GEOMETRY(ii)

 

Contents:

 

Circles: A shape designed of fixed points of equal distance from the centre, this distance is known as the radius. The circumference is the perimeter. An arc is a section of the circumference. A chord is a line joining two points on the circumference. The diameter is a chord which passes through the centre. A segment is part of the circle that is separated by a line. A sector is part of the circle separated by two lines of radius. A tangent is an outside line that just touches the perimeter at one point only.

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Pi: Circumference is calculated from the value of the radius or the diameter. Pi or is 227 or approximately 314 and is necessary for many calculations.

Circumference = 2radius or diameter.

Radius = circumference2 or diameter2.

Diameter = circumference or 2radius.

Area = radius.

Exercise with (Note: please use whole numbers only).

             

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Properties and Rules:

(i) The angle of a triangle on the circumference which includes a line through the diameter is always 90, (see (i) angle a at points A, B, C,) Right Angle.

(ii) An isosceles triangle is formed by 2 radii, the angles on the circumference are identical (see(ii) angle a at points M and N). A perpendicular line from the base of the triangle always (i) cuts the line in half and (ii) passes through the centre of the circle.

(iii) The opposide angle from a chord is twice the size at the centre than at the circumference (see (iii) angle at point O is twice the angle at point P). Note angles b are identical.

(iv) The angles around the circumference are identical if they are from a triangle from the same chord (see (iv) angle a is equal to angle a), and if a triangle is added to the opposite side of the chord, angle b the two opposite angles a + b sum to 180.

(v) The angle of a triangle on the circumference which includes a chord for the base is always the same (see (v) angles a at points C and D).

(vi) A cyclic quadrilateral is one inside the circle in which all of its corners touch the circumference. The opposite angles of a cyclic quadrilateral add up to 180(see (vi) angles a and c add up to 180; also angles b and d add up to 180). If a line is extended from a corner, that angle is equal to the opposite angle (see (vi) angle e is equal to angle a).

(vii) The angle between a radius and a tangent is a right angle (see (vii) where angles at points M and N are right angles). Tangents that meet at an external point are equal in length (see(vii) line MP is equal to line NP).

(vii) A tangent (see (viii) tangent perpendicular to OP) and an adjoined chord PR have an identical angle a at point P to the angle in the opposite segment a at point Q.

(ix) Intersecting chords, if two chords are extended outside the circle and meet at a point X then the lengths of these chords equate to (see (ix) AX x BX = CX x DX). This also applies if the meeting point X is a cross over inside the circle.

(x) If a chord is extended outside the circle to meet a tangent the complete length of the chord times its extension is equal to the square of the length of the tangent (see (x) AX x BX = TX).

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Examples:

Example: Find the angles b and c if AC is the diameter and angle a is 30.

Answer: b is 90 AC is a diameter. Therefore c is 180 - 90 - 30 = 60.

 

Example: Find angles b and c if point O is centre and angle a is 30.

Answer: Drop a line from point O perpendicular to the chord PQ. This gives two right angled triangles. Angle a is equal to angle c at 30(isosceles triangle). Angle b is shared equally between the two triangles. From any one triangle b is 180 - 90 - 30 = 60. Therefore b is 60 times 2 = 120.

Example: Find angles a and b and c if triangle OPQ is equilateral.

Answer: Each angle in the equilateral triangle is 60, angle a is on a straight line, 180- 60making a = 120. Angles b and c are in a base of an isosceles triangle and are equal. Therefore 180- 120= 60 shared between two angles, b and c are 30.

Example: Find the reflex angle (reflex angle is an angle greatre than 180) at O, and angles s and q.

Answer: 360around a point, minus 80= 280. Angle s is half of 80, equals 40. The points PQRS make up a cyclic quadrilateral where opposite corners make up 180. Therefore angle q is 180- angle s of 40= 140.

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Example: Find the angle b if angle a is 60by drawing in the missing chord.

Answer: Join DC this chord now becomes the base for both triangles ADC and BDC. Therefore angles a and b are identical at 60.

 

Example: AB and DC are parallel, find the angles x and b if angle a is 30.

Answer: The triangle ABx is isosceles, base on a chord, angle b is equal to angle a = 30. This leaves 180- 30- 30= 120angle at the crossover (opposite angles are identical). Therefore 360(at the point) - 2 times 120= 120. Both angles x = 120. x = 1202 = 60.

Example: PR is a tangent to the circle at P, if a chord is drawn between P and Q find angles o and p and q if r is 15.

Answer: Angle o is found by OPR being a right hand triangle with 15also known at r. 180- 90- 15= 75. Because OP and OQ are radii and equal, the base angles on the chord p and q are also equal, o is known as 75, therefore 180- 75= 105shared between p and q. 1052 = 52each.

Example: AB and AC are tangents to the circle at a chord, AB is the chord, angle c is 50.

Answer: As AB and AC are tangents the triangle ABC is isosceles, therefore angles a and b are equal. 180- 50= 130. 130shared between a and b is 1302 = 65each.

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Example: AB is the diameter of a circle and CD is a tangent at C, a chord connects A to C, find the internal angles of the triangle if the angle c to the tangent is 75.

Answer: Angle c is 90due to the angle being on the circumference with the base being the diameter. The angle between the chord and the tangent is equal to the angle in the opposite segment, therefore the remaining unknown angle at C is 180- 90- 75= 15. The opposite angle is angle a, therefore this is also 15. The remaining angle b is 180- 90- 15= 75. Note b is also an opposite angle to the 75angle given between a chord and a tangent, and so is equal.

Example: The chords AB and CD crossover at X, if the lengths: AX is 4, BX is 6, CX is 5, find the length DX.

Answer: AX x Bx = CX x DX, or rearranged to find DX: DX = AX x BXCX. Therefore DX = 4 x 65 = 48.

 

Example: Find the length of chord AB where the extension to a point X is 5 and the tangent to the same point X is 8.

Answer: AX x BX = 8, or rearranged to find AX: AX = 64BX. AX is therefore 12.8. AB is AX - BX or 128 - 5. AB = 78.

 

Example: How far away is the horizon seen from the crow's nest of a ship 50 metres high. (Assume the earth's radius is 6,400,000 metres).

Answer: Using the principle of the above example, AB = 12,800,000 (the diameter), AX = 12800050 and BX = 50. AX x BX = horizon. Rearranged: horizon = 12800050 x 50. Horizon = 25,298 metres.

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