Geometry of Circles Chapter 21 Review Set Solutions PDF

## REVIEW SET 21 ### 1 #### a - **BOC** = 2 × **BAC** {angle at the centre} - **BOC** = 2 × 36° = 72° - **OB** = **OC** {radii} - **AOB** is isosceles. - **OBC** = **OCB** = a° {base angles of isosceles triangle} - a + a + 72 = 180 {angles in a triangle} - 2a + 72 = 180 - 2a = 108 - a = 54 #### b -...

## REVIEW SET 21 ### 1 #### a - **BOC** = 2 × **BAC** {angle at the centre} - **BOC** = 2 × 36° = 72° - **OB** = **OC** {radii} - **AOB** is isosceles. - **OBC** = **OCB** = a° {base angles of isosceles triangle} - a + a + 72 = 180 {angles in a triangle} - 2a + 72 = 180 - 2a = 108 - a = 54 #### b - **ABC** = 90° {angle in a semi-circle} - 28 + 90 + a = 180 {angles in a triangle} - a + 118 = 180 - a = 62 ### 2 - **ACD** = **ABD** = 41° {angles on same arc} - 41 + a = 102 {exterior angle of AECD} - a = 61 - **CBD** = β {angle between a tangent and a chord} - **BCD** = α - α + β + γ = 180° {angles in a triangle} ### 3 #### a - **ABC** = 90° {angle in a semi-circle} - x + (x + 14) + 90 = 180 {angles in a triangle} - 2x + 104 = 180 - 2x = 76 - x = 38 #### b - **AOC** + 240° = 360° {angles at a point} - **AOC** = 120° - **ABC** = **AOC** {angle at the centre} - **ABC** = 1/2 × 120° - = 60° - x = 80 + 60 {exterior angle of a triangle} - x = 140 #### c - **CBA** = **CAT** = 38° {angle between a tangent and a chord} - **CB** = **CA** {given} - **AABC** is isosceles. - **CAB** = **CBA** = 38° {base angles of isosceles triangle} - x + 38 + 38 = 180 {angles in a triangle} - x+76= 180 - x = 104 ### 4 - **AM** = **CM** {tangents from an external point} - **CM** = **BM** - **AM** = **BM** - M is the midpoint of [AB]. #### b - **MAC** = α and **MBC** = β {see above} - **AM** = **CM**, ΔAMC is isosceles. - **ACM** = **CAM** = α {base angles of isosceles triangle} - **CM** = **BM**, ΔBMC is isosceles. - **MCB** = **CBM** = β {base angles of isosceles triangle} - In ΔABC, α + β + β + α = 180° {angles in a triangle} - 2α + 2β = 180° - α + β = 90° - **ACB** = α + β - **ACB** = 90° ### 5 - Construct [XB] and [PB]. - Draw the tangent to the circle at X. - **BXS** = **XAB** {angle between a tangent and a chord} - **PAX** = **PBX** (angles subtended by the same arc) - **XAB** = **PAX** {given} - **PBX** = **BXS** - These are alternate angles between [PB] and [RXS]. - The tangent at X is parallel to [PB]. - Join [BC]. - **AOB** = α - **AOB** = 2 × **ACB** {angle at the centre} - **ACB** = α / 2 - **DOC** = β - **DOC** = 2 × **DBC** {angle at the centre} - **DBC** = β / 2 - In ABEC, α/2 + β/2 + 90 = 180 {angles in a triangle, BEC = 90° angles on a line} - α/2 + β/2 = 90 - α + β = 180 ### 6 - We construct [BQ]. - Let **QAC** be α - **PAB** = α {AP bisects CAB} - Let **ACB** = β - **AQB** β {angles on same arc} - In AAPC, **APC** + α + β = 180° {angles in a triangle} - **APC** = 180° - α - β - In AABQ, **ABQ** + α + β = 180° {angles in a triangle} - **ABQ** = 180° - α - β - **APC** = **ABQ** ## PRACTICE TEST 21B ### 1 #### a - **DBC** = **DAC** = x° {angles on the same arc} - 56 + x = 142 {exterior angle of a △} - x = 86 #### b - We construct [OB]. - M bisects [BC] {chord of a circle} - **BM** = 5 cm - **OB** = **OA** = x {radii} - 3² + 5² = x² {Pythagoras} - x² = 34 - x = √34 {as x > 0} - x ≈ 5.83 ### 2 - We construct [OA]. - **OA** = **OB** = 8 cm {radii} - **OAC** = 90° {radius-tangent theorem} - OA² + AC² = OC² {Pythagoras} - 8² + 15² = (x + 8)² - 64 + 225 = x² + 16x + 64 - x² + 16x - 225 = 0 - (x - 9)(x + 25) = 0 - x = 9 {as x > 0} #### a - **obtuse BOD** = 2 × **BAD** {angle at the centre} - = 2α° #### b - **reflex BOD** = 2 × **BCD** {angle at the centre} - = 2β° - **obtuse BOD** + **reflex BOD** = 360° {angles at a point} - 2α + 2β = 360 - 2(α + β) = 360 - α + β = 180 #### c - **AOC** + 250° = 360° {angles at a point} - **AOC** = 110° - **ABC** = **AOC** {angle at the centre} - x = 1/2 × 110 - x = 55 ### 3 #### a - **OCT** = 90° {radius-tangent theorem} - **COT** = 70° {angles in △COT} - **OA** = **OC** {radii} - **AOAC** is isosceles. - **OCA** = **OAC** = x° {base angles of isosceles triangle} - x + x + 70 = 180 {angles in △AOC} - 2x + 70 = 180 - 2x = 110 - x = 55 #### b - Join [OT]. - In ΔΑΟΒ, **ABO** + 52° + 90° = 180° {angles in a triangle} - **ABO** = 38° - In AOTB, **OTB** = 90° {radius-tangent} - **TB** = 6 / tan 38° - **TB** ≈ 7.68 cm - **TB** = **BS** {tangents from an external point} - x ≈ 7.68 ### 4 - In ∆ABC, x + 16 + 2x + 20 + x = 180 {angles in a triangle} - 4x + 36 = 180 - 4x = 144 - x = 36 - x + 16 = 52 and 2x + 20 = 92 - In △DEF, y - 22 + y + 34 + y = 180 {angles in a triangle} - 3y + 12 = 180 - 3y = 168 - y = 56 - y - 22 = 34 and y + 34 = 90 - As **FED** = 90°, FD is a diameter of the circle {angle in a semi-circle} - sin 56° = 5.5 / FD - **FD** = 5.5 / sin 56° - **FD** ≈ 6.63 m - The diameter of the circle is approximately 6.63 m. ### 5 - **OD** = **OB** {radii} - ΔODB is isosceles. - **OBD** = **ODB** = α {base angles of isosceles triangle} - **ACD** = **ABD** = α {angles on same arc} - But **BDO** = α {given} - **BDO** = **ACD** - **ABD** = **ACD** = α - **BAX** = **CAD** {angles subtended by the same arc} - **ABX** and **ACD** each have an angle marked α and an angle marked with a - The third angles of these triangles must be equal - **AXB** = **ADC**