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21 Nov 2021, 10:24
Kudos
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Attachment:
Hi All,
I have posted a video on YouTube to discuss about Circles : Basics and Properties
Attached pdf of this Article as SPOILER at the top! Happy learning!
Following is Covered in the Video
Theory
- ¤ What is a Circle?
¤ Circle Geometry Definitions
¤ Circle : Area and Circumference
¤ Semicircle : Area and Circumference
¤ Arc of a Circle
¤ Sector of Circle
¤ Properties of Circles
What is a Circle?
A Circle is a 2D figure which is formed by joining all the points in a 2D plane which are at a fixed distance (i.e radius) from a single point. (i.e center of the circle)
Circle Geometry Definitions
¤ Radius – A line segment joining the center of the circle to any point on the circle. (Ex: OA)
¤ Chord– A line segment whose two end points lie on the circle (Ex: BC)
¤ Diameter– A chord which passed through the center. (Ex: DE)
( Diameter = 2* Radius )
¤ Secant– A line which cuts the circle at two points. (Ex: line s)
¤ Tangent– A line which touches circle at only one point. (Ex: line t)
Circle : Area and Circumference
Area of a Circle with radius r, A = ∏ \(r^2\)
Circumference of a Circle with radius r, C = 2 ∏ r
Central Angle = 360 ̊
Semicircle : Area and Circumference
Area of a Semicircle with radius r, A = \(\frac{∏r^2}{2}\)
Circumference of a Semicircle with radius r, C = ∏ r + 2r
Central Angle = 180 ̊
Arc of a Circle
Arc of a circle is a part of the Circumference of the circle.
Length of Arc AB, which subtends angle Θ at the center,
AB = \(\frac{Θ}{360 ̊}\)∗ 2 ∏ r
Sector of Circle
Sector of a circle is a part of the circle made by the arc of the circle and the two radii connecting the arc to the center of the circle.
Area of sector OACB, which subtends angle Θ at the center,
Area of OACB =\( \frac{Θ}{360 ̊}∗ ∏ r^2\)
Circumference of Sector OACB is given by
Circumference of OACB = \(\frac{Θ}{360 ̊}\)∗ 2 ∏ r + 2r
Properties of Circles
PROP 1: A Chord subtends same angle at any point on the circle.
PROP 2: Angle subtended by the chord at the center is twice the angle subtended by the chord at any other point on the circle.
PROP 3: Diameter subtends 90 ̊ at any point on the circle
PROP 4: From an external point there are only two tangents which can be drawn to a circle and the length of these tangents is equal.
PROP 5: A tangent always makes 90 ̊with the line joining the point of tangency (point of intersection of the tangent with the circle) to the center of the circle.
PROP 6: Cyclic Quadrilateral
A quadrilateral whose all 4 vertices lie on the circumference of the circle is called a Cyclic Quadrilateral.
Sum of all the angles of Cyclic Quadrilateral = 360 ̊
∠ A + ∠ B + ∠ C + ∠ D = 360 ̊
Sum of diagonally opposite angles = 180 ̊
∠ A + ∠ C = 180 ̊
∠ B + ∠ D = 180 ̊
PROP 7: Perpendicular drawn from the center of the circle to a chord bisects the chord.
PROP 8: Equal chords are equidistant from the center. Or
Chords which are equidistant from the center are equal.
PROP 9: Equal chords subtend same angle at the center of the circle. Or
Chords which subtend same angle at the center of the circle are equal.
Hope it Helps!
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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04 Dec 2021, 11:19
Kudos
Bookmarks
Hi All,
I have posted a video on you-tube to discuss about Equilateral Triangle Inscribed in A Circle
And, Length of Major and Minor Arc formed by a Side of an Equilateral Triangle Inscribed in a Circle
Following is Covered in the Video
Measure of Center Angle
Measure of Center Angle ∠AOB = ∠AOC = ∠BOC = 120° ( Watch this video to understand how )
Hint: ∠ACB = 60°, so center angle will be twice of it
Also, ∠OAB = ∠OBA = 30°
∠OBC = ∠OCB = 30°
∠OAC = ∠OCA = 30°
Height of Triangle
Height of triangle, h (AD) = \(\frac{3r}{2}\)
( Watch this video to understand how )
Base of Triangle
Base of triangle, BC = √𝟑 r ( Watch this video to understand how )
Area of Triangle
Area of triangle ABC = \(\frac{3√𝟑 r^2}{4}\)
( Watch this video to understand how )
Length of Minor Arc formed by one side of the triangle
Length of Minor Arc AB = \(\frac{120˚}{360˚}* 2 \pi \) r = \(\frac{2\pi r}{3}\) (as Minor angle ∠AOB = 120˚)
( Watch this video to understand how )
Length of Major Arc formed by one side of the triangle
Length of Major Arc AB = \(\frac{240˚}{360˚} * 2\pi r\) = \(\frac{4\pi r}{3}\) (as Major angle ∠AOB = 240˚)
( Watch this video to understand how )
Hope it helps!
I have posted a video on you-tube to discuss about Equilateral Triangle Inscribed in A Circle
And, Length of Major and Minor Arc formed by a Side of an Equilateral Triangle Inscribed in a Circle
Following is Covered in the Video
- ¤ Equilateral Triangle Inscribed in a Circle : Properties
-> Measure of Center Angle
-> Height of Triangle
-> Base of Triangle
-> Area of Triangle
¤ Length of Minor Arc formed by one side of the triangle
¤ Length of Major Arc formed by one side of the triangle
Measure of Center Angle
Measure of Center Angle ∠AOB = ∠AOC = ∠BOC = 120° ( Watch this video to understand how )
Hint: ∠ACB = 60°, so center angle will be twice of it
Also, ∠OAB = ∠OBA = 30°
∠OBC = ∠OCB = 30°
∠OAC = ∠OCA = 30°
Height of Triangle
Height of triangle, h (AD) = \(\frac{3r}{2}\)
( Watch this video to understand how )
Base of Triangle
Base of triangle, BC = √𝟑 r ( Watch this video to understand how )
Area of Triangle
Area of triangle ABC = \(\frac{3√𝟑 r^2}{4}\)
( Watch this video to understand how )
Length of Minor Arc formed by one side of the triangle
Length of Minor Arc AB = \(\frac{120˚}{360˚}* 2 \pi \) r = \(\frac{2\pi r}{3}\) (as Minor angle ∠AOB = 120˚)
( Watch this video to understand how )
Length of Major Arc formed by one side of the triangle
Length of Major Arc AB = \(\frac{240˚}{360˚} * 2\pi r\) = \(\frac{4\pi r}{3}\) (as Major angle ∠AOB = 240˚)
( Watch this video to understand how )
Hope it helps!
13 Jan 2023, 10:26
Kudos
Bookmarks
Show SpoilerDOWNLOAD PDF: How To Solve: Similar Triangles
Attachment:
How To Solve: Similar Triangles
Attached pdf of this Article as SPOILER at the top! Happy learning!
Hi All,
I have recently uploaded a video on YouTube to discuss Similar Triangles in Detail:
Following is covered in the video
- ¤ Definition of Similar Triangles
¤ Properties of Similar Triangles
¤ Relationship of Perimeter of two Similar Triangles
¤ Relationship of Area of two Similar Triangles
Definition of Similar Triangles
Two triangles are similar if at least two of their corresponding angles are equal.
- => If two angles are equal then the third angle will also be equal (As sum of the angles is 180°)
=> If all three corresponding angles of two triangles are equal then they are similar triangles
Attachment:
Image-1.jpg [ 10.69 KiB | Viewed 3870 times ]
In above Figure △ ABC and △ DEF are similar because ∠A = ∠D, ∠B = ∠E and ∠C = ∠F
Properties of Similar Triangles
If two triangles are similar, then their corresponding sides will be in the same ratio.
Attachment:
Image-1.jpg [ 10.69 KiB | Viewed 3870 times ]
In above Figure △ ABC and △ DEF are similar
=> \(\frac{AB}{DE}\) = \(\frac{BC}{EF}\) = \(\frac{AC}{DF}\)
Relationship of Perimeter of two Similar Triangles
Ratio of Perimeter of two similar triangles is equal to the ratio of their sides.
Attachment:
Image-1.jpg [ 10.69 KiB | Viewed 3870 times ]
In above Figure △ ABC and △ DEF are similar
=> \(\frac{AB}{DE}\) = \(\frac{BC}{EF}\) = \(\frac{AC}{DF}\) = k (assume)
=> AB = k*DE
=> BC = k*EF
=> AC = k*DF
=> Perimeter of △ ABC / Perimeter of △ DEF = \(\frac{AB + BC + AC }{ DE + EF + DF}\) = \(\frac{k*DE + k*EF + k*DF }{ DE + EF + DF}\)
= \(\frac{k * ( DE + EF + DF ) }{ DE + EF + DF}\) = k = \(\frac{AB}{DE}\) = \(\frac{BC}{EF}\) = \(\frac{AC}{DF}\)
Relationship of Area of two Similar Triangles
Ratio of Area of two similar triangles is equal to square of ratio of their sides.
Attachment:
Image-2.jpg [ 11.85 KiB | Viewed 3804 times ]
In above Figure △ ABC and △ DEF are similar and AG is perpendicular(⊥) to BC and DH ⊥ EF
If we consider △ AGB and △ DHE, then ∠B = ∠E, ∠G = ∠H = 90° => ∠GAB = ∠HDE
=> △ AGB and △ DHE
=> Their sides will be in the same ratio
=> \(\frac{AG}{DH}\) = \(\frac{GB}{HE}\) = \(\frac{AB}{DE}\) ...(1)
And we already know that △ ABC and △ DEF
=> \(\frac{AB}{DE}\) = \(\frac{BC}{EF}\) = \(\frac{AC}{DF}\) = k ...(2)
From (1) and (2) we get
\(\frac{AG}{DH}\) = \(\frac{GB}{HE}\) = \(\frac{AB}{DE}\) = \(\frac{AB}{DE}\) = \(\frac{BC}{EF}\) = \(\frac{AC}{DF}\) = k
=> Area of △ ABC / Area of △ DEF = (\(\frac{1}{2}\) * BC * AG) / (\(\frac{1}{2}\) * EF * DH) = \(\frac{BC * AG }{ EF * DH}\) = \(\frac{BC}{EF}\) * \(\frac{AG}{DH}\) = k * k = \(k^2\)
Hope it helps!
Good Luck!
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
