Class 10: Circles – Exercise 18(c) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

18c10 Question 1: In the given diagram, $chord AB = chord BC$ and AB = AC.

(i) What is the relation between $\widehat{AB}$ and $\widehat{BC}$ ?

(ii) What is the relation between $\angle AOB$ and $\angle BOC$ ?

(iii) If $\widehat{AD}$ is greater than $\widehat{ABC}$ , then what is the relation between the chord $AD$ and $AC$ ?

(iv) If $\angle AOB = 50^o$ , find the measure of $\angle BAC$ .

Answer:

Given: $AB = BC$

(i) $\widehat{AB}=\widehat{BC}$ (Since equal chords subtend equal arcs)

(ii) $\angle AOB = \angle BOC$ (equal arcs will subtend equal angles at the center)

(iii) If $\widehat{AD} > \widehat{ABC}$

Therefore $AD > AC$

(iv) Given $\angle AOB = 50^o$

$\displaystyle \angle ABC = \frac{1}{2} Reflex \ \angle AOC = \frac{1}{2} . 260 = 130^o$

$\displaystyle \text{Therefore } \angle BAC = \frac{1}{2} (180-130) = 25^o$

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Question 2: In $\triangle ABC$ , the $\perp$ from vertices $A$ and $B$ on their opposite sides meet (when produced) the circumference of the triangle at points $D$ and $E$ respectively. Prove that: $\widehat{CD} = \widehat{CE}$

Answer:

Consider $\triangle AOM$ & $\triangle BPN$

$\angle AMO = \angle BNO = 90^o$ (given)

$\angle NBO = \angle AOM$ (vertically opposite angles)

$\therefore \triangle AOM \sim \triangle BPN$ (AAA postulate)

Therefore $\angle NBO = \angle MAO$

Therefore Arcs that subtend equal angle at circumference are equal

Therefore $\widehat{CD} = \widehat{CE}$

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Question 3: In a cyclic trapezium, prove that non parallel sides are equal and the diagonals are also equals.

Answer:

$ABCD$ is a cyclic trapezium. $AB \parallel DC$

Chord $AD$ subtends $\angle ABD$ at circumference

Chord $BC$ subtends $\angle BDC$ at circumference

But $\angle ABD = \angle BDC$ (vertically opposite angles)

Therefore $Chord \ AD = Chord \ BC$

$\Rightarrow AD = BC$

Now in $\triangle ADC$ & $\triangle BDC$

$DC$ is common

$\angle CAD = \angle DBC$ (angles in the same segment)

$AD = BC$ (proved above)

$\therefore \triangle ADC \cong \triangle BDC$ (SAS axiom)

$\therefore AC = DB$

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18c9 Question 4: In the given diagram, $AD$ is the diameter of the circle with center $O$ . Chords $AB,=BC = CD$ . If $\angle DEF = 110^o$ , find (i) $\angle AEF$ (ii) $\angle FAB$

Answer:

Given $AB = BC = CD$

(i) $AOD$ is the diameter

$\angle AED = 90^o$ (angle in semicircle)

$\angle AEF + \angle AED = \angle FED$

$\angle AEF = 110-90=20^o$

(ii) Since $AB = BC = CD$ (given)

$\angle AOB = \angle BOC = \angle COD$ (equal arcs subtend equal angles at center)

We know $\angle AOD = 180^o$

$\Rightarrow \angle AOB = \angle BOC = \angle COD = 60^o$

In $\triangle AOB$

$OA = OB$ (radius of the same circle)

Therefore $\angle OAB = \angle OBA$

$\angle OAB + \angle OBA = 180-\angle AOB = 180-60 = 120^o$

$\Rightarrow \angle OAB = \angle OBA = 60^o$

$ADEF$ is a cyclic quadrilateral

$\angle DEF + \angle DAF = 180^o$ (opposite angles are supplementary)

$\angle DAF = 180-110=70^o$

Now $\angle FAB = \angle DAF + \angle OAB = 70+60=130^o$

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18c8 Question 5: In the given diagram, if $\widehat{AB} = \widehat{CD}$ of circle with center $O$ , prove that quadrilateral $ABCD$ is an isosceles trapezium.

