MATHEMATICS
MORE EXERCISES
Perpendicular lines to a chord.

Question  1

O is the centre of the circle.
OQ ⊥ PR.
1.1   If OP = 5 cm and PR = 8 cm, calculate OQ.
[ A 1.1 ]

1.2   If OP = 13 cm and OQ = 5 cm, find PR.
[ A 1.2 ]

1.3   If OQ = 8 cm and PR = 15 cm, find OP.
[ A 1.3 ]

Question  2
Find, giving reasons, the value of x and y
in each diagram :
2.1
OM ⊥ AB;  AB = 8 cm;  OA = 5 cm;
AM = x;  OM = y [ A 2.1 ]
2.2
ME ⊥ CD;  ME = 5 mm;  MC = 13 mm;
CE = x;  CD = y [ A 2.2 ]
2.3
OM ⊥ AB;  ON ⊥ CD;  AB = 18 cm;
AB ǁ CD;  OC = 15 cm;  CD = 24 cm;
OM = x; and MON = y [ A 2.3 ]

Question  3

In the diagram O is the centre of the circle.
AO = 12 cm;  AB = 14 cm;  ON = √44;
MO ⊥ AB;  OP ⊥ AC;  ON ⊥ BC
3.1   Calculate OM. [ A 3.1 ]
3.2   Caclulate BC. [ A 3.2 ]
3.3   If AP = 4 cm and OP = √128 cm,
prove that OP ⊥ AC. [ A 3.3 ]

Question  4

In the diagram O is the centre of the circle.
OR = 5 mm;  OS = 7 mm;
MN = 48 mm;  OR ⊥ PQ;  OS ⊥ MN
Determine the length of PQ. [ A 4. ]

Question  5

In the diagram M is the centre of the circle.
AC = 15 mm;  AB = 15 mm;
BC = 18 mm;  AMD is a straight line and
AMD ⊥ BC
Determine the length of AM. [ A 5. ]

Question  6

In die diagram O is the centre of two circles.
Chord AD of the bigger circle intersects
the smaller circle in B and C.
OB = 5 cm;  BC = 6 cm;  OD = 8,75 cm.
Determine the length of AD and AB. [ A 6. ]

#### 1.1

\begin{align*} OQ \perp PR \kern4mm\ .\ .\ .\ given \\ \therefore PQ &= QR \\ \tag{line fron centre \perp chord} \\ PQ &= QR = 4\ cm \\ \tag{PR = 8 cm . . . given} \\ OP &= 5\ cm\ and\ PQ &= 4\ cm \\ \tag{given} \\ In\ \Delta OPQ\ \:\ OP^2 &= OQ^2 + PQ^2 \\ \tag{Pythagoras} \\ 5^2 &= OQ^2 + 4^2 \\ OQ^2 &= 25 - 16 \\ &= 9 \\ OQ &= \sqrt{9} \\ &= 3 \\ \end{align*}
[ Q 1.1 ]

#### 1.2

\begin{align*} \text{OQ ⊥ PR . . . given} \\ \text{∴ PQ = QR . . . line from centre ⊥ chord} \\ \text{OP = 13 cm en OQ = 5 cm . . . given} \\ In\ \Delta OPQ\ OP^2 &= OQ^2 + PQ^2 \\ \tag{Pythagoras} \\ 13^2 &= 5^2 + PQ^2 \\ PQ^2 &= 169 - 25 \\ &= 144 \\ PQ &= \sqrt{144} \\ &= 12 \\ PR &= 24\ cm \\ \end{align*}
[ Q 1.2 ]

#### 1.3

\begin{align*} \text{OQ ⊥ PR . . . given} \\ \text{∴ PQ = QR . . . line from centre ⊥ chord} \\ \text{PR = 15 cm en PQ = 7,5 cm . . . given} \\ In\ \Delta OPQ\ OP^2 &= OQ^2 + PQ^2 \\ \tag{Pythagoras} \\ &= 8^2 + 7,5^2 \\ &= 64 + 56,25 \\ &= 120,25 \\ OP &= \sqrt{120,25} \\ &= 10,966\ cm \\ \end{align*}
[ Q 1.3 ]

#### 2.1

\begin{align*} \text{OM ⊥ AB . . . given} \\ \text{∴ AB = MB . . . line from centre ⊥ chord} \\ \text{AB = 8 cm and OA = 5 cm . . . goven} \\ \text{∴ AB = 4 cm} \\ \text{x = 4} \\ In\ \Delta OAM\ OA^2 &= OM^2 + AM^2 \\ \tag{Pythagoras} \\ 5^2 &= y^2 + 4^2 \\ y^2 &= 25 - 16 \\ &= 9 \\ y &= \sqrt{9} \\ &= 3\ cm \\ \end{align*}
[ Q 2.1 ]

#### 2.2

\begin{align*} \text{ME ⊥ CD . . . given} \\ \text{∴ CE = ED . . . line from centre ⊥ chord} \\ \text{ME = 5 mm en MC = 13 mm . . . given} \\ In\ \Delta CEM\ MC^2 &= ME^2 + CE^2 \\ \tag{Pythagoras} \\ 13^2 &= 5^2 + x^2 \\ x^2 &= 169 - 25 \\ &= 144 \\ x &= \sqrt{144} \\ &= 12\ mm \\ CD &= 2 \times CE \\ &= 24 \\ y &= 24\ mm \\ \end{align*}
[ Q 2.2 ]

#### 2.3

\begin{align*} \text{MO ⊥ AB . . . given} \\ \text{∴ AM = MB . . . line from centre ⊥ chord} \\ \text{AB = 18 cm en OC = 15 cm . . . given} \\ \text{OM = x en MON = y . . . given} \\ \text{AB = 18 cm and thus AM = 9 cm} \\ \text{OA = OC = 15 cm} \\ In\ \Delta OAM\ OA^2 &= OM^2 + AM^2 \\ \tag{Pythagoras} \\ 15^2 &= x^2 + 9^2 \\ x^2 &= 225 - 81 \\ &= 144 \\ x &= \sqrt{144} \\ &= 12\ cm \\ \text{NO ⊥ CD . . . given} \\ \text{∴ CN = ND . . . line from centre ⊥ chord} \\ \text{CD = 24 cm en CN = 12 cm . . . given} \\ In\ \Delta OCN\ : OC^2 &= ON^2 + CN^2 \\ \tag{Pythagoras} \\ 15^2 &= ON^2 + 12^2 \\ ON^2 &= 225 - 144 \\ &= 81 \\ ON &= \sqrt{81} \\ &= 9\ cm \\ y &= MON = MO + ON \\ y &= 12 + 9 cm \\ y &= 21\ cm \\ \end{align*}
[ Q 2.3 ]

#### 3.1

\begin{align*} \text{OM ⊥ AB . . . given} \\ \text{∴ AM = MB . . . line from centre ⊥ chord} \\ \text{AB = 14 cm en OA = 12 cm . . . given} \\ \text{∴ AM = 7 cm.} \\ In\ \Delta OAM\ OA^2 &= AM^2 + OM^2 \\ \tag{Pythagoras} \\ 12^2 &= 7^2 + OM^2 \\ OM^2 &= 144 - 49 \\ &= 95 \\ OM &= \sqrt{95} \\ &= 9,747\ cm \\ \end{align*}
[ Q 3. ]

#### 3.2

\begin{align*} \text{OB = AO = 12 cm . . . radii and given} \\ \text{ON ⊥ BC . . . given} \\ \text{∴ BN = NC . . . line from centre ⊥ chord} \\ In\ \Delta ONB\ : OB^2 &= BN^2 + ON^2 \\ \tag{Pythagoras} \\ 12^2 &= BN^2 + (\sqrt{44})^2 \\ BN^2 &= 144 - 44 \\ &= 100 \\ BN &= \sqrt{100} \\ &= 10\ cm \\ BC &= 2 \times BN \\ &= 2 \times 10 \\ &= 20\ cm \\ \end{align*}
[ Q 3. ]

