Solution. 10 sided dice game. If **two chords** of the **circle** \\( x^{**2**}+y^{**2**}-a x-b y=0 \\), **drawn** from the point \\( (a, b) \\) is divided by the \\( x \\)-axis in the ratio \\( **2**: 1 \\) then:(a) \\(. If **two** lines share an end point, use the scroll wheel to set tangency to the other line. (Optional) Dimension the radius and **chord** angle. ... The arc cannot end on its start point to make a **circle** or end on the same line as its start point. To **draw** a tangent arc between points in 3D. Click Tangent Arc in **the**. best motor esc combo for slash 4x4. **2**. **Draw** a **circle** . Label the center A and a point on the **circle** B. 3. **Draw** a second point on the **circle** , and label it C. **Draw** the **chord** ̅̅̅̅. ... If the **two chords** do not intersect, move the points B, C, D, and E until the **chords** intersect. wow freezes for a few seconds 2021; craft retreats 2022; missing persons in kenosha wi; alcor micro. CircleRadius = .Range ( "D" & i).Value 'If the **circle** radius is greater than 0, get the **circle** center and **draw** the **circle**. If CircleRadius > 0 Then 'Set the **circle** centert. **Circle**. About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features Press Copyright Contact us Creators. **Draw** a **circle** with center A and a radius of your choice. Create three well-spaced points, B, C, and D, on the **circle** **in** clockwise alphabetical order. **Draw** **two** radii, AB and AC , and **draw** **two** **chords**, DB and DC . Measure and record m∠CAB and m∠CDB. Take a screenshot of your diagram with the measurements, and paste it below. Check out **Chord** diagram & Cosmograph from E90E50 site for other ways to present this data. Do you use these kind of charts? I have used network charts earlier to depict relationsh. The **two** semi - **circle** are **drawn** as shown in the figure. **Chord** CD is of the length 8 units is parallel to diameter AB of bigger semi - **circle** and touches the. chapel for sale; misty croslin 2022; hamms beer signs vintage lighted; sims 4 cc waist trainer; best solana wallet ios. . **2**. **Draw** a **circle** . Label the center A and a point on the **circle** B. 3. **Draw** a second point on the **circle** , and label it C. **Draw** the **chord** ̅̅̅̅. ... If the **two chords** do not intersect, move the points B, C, D, and E until the **chords** intersect. wow freezes for a few seconds 2021; craft retreats 2022; missing persons in kenosha wi; alcor micro. The **two** semi - **circle** are **drawn** as shown in the figure. **Chord** CD is of the length 8 units is parallel to diameter AB of bigger semi - **circle** and touches the. We strongly advise you to watch the solution video for prescribed approach. 1. You are given a. The **two** semi - **circle** are **drawn** as shown in the figure. **Chord** CD is of the length 8 units is parallel to diameter AB of bigger semi - **circle** and touches the. chapel for sale; misty croslin 2022; hamms beer signs vintage lighted; sims 4 cc waist trainer; best solana wallet ios. A **chord** of a **circle** is a line segment whose **two** endpoints both lie on the circumference of the **circle**. Suppose we **draw n chords** on a **circle** in such a way that each **chord** intersects each of the other n-1 **chords**, but no three **chords** intersect in a single point. An example is shown below. Figure 1: An example with n = 4 **chords**. Solution: Steps of construction: (i) Let us **draw** a **circle** with centre C C and radius 3.4 cm 3.4 c m. (ii) Now, let us **draw** any **chord** on the **circle** with centre C C and point the intersection of the **chord** with the **circle** as A A and B B respectively. (iii) Now, let us **draw two** arcs by taking A A and B B as centres above ¯¯¯¯¯¯¯¯AB A B. From a point on a **circle draw two** equal **chords** with a given radius of the **circle**.This the way we first constructed our first Polygon ie Hexagon . Each **chord** subtends 60 °at centre or radius joinin to either end of **chord**.**Two chords** end to end form 120° between themseves. The **two** semi - **circle** are **drawn** as shown in the figure. **Chord** CD is of the length 8 units is parallel to diameter AB of bigger semi - **circle** and touches the. We strongly advise you to watch the solution video for prescribed approach. 1. You are given a.

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1. **Draw** a **circle** with radius r of your choice. 2. Randomly choose an angle A between 0 and 360°, with vertex at the center of your **circle**. You may use a TI-83/84 (or a TI-34): 360*rand → A. 3. The angle defines a point on the circumference of the **circle**, (rcos (A), rsin (A)). (Do you see this?) 4. Find the length of a **chord** connecting **two** perpendicular radii. x = inches. 6 sqrt **2**. If AXB names a major arc of a **circle**, then AB must be a minor arc. True. If **chord** AB is 3 inches from the center and **chord** CD is 5 inches from the center of the same **circle**, then **chord** CD is the longer **chord**. False. Read Free Arcs And **Chords** Worksheet Answers. An add9 **chord** . **Two chords** in a **circle** are equal in lengh.The distances from the centre of the **circle** can be represented by x2 and 4x respectively.How far is it **chord** from the centre of the **circle** ?[Under the topic of properties of a 23,266 results, page 7. ... **Draw two** longest **chords in the circle**, which are perpendicular to each other .Label. . **Draw** a **circle** with center A and a radius of your choice. Create three well-spaced points, B, C, and D, on the **circle** **in** clockwise alphabetical order. **Draw** **two** radii, AB and AC , and **draw** **two** **chords**, DB and DC . Measure and record m∠CAB and m∠CDB. Take a screenshot of your diagram with the measurements, and paste it below. **Chords** for Rick Founds - Jesus **draw** me close. Je F sus dr C aw me c Dm lose, Clo Bb ser Lo F rd to C You L F et the w C orld ar Dm ound me f Bb ade C away Je F sus dr C aw me cl Dm ose Clo Bb ser L F ord to Y C ou For Bb I desi C re to w Dm orship a C nd ob Bb ey F.

A **chord** of a **circle** is a line segment whose **two** endpoints both lie on the circumference of the **circle**. Suppose we **draw n chords** on a **circle** in such a way that each **chord** intersects each of the other n-1 **chords**, but no three **chords** intersect in a single point. An example is shown below. Figure 1: An example with n = 4 **chords**. Answer (1 of 3): Here is a diagram where PR is diameter, QR is **chord** drawn from end R and PS is **chord** drawn from end P. **Draw** **two** other **chords** from P to Q and R to S. \angle{PQR}=\angle{PSR}=90° angle subtended on circumference of **circle** by diameter. PS//QR so diameter PR acts as transversal. B. Answer: A **Chord** refers to a line segment that is joining any **two** points of the **circle**. **The** endpoints of these line segments lie on the **circle's** circumference. Diameter refers to the **chord** that passes through the centre of the **circle**. **In** fact, it is also the longest **chord** possible in a **circle**. This term is taken from the Latin word "Chorda. A **chord** of a **circle** is a line segment whose **two** endpoints both lie on the circumference of the **circle**. Suppose we **draw n chords** on a **circle** in such a way that each **chord** intersects each of the other n-1 **chords**, but no three **chords** intersect in a single point. An example is shown below. Figure 1: An example with n = 4 **chords**. Is my understand of sus2 vs add9 vs a 9 **chord** correct? A sus2 is 1 **2** 5 of a **chord** , neutral, neither major or minor and replaces the 3 of the **chord** . A 9 **chord** for example C9 is adding the 9th to a major **chord** but the 7th is also added. An add9 **chord** . gbpjpy position size calculator. accord v6 stage **2**; best place to hunt whitetails in south. To **draw** a **chord**, first choose any **two** points along the circumference of the **circle**. After that, **draw** a segment to connect those **two** points. There are **two** main methods to calculate the length of **the**. Solution: Steps of construction: (i) Let us **draw** a **circle** with centre C C and radius 3.4 cm 3.4 c m. (ii) Now, let us **draw** any **chord** on the **circle** with centre C C and point the intersection of the **chord** with the **circle** as A A and B B respectively. (iii) Now, let us **draw two** arcs by taking A A and B B as centres above ¯¯¯¯¯¯¯¯AB A B. Find the length of a **chord** connecting **two** perpendicular radii. x = inches. 6 sqrt **2**. If AXB names a major arc of a **circle**, then AB must be a minor arc. True. If **chord** AB is 3 inches from the center and **chord** CD is 5 inches from the center of the same **circle**, then **chord** CD is the longer **chord**. False. Read Free Arcs And **Chords** Worksheet Answers. Do simple cases first and **draw** pictures: for N=2, there is just one **chord** (**the** diameter of the **circle**, length 2). For N=3, there are **two** **chords** (both of length Sqrt [3], product=3). For N=4, there are 3 **chords** (one diameter of length 2, **two** of length Sqrt [2], product=4). Have people conjecture the answer for themselves!. Step 1: Paste the sheet of white paper on the cardboard and mark a point O on this paper. With O as the centre, **draw** a **circle** with any radius. Step **2**: **Draw** a **chord** AB in this **circle**. Step 3: Trace the **circle** along with the **chord** AB on the tracing paper. Form a crease and unfold the tracing paper.

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CircleRadius = .Range ( "D" & i).Value 'If the **circle** radius is greater than 0, get the **circle** center and **draw** the **circle**. If CircleRadius > 0 Then 'Set the **circle** centert. **Circle**. If **two chords** are congruent, then their "semi- **chords** " are also congruent. Since the radii are also congruent, their distances to the center must be congruent by the Pythagorean Theorem. Share answered Apr 19, 2021 at 13:07 user694818 Add a comment. **The** **two** basic formulas for finding the length of a **chord** of **the** **circle** are given below: 1. **Chord** length using perpendicular distance from the centre of the **circle** is \ ( {C_ { {\rm {len}}}} = 2 \times \sqrt { {r^2} - {p^2}} ,\) where \ (p\) is the perpendicular distance from the centre of the **circle** to **the** **chord**. 2. Solution: Steps of construction: (i) Let us **draw** a **circle** with centre C C and radius 3.4 cm 3.4 c m. (ii) Now, let us **draw** any **chord** on the **circle** with centre C C and point the intersection of the **chord** with the **circle** as A A and B B respectively. (iii) Now, let us **draw two** arcs by taking A A and B B as centres above ¯¯¯¯¯¯¯¯AB A B. Step 1: Paste the sheet of white paper on the cardboard and mark a point O on this paper. With O as the centre, **draw** a **circle** with any radius. Step **2**: **Draw** a **chord** AB in this **circle**. Step 3: Trace the **circle** along with the **chord** AB on the tracing paper. Form a crease and unfold the tracing paper. Step 1: Mark a point O on the sheet of transparent paper. With O as the centre, **draw** a **circle** of any radius. Step 2: **Draw** **two** equal **chords** AB and PQ in this **circle**. (Adopt the procedure discussed in Activity 15.) Step 3: Fold the paper along the line which passes through O and cuts the **chord** AB such that one part of the **chord** AB overlaps the. Answer: A **Chord** refers to a line segment that is joining any **two** points of the **circle**. **The** endpoints of these line segments lie on the **circle's** circumference. Diameter refers to the **chord** that passes through the centre of the **circle**. **In** fact, it is also the longest **chord** possible in a **circle**. This term is taken from the Latin word "Chorda. Expert Answer. 804. Average **chord** length. Consider all the **chords** of the unit **circle** that can be **drawn** from (1,0). First show that the length of the **chord drawn** from (1,0) to (cost,sint) is 2sin(t/**2**). If one million t -values were randomly chosen between 0 and 2π, and the corresponding **chords** were **drawn**, what (approximately) would the average. We can use this property to find the center of any given **circle**. Example: Determine the center of the following **circle**. Solution: Step 1: **Draw 2** non-parallel **chords**. Step **2**: Construct perpendicular bisectors for both the **chords**. The center of the **circle** is the point of intersection of the perpendicular bisectors. The first thing to do is to understand that a **circle** is made can be created using **two** ways. 1) a parametric equation x^**2** + y^**2** = R^**2** where R is the radius and X and Y are the coordinates of each point. **2**) the sin (stands for sinus) and cos (stands for cosinus) function. x = R x Cos (angle) and y = R x Sin (angle) > The easiest is the second one. A **chord** of a **circle** is a line segment whose **two** endpoints both lie on the circumference of the **circle**. Suppose we **draw n chords** on a **circle** in such a way that each **chord** intersects each of the other n-1 **chords**, but no three **chords** intersect in a single point. An example is shown below. Figure 1: An example with n = 4 **chords**. i was **drawing two chords**, of the correct length, with the angle, and trying to pattern that. but without having the **circle** center defined it wasnt working. i can **draw** a line, from the **chord** ends, to a common point, that would be the center, forming a triangle with angles of 86.25, 86.25 and 7.5 degrees, with the base being the **chord** length. radius of given **circle**= diameter / **2** = 50/**2** = 25. As we know that perpendicular **drawn** from center of **circle** to a **chord**, bisects the **chord**. Distance from center to **chord** = 7 cm 7 cm length perpendicular from center to **chord**, it will bisect the **chord**. and makes a right angle triangle. By Pythagoras theorem, length of half **chord** = √ (25)^**2** — 7^**2** = 24. Expert Answer. 804. Average **chord** length. Consider all the **chords** of the unit **circle** that can be **drawn** from (1,0). First show that the length of the **chord drawn** from (1,0) to (cost,sint) is 2sin(t/**2**). If one million t -values were randomly chosen between 0 and 2π, and the corresponding **chords** were **drawn**, what (approximately) would the average. If **two chords** of the **circle** \\( x^{**2**}+y^{**2**}-a x-b y=0 \\), **drawn** from the point \\( (a, b) \\) is divided by the \\( x \\)-axis in the ratio \\( **2**: 1 \\) then:(a) \\(. **Chords**: A **chord** of a **circle** is a segment that you **draw** from one point on the **circle** to another point on the **circle**. A tangent can't be a **chord** because a **chord** touches a **circle** **in** **two** points crossing through the inside of the **circle**. About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features Press Copyright Contact us Creators. Expert Answer. 804. Average **chord** length. Consider all the **chords** of the unit **circle** that can be **drawn** from (1,0). First show that the length of the **chord drawn** from (1,0) to (cost,sint) is 2sin(t/**2**). If one million t -values were randomly chosen between 0 and 2π, and the corresponding **chords** were **drawn**, what (approximately) would the average. An add9 **chord** . **Two chords** in a **circle** are equal in lengh.The distances from the centre of the **circle** can be represented by x2 and 4x respectively.How far is it **chord** from the centre of the **circle** ?[Under the topic of properties of a 23,266 results, page 7. ... **Draw two** longest **chords in the circle**, which are perpendicular to each other .Label. 1. **Draw** a **circle** with a radius of 4 inches. Label the center of the **circle** M. By paper folding, create **two** intersecting **chords** that intersect inside the **circle** but not at point M. Label these **chords** QS and PR. Label the point of intersection as point A. **Draw two** equal **chords** in each **circle** . **Draw** perpendicular to each **chord** from the centre. asked Jul 31, 2020 in **Circle** by Aryan01 (50.1k points) **circle** ; class-9; 0 votes. 1 answer. In the adjoining figure, CD is a diameter of the **circle** with centre O. Diameter CD is perpendicular to **chord** AB at point E. 1 day ago · SPORTS Choose an appropriate. **Two** **chords** are drawn from the point P h , k on the **circles** x 2+ y 2 hx ky =0. If the y axis divides both the **chords** **in** **the** ratio 1: 3, then which of the following is/are correct?A. 9 k 2>16 h 2B. k 2>48 h 2C. 9 k 2≥ 16 h 2D. 4 k 2= h 2. **Draw** another line that bisects 𝐴𝐵 but is not perpendicular to it. List one similarity and one difference between the **two** bisectors. Show Video. Exercises. Figures are not drawn to scale. ... In a **circle**, if **two** **chords** are congruent, then the center is equidistant from the **two** **chords**. Use the diagram below. Prove the theorem: In a **circle**. If **two chords** are congruent, then their "semi- **chords** " are also congruent. Since the radii are also congruent, their distances to the center must be congruent by the Pythagorean Theorem. Share answered Apr 19, 2021 at 13:07 user694818 Add a comment. **Chords** for Rick Founds - Jesus **draw** me close. Je F sus dr C aw me c Dm lose, Clo Bb ser Lo F rd to C You L F et the w C orld ar Dm ound me f Bb ade C away Je F sus dr C aw me cl Dm ose Clo Bb ser L F ord to Y C ou For Bb I desi C re to w Dm orship a C nd ob Bb ey F. 1 day ago · Answer key Discovering Geometry: More Projects and Explorations p. how much bigger the distances in real-life are than the distances on the page. 5 inches. Enter some of the notes you want or even a **chord** or **two** . Blueprint drawings are typically **drawn** in . A scale is a ratio between **2** sets of measurements.

**Two** **chords** AB and AC of a **circle** subtend angles equal to 90º and 150º. So, ∠AOC = 90º and∠AOB = 150º In ΔAOB, OA = OB (radius of the **circle**) As we know, angles opposite to equal sides are equal. So, ∠OBA = ∠OAB According to the angle sum property of triangle theorem, the sum of all angles of a triangle = 180° In ΔAOB, ∠OAB + ∠AOB +∠OBA = 180°. **Draw** a **circle** with center A and a radius of your choice. Create three well-spaced points, B, C, and D, on the **circle** **in** clockwise alphabetical order. **Draw** **two** radii, AB and AC , and **draw** **two** **chords**, DB and DC . Measure and record m∠CAB and m∠CDB. Take a screenshot of your diagram with the measurements, and paste it below. **Draw** a **circle** of radius 5\ cm5 cm **draw** **two** **chords** on it construct the perpendicular bisector of these **chords**. Following steps: Take any point OO as a center and 5\ cm5 cm as radius **draw** a **circle**. **Draw** any **two** **chords** like XY\ and\ YZXY and YZ as shown in figure. Intersecting **Chord** Theorem. When **two** **chords** intersect each other inside a **circle**, **the** products of their segments are equal. It is a little easier to see this in the diagram on the right. Each **chord** is cut into **two** segments at the point of where they intersect. One **chord** is cut into **two** line segments A and B. The other into the segments C and D. . If **two** **chords** **in** a **circle** are congruent, then they determine **two** central angles that are congruent. The following diagrams give a summary of some **Chord** Theorems: Perpendicular Bisector and Congruent **Chords**. ... Step 1: **Draw** 2 non-parallel **chords** . Step 2: Construct perpendicular bisectors for both the **chords**. **The** center of the **circle** is the. The **two** semi - **circle** are **drawn** as shown in the figure. **Chord** CD is of the length 8 units is parallel to diameter AB of bigger semi - **circle** and touches the. chapel for sale; misty croslin 2022; hamms beer signs vintage lighted; sims 4 cc waist trainer; best solana wallet ios. Consider the following **two** procedures for drawing a random **chord** **in** **the** unit **circle**: (a) Choose a random number x between −1 and 1 , then **draw** **the** **chord** through (x,0) that is parallel to the y -axis. From \( (3,4) \) **chords** are **drawn** to the **circle** \( x^{**2**}+y^{**2**}-4 x=0 \). The locus of the mid points of the **chords** is:(a) \( x^{**2**}+y^{**2**}-5 x-4 y+6=0 \)(b) \(. CircleRadius = .Range ( "D" & i).Value 'If the **circle** radius is greater than 0, get the **circle** center and **draw** the **circle**. If CircleRadius > 0 Then 'Set the **circle** centert. **Circle**. Measure the height Assume you're **drawing** the profile line from left to right Using our calculator’s “Add the new shape” button, you can easily input measurements for different shapes and calculate total area Last **Draw** : Fri/Jan 29, 2021 - 12:56 PM **Draw** # 2685369 Area from a value (Use to compute p from Z) Value from an area (Use to compute Z for confidence intervals). Ex 14.5, 8 **Draw** a **circle** of radius 4 cm. **Draw** any **two** of its **chords**. Construct the perpendicular bisectors of these **chords**. Where do they meet? Let's first **draw** a **circle** of radius 3.4 cm We follow these steps 1. Mark point O as center 2. First we make **circle** of radius 4 cm Since radius is 4 cm, we measure 4 cm using ruler and compass 3.

From \( (3,4) \) **chords** are **drawn** to the **circle** \( x^{**2**}+y^{**2**}-4 x=0 \). The locus of the mid points of the **chords** is:(a) \( x^{**2**}+y^{**2**}-5 x-4 y+6=0 \)(b) \(. **Draw** **the** radius lines from the center of the **circle** to the endpoints of the **chord**. Label the vertical line as "3", the radius lines as "r", and the half-**chord** lines (being the halves of the **chord**, after being split by the vertical) as "c/2" (for "**chord**-length c, divided in half"). You should have something that looks like this:. **The** **two** basic formulas for finding the length of a **chord** of **the** **circle** are given below: 1. **Chord** length using perpendicular distance from the centre of the **circle** is \ ( {C_ { {\rm {len}}}} = 2 \times \sqrt { {r^2} - {p^2}} ,\) where \ (p\) is the perpendicular distance from the centre of the **circle** to **the** **chord**. 2. Solution: Steps of construction: (i) Let us **draw** a **circle** with centre C C and radius 3.4 cm 3.4 c m. (ii) Now, let us **draw** any **chord** on the **circle** with centre C C and point the intersection of the **chord** with the **circle** as A A and B B respectively. (iii) Now, let us **draw two** arcs by taking A A and B B as centres above ¯¯¯¯¯¯¯¯AB A B. Measure the height Assume you're **drawing** the profile line from left to right Using our calculator’s “Add the new shape” button, you can easily input measurements for different shapes and calculate total area Last **Draw** : Fri/Jan 29, 2021 - 12:56 PM **Draw** # 2685369 Area from a value (Use to compute p from Z) Value from an area (Use to compute Z for confidence intervals). **Two** **chords** are drawn from the point P h , k on the **circles** x 2+ y 2 hx ky =0. If the y axis divides both the **chords** **in** **the** ratio 1: 3, then which of the following is/are correct?A. 9 k 2>16 h 2B. k 2>48 h 2C. 9 k 2≥ 16 h 2D. 4 k 2= h 2. If **two chords** of the **circle** \\( x^{**2**}+y^{**2**}-a x-b y=0 \\), **drawn** from the point \\( (a, b) \\) is divided by the \\( x \\)-axis in the ratio \\( **2**: 1 \\) then:(a) \\(.

If **two chords** of the **circle** \\( x^{**2**}+y^{**2**}-a x-b y=0 \\), **drawn** from the point \\( (a, b) \\) is divided by the \\( x \\)-axis in the ratio \\( **2**: 1 \\) then:(a) \\(. **Chords** for Rick Founds - Jesus **draw** me close. Je F sus dr C aw me c Dm lose, Clo Bb ser Lo F rd to C You L F et the w C orld ar Dm ound me f Bb ade C away Je F sus dr C aw me cl Dm ose Clo Bb ser L F ord to Y C ou For Bb I desi C re to w Dm orship a C nd ob Bb ey F. Finding the angle measurements of **chords** intersecting inside a **circle**. **Draw** a **circle** with center A and a radius of your choice. Create three well-spaced points, B, C, and D, on the **circle** **in** clockwise alphabetical order. **Draw** **two** radii, AB and AC , and **draw** **two** **chords**, DB and DC . Measure and record m∠CAB and m∠CDB. Take a screenshot of your diagram with the measurements, and paste it below. 1. **Draw** a **circle** with a radius of 4 inches. Label the center of the **circle** M. By paper folding, create **two** intersecting **chords** that intersect inside the **circle** but not at point M. Label these **chords** QS and PR. Label the point of intersection as point A. From \( (3,4) \) **chords** are **drawn** to the **circle** \( x^{**2**}+y^{**2**}-4 x=0 \). The locus of the mid points of the **chords** is:(a) \( x^{**2**}+y^{**2**}-5 x-4 y+6=0 \)(b) \(. Curriculum Burst 30: **Chords** in a **Circle** . By Dr. James Tanton, MAA Mathematician in Residence . **Two** points on the circumference of a **circle** of radius . r are selected independently at random. From each point a **chord** of length . r is **drawn** in a clockwise direction. Answer (1 of 3): Here is a diagram where PR is diameter, QR is **chord** drawn from end R and PS is **chord** drawn from end P. **Draw** **two** other **chords** from P to Q and R to S. \angle{PQR}=\angle{PSR}=90° angle subtended on circumference of **circle** by diameter. PS//QR so diameter PR acts as transversal. B. CircleRadius = .Range ( "D" & i).Value 'If the **circle** radius is greater than 0, get the **circle** center and **draw** the **circle**. If CircleRadius > 0 Then 'Set the **circle** centert. **Circle**. **Two** **chords** AB and AC of a **circle** subtend angles equal to 90º and 150º. So, ∠AOC = 90º and∠AOB = 150º In ΔAOB, OA = OB (radius of the **circle**) As we know, angles opposite to equal sides are equal. So, ∠OBA = ∠OAB According to the angle sum property of triangle theorem, the sum of all angles of a triangle = 180° In ΔAOB, ∠OAB + ∠AOB +∠OBA = 180°. **Draw** a diagram. The **two** **chords** **in** **the** diagram are of length 6 and 12 units so units and |AB| = 6 units. Suppose the radius of the **circle** is r units and |BC| = x units then |EC| = 2x units. Triangles CAB and CDE are right triangles. Pythagoras can help at this point. Penny. If **two chords** of the **circle** \\( x^{**2**}+y^{**2**}-a x-b y=0 \\), **drawn** from the point \\( (a, b) \\) is divided by the \\( x \\)-axis in the ratio \\( **2**: 1 \\) then:(a) \\(. Paths always start with an M (move) command. You'll also need an A (arc) command, and probably an L line command for the line that bisects the **circle** . For the arc command, you just need to know the X and Y coordinates of the start and end points (B. (i) **Draw** a **circle** with centre O and a suitable radius. (ii) **Draw** **two** **chords** AB and CD which are not parallel to each other. (iii) **Draw** **the** perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these **two** **chords** intersect each other at the centre O, of the **circle**.