Answer:

Given $\widehat{AB}=\widehat{CD}$

$\angle ADB = \angle DBC$

(Equal arcs subtend equal angles at the circumference)

$\Rightarrow AD \parallel BC$ (alternate angles are equal)

Therefore $ABCD$ is a trapezium

Also $Chords AB = Chard CD \ (since \ \widehat{AB}=\widehat{CD} )$

Therefore $ABCE$ is an Isosceles trapezium.

$\\$ 18c7

Question 6: In a given figure $\triangle ABC$ is an isosceles triangle and $O$ is the center of the circumcircle. Prove $AP$ bisects $\angle BPC$ .

Answer:

Given $AB = AC$ (Since $\triangle ABC$ is an isosceles triangle)

$\angle APB = \angle APC$ (equal chords subtend equal angles on the circumference)

Therefore $AP$ bisects $\angle BPC$ .

$\\$

Question 7: If two sides of a cyclic quadrilateral are parallel, prove that:

(i) its other two sides are equal

(ii) its diagonals are equal

Answer:

(i) Given $AB \parallel DC$

Therefore $\angle BAC = \angle DCA$ (alternate angles)

Chord $AD$ subtends $\angle DCA$ on circumference.

Chord $BC$ subtends $\angle CAB$ on circumference.

Therefore $AD = BC$ (since equal chords subtend equal angles on the circumference)

(ii) Consider $\triangle ADB \ \ \& \ \ \triangle ABC$

$AB$ is common

$AD = CB$ (proved above)

$\angle ACB = \angle ADB$ (angles in the same segment)

Therefore $\triangle ADC \cong \triangle ABC$ (by SAS axiom)

Therefore $AC = DB$ (corresponding parts of congruent triangles are equal)

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18c6 Question 8: In the given diagram, circle with center $O$ . $PQ=QR=RS$ and $\angle PTS=75^o$ . Calculate:

(i) $\angle POS$

(ii) $\angle QOR$

(iii) $\angle PQR$

Answer:

Given $PQ = QR = RS$

$\Rightarrow \angle POQ = \angle QOR = \angle ROS$ (Equal chords subtend equal angles at the center of a circle)

$\angle POS = 2 \ \angle PTS$ (angle subtended at the center is twice that of the one subtended on the circumference)

$\displaystyle \angle POQ = \angle QOR = \angle ROS = \frac{150}{3} = 50^o$

In $\triangle OPQ$ ,

$OP = OQ$ (radius of the same circle)

$\angle OPQ = \angle OQP$

Therefore $\angle OPQ + \angle OQP + \angle POQ = 180^o$

$\angle OPQ + \angle OQP = 180-50=130^o$

$2 \angle OPQ = 130$

$\angle OPQ = 65^o$

Similarly, in $\triangle OQR, \angle OQR = \angle ORQ = 65^o$

and in $\triangle ORS, \angle ORS = \angle OSR = 65^o$

Therefore

(i) $\angle POS = 150^o$

(ii) $\angle QOR = 50^o$

(iii) $\angle PQR = \angle PQO + \angle OQR = 65+65 = 130^o$

$\\$

18c5 Question 9: In the given diagram, $AB$ is the side of regular six-side polygon and $AC$ is a side of a regular eight-sides polygon inscribed in a circle with center $O$ . Calculate:

(i) $\angle AOB$

(ii) $\angle ACB$

(iii) $\angle ABC$

Answer:

(i) Since $AB$ is the side of a regular hexagon

$\angle AOB = 60^o$

(ii) $\widehat{AB}$ subtends $\angle ACB$ at the circumference and $\angle AOB$ at the center

$\displaystyle \text{We know } \angle ACB = \frac{1}{2} \angle AOB = \frac{60}{2} = 30^o$

(iii) Since $AC$ is the side of a regular octagon

$\displaystyle \angle AOC = \frac{360}{8} = 45^o$

$\widehat{AC}$ subtends $\angle ABC$ at the circumference and $\angle AOC$ at the center

$\displaystyle \text{Therefore } \angle ABC = \frac{1}{2} \angle AOC = \frac{45}{2} = 22.5^o$

$\\$

Question 10: In a regular pentagon $ABCDE$ inscribed in a circle, find the ratio of the $\angle EDA : \angle ADC$ [1990]