#### 3.3

\begin{align*} \text{OP ⊥ AC if ∠AQO = 90° } \\ In\ \Delta OAP\ : OA^2 &= 12^2 = 144 \\ AP^2 &= 4^2 = 16 \\ OP^2 &= (\sqrt{128})^2 = 128 \\ AP^2 + OP^2 &= 16 + 128 \\ &= 144 = OA^2 \\ \text{∴ OAP is a rectangular triangle} \\ \text{∴ ∠OPA = 90°} \\ \text{∴ OP ⊥ AP} \\ \end{align*}
[ Q 3. ]

\begin{align*} \text{OS ⊥ MN . . . given} \\ \text{MS = SN en MN = 48 mm . . . given} \\ \text{MS = SN = 24 mm . . . given} \\ In\ \Delta OMS\ : OM^2 &= MS^2 + MO^2 \\ \tag{Pythagoras} \\ &= 24^2 + 7^2 \\ &= 625 \\ OM &= \sqrt{625} \\ &= 25 \\ In\ \Delta OPR\ : OP^2 &= PR^2 + OR^2 \\ \tag{Pythagoras} \\ 25^2 &= PR^2 + 5^2 \\ \tag{OP = om } \\ PR^2 &= 625 - 25 \\ &= 600 \\ &= 10\sqrt{6} \\ PQ &= 2 \times PR \\ \tag{OR ⊥ PQ} \\ &= 2 \times 10\sqrt{6} \\ &= 20\sqrt6 \\ &= 48,90\ mm \\ \end{align*}
[ Q 4. ]

\begin{align*} \text{AB = AC = 15 mm . . . given} \\ \text{∴ Δ ABC is an isosceles triangle} \\ \text{AMD ⊥ BC . . . given} \\ \text{BD = DC = 9 . . . line from centre ⊥ chord} \\ In\ \Delta ABD\ : AB^2 &= AD^2 + BD^2 \\ \tag{Pythagoras} \\ 15^2 &= AD^2 + 9^2 \\ AD^2 &= 225 - 81 \\ &= 144 \\ AD &= \sqrt{144} \\ &= 12 \\ AM &= AD - MD \\ Let\ MD &= x \\ In\ \Delta MBD\ : MB &= AM\ \ \ .\ .\ .\ \ \ radii \\ MB^2 &= MD^2 + BD^2 \\ \tag{Pythagoras} \\ (12 - x)^2 &= x^2 + 9^2 \\ 144 - 24x + x^2 &= x^2 + 81 \\ 24x &= 144 - 81 \\ x &= \frac{63}{24} \\ MD &= \frac{21}{8} \\ AM &= 12 - \frac{21}{8} \\ &= \frac{75}{8} \\ &= 9,375 \\ \end{align*}
[ Q 5. ]

\begin{align*} \text{Draw OE ⊥ ABCD} \\ \text{OE ⊥ ABCD and thus BE = EC and AE = ED} \\ \text{BE = EC = 3 . . . BC = 6} \\ In\ \Delta OBE\ : OB^2 &= OE^2 + BE^2 \\ \tag{Pythagoras} \\ 5^2 &= OE^2 + 3^2 \\ OE^2 &= 25 - 9 \\ &= 16 \\ OE &= \sqrt{16} \\ &= 4 \\ In\ \Delta OED\ : OD^2 &= ED^2 + OE^2 \\ \tag{Pythagoras} \\ 8,5^2 &= ED^2 + 4^2 \\ ED^2 &= 72,25 - 16 \\ &= 56,25 \\ ED &= \sqrt{56,25} \\ &= 7,5 \\ AD &= 2 \times ED \\ &= 2 \times 7,5 \\ AD &= 15\ cm \\ \end{align*}
[ Q 6. ]