Answer:

Consider $\widehat{AE}$

It subtends $\angle AOE$ at the center and $\angle ADE$ at the circumference

$\displaystyle \text{Therefore } \angle ADE = \frac{1}{2} \angle AOE = \frac{1}{2} \times \frac{360}{5} = 36^o$

Similarly, for $\widehat{BC}$ we have $\angle BDC = 36^o$

$\angle ADC = \angle ADB + \angle BDC = 36+36=72^o$

$\angle ADE : \angle ADC = 36 :72 = 1:2$

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18c4 Question 11: In the given diagram, $AB=BC=CD$ and $\angle ABC = 132^o$ . Find:

(i) $\angle AEB$

(ii) $\angle AED$

(iii) $\angle COD$ [1993]

Answer:

Given $AB = BC = CD, ABC = 32^o$

$\Rightarrow \angle AOB = \angle BOC = \angle COD$ (equal arcs subtend equal angles at the center of a circle)

(i) In cyclic quadrilateral $ABCE$

$\angle ABC + \angle AEC = 180^o$ (opposite angles in a cyclic quadrilateral are supplementary)

$132+ \angle AEC = 180 \Rightarrow \angle AEC = 48^o$

Since $AB = BC \Rightarrow \angle AEB = \angle BEC$ (equal chords subtend equal angles on the circumference)

$\displaystyle \therefore \angle AEB = \frac{1}{2} \angle AEC = 24$

(ii) Since $AB = BC = CD$

$\angle AEB = \angle BEC = \angle CED$

$\angle AED = 24+24+24=72$

(iii) $\widehat{CD}$ subtends $\angle COD$ at the center and $\angle CED$ on the circumference

$\therefore \angle COD = 2 \angle CED = 2 \times 24 = 48^o$

$\\$

Question 12: In the diagram, $O$ is the center of the circle and the length of $\widehat{AB} = 2 \times \widehat{BC}$ . If $\angle AOB=108^o$ find:

(i) $\angle CAB$

(ii) $\angle ADB$ [1996]

Answer:

Given $\widehat{AB} = 2 \times \widehat{BC}, \angle AOB = 108^o$

(i) $\angle AOB = 2 \angle BOC$

$\displaystyle \Rightarrow \angle BOC = \frac{108}{2} = 54$

(ii) $\displaystyle \angle BAC = \frac{1}{2} \angle BOC$ (angle subtended at the center is twice that subtended at the circumference by a chord)

$\displaystyle \angle BAC = \frac{1}{2} \times 54 = 27^o$

(iii) Similarly, $\displaystyle \angle ACB = \frac{1}{2} \angle AOB = \frac{108}{2} = 54$

In cyclic quadrilateral $ADBC$

$\angle ADB + \angle ACB = 180^o$

$\Rightarrow \angle ADB = 180-54=126^o$

$\\$ c13

Question 13: In the diagram given below, $O$ is the center of the circle. $AB$ is the side of regular pentagon and $AC$ is the side of a regular hexagon. Find the angles of the $\triangle ABC$ .

Answer:

$AB$ is the side of a regular pentagon

$\displaystyle \Rightarrow \angle AOB = \frac{360}{5} = 72^o$

$AC$ is a side of regular hexagon

$\displaystyle \Rightarrow \angle AOC = \frac{360}{6} = 60^o$

Now $\angle AOB + \angle AOC + reflex \angle BOC = 360^o$

$\angle BOC = 360 - 72-60 = 228^o$

$\widehat{BC}$ subtends $\angle BOC$ at the center and $\angle BAC$ on the circumference

$\displaystyle \angle BAC = \frac{1}{2} \angle BOC = \frac{228}{2} = 114^o$

Similarly $\displaystyle \angle ABC = \frac{1}{2} \angle AOC = \frac{60}{2} = 30^o$

$\displaystyle \angle ACB = \frac{1}{2} \angle AOB = \frac{1}{2} \times 72 = 36^o$

Therefore $\angle ABC = 30^o, \angle ACB = 36^o and \angle BAC = 114^o$

$\\$

18c1 Question 14: In the given diagram, $BD$ is the side of a regular hexagon, $DC$ is the side of a regular pentagon and $AD$ is a diameter. Calculate: