3 Mark Question

Question 13 Marks

In the adjoining figure, CD is a diameter of the circle with centre O. Diameter CD is perpendicular to chord AB at point E. Show that ∆ABC is an isosceles triangle.

Given: O is the centre of the circle.
diameter CD ⊥ chord AB, A-E-B
To prove: ∆ABC is an isosceles triangle.

Answer

Proof:
diameter CD ⊥ chord AB [Given]
∴ seg OE ⊥ chord AB [C-O-E, O-E-D]
∴ seg AE ≅ seg BE ……(i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord]
In ∆CEA and ∆CEB,
∠CEA ≅ ∠CEB [Each is of 90°]
seg AE ≅ seg BE [From (i)]
seg CE ≅ seg CE [Common side]
∴ ∆CEA ≅ ∆CEB [SAS test]
∴ seg AC ≅ seg BC [c. s. c. t.]
∴ ∆ABC is an isosceles triangle.

View full question & answer→

Question 23 Marks

In the adjoining figure, P is the centre of the circle. Chord AB and chord CD intersect on the diameter at the point E. If ∠AEP ≅ ∠DEP, then prove that AB = CD.

Given: P is the centre of the circle.
Chord AB and chord CD intersect on the diameter at the point E. ∠AEP ≅ ∠DEP
To prove: AB = CD
Construction: Draw seg PM ⊥ chord AB, A-M-B
seg PN ⊥ chord CD, C-N-D

Answer

Proof:

∠AEP ≅ ∠DEP [Given]
∴ Seg ES is the bisector of ∠AED.
PoInt P is on the bisector of ∠AED.
∴ PM = PN [Every point on the bisector of an angle is equidistant from the sides of the angle.]
∴ chord AB ≅ chord CD [Chords which are equidistant from the centre are congruent.]
∴ AB = CD [Length of congruent segments]

View full question & answer→

Question 33 Marks

In the adjoining figure, C is the centre of the circle, seg QT is a diameter, $CT =13, CP =5$. Find the length of chord RS.

Given: In a circle with centre C, QT is a diameter, CT $=13$ units, $C P=5$ units
To find: Length of chord RS
Construction: Join points R and C .

Answer

i. $C R=C T=13$ units $\qquad$ (i) [Radii of the same circle]
$\text { In } \triangle C P R, \angle C P R=90^{\circ}$
$\therefore C R^2=C P^2+R P^2[\text { Pythagoras theorem }]$
$\therefore 13^2=5^2+R P^2[\text { From (i)] }$
$\therefore 169=25+R P^2[\text { From (i) }]$
$\therefore R P^2=169-25=144$
$\therefore R P=\sqrt{144}[\text { Taking square root on both sides }]$
$\therefore R P=12 cm . . . \text { (ii) }$
ii. Now, seg CP_L chord RS [Given]
$\therefore R P=\frac{1}{2} R S$ [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
$\therefore 12=\frac{1}{2} RS[\text { From (ii)] }$
$\therefore R S=2 \times 12=24$
$\therefore$ The length of chord RS is $24$ units.

View full question & answer→

Question 43 Marks

Radius of a circle is 5 cm. The length of a chord of the circle is 8 cm. Find the distance of the chord from the centre.

View full question & answer→

Question 53 Marks

Prove that, if a diameter of a circle bisects two chords of the circle then those two chords are parallel to each other.
Given: O is the centre of the circle.
seg PQ is the diameter.
Diameter PQ bisects the chords AB and CD in points M and N respectively.
To prove: chord AB || chord CD.

Answer

Proof:
Diameter PQ bisects the chord AB in point M [Given]
∴ seg AM ≅ seg BM
∴ seg OM ⊥ chord AB [Segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord, P-M-O, O-N-Q]
∴ ∠OMA = 90° …..(i)
Also, diameter PQ bisects the chord CD in point N [Given]
∴ seg CN ≅ seg DN
seg ON ⊥ chord CD [Segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord, P-M-O, O-N-Q]
∴ ∠ONC = 90° …..(ii)
Now, ∠OMA + ∠ONC = 90° + 90° [From (i) and (ii)]
= 180°
But, ∠OMA and ∠ONC form a pair of interior angles on lines AB and CD when seg MN is their transversal.
∴ chord AB || chord CD [Interior angles test]

View full question & answer→

Question 63 Marks

n the adjoining figure, centre of two circles is O. Chord AB of bigger circle intersects the smaller circle in points P and Q. Show that AP = BQ.

Given: Two concentric circles having centre O.
To prove: AP = BQ
Construction: Draw seg OM ⊥ chord AB, A-M-B

Answer

Proof:
For smaller circle,
seg OM ⊥ chord PQ [Construction, A-P-M, M-Q-B]
∴ PM = MQ …..(i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
For bigger circle,
seg OM ⊥ chord AB [Construction]
∴ AM = MB [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
∴ AP + PM = MQ + QB [A-P-M, M-Q-B]
∴ AP + MQ = MQ + QB [From (i)]
∴ AP = BQ

View full question & answer→

Question 73 Marks

Radius of a circle with centre $O$ is $41$ units. Length of a chord PQ is $80$ units, find the distance of the chord from the centre of the circle.
Given: In a circle with centre O ,
$O P$ is radius and PQ is its chord,
seg $OM \perp$ chord PQ , $P - M - Q$
$O P=41$ units, $P Q=80$ units,
To Find: Distance of the chord from the centre of the circle(OM)

Answer

i. $\frac{1}{2} P M=(P Q)$ [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
$=\frac{1}{2}(80)=40$ Units
ii. In $\triangle O M P, \angle O M P=90^{\circ}$
$\therefore OP ^2= OM ^2+ PM ^2$ [Pythagoras theorem]
$\therefore 41^2= OM ^2+40^2$ [From (i)]
$\therefore OM ^2=41^2-40^2$
$=(41-40)(41+40)\left[a^2-b^2=(a-b)(a+b)\right]$
$=(1)(81)$
$\therefore OM ^2=81 OM =\sqrt{ } 81=9$ units [Taking square root on both sides] [From (i)]
$\therefore$ The distance of the chord from the centre of the circle is $9$ units.

View full question & answer→

Question 83 Marks

Radius of a circle is $34\ cm$ and the distance of the chord from the centre is $30\ cm$ , find the length of the chord.
Given: in a circle with centre $A$,
$P A$ is radius and PQ is chord,
seg $AM \perp$ chord PQ , $P - M - Q$
$AP =34 cm, AM =30 cm$
To Find: Length of the chord (PQ)

Answer

$\text { I. In } \triangle A M P, \angle A M P=90^{\circ}$
$\therefore AP^2=AM^2+PM^2[\text { Pythagoras theorem }]$
$34^2=30^2+PM^2$
$\therefore PM^2=34^2-30^2$
$\therefore PM^2(34-30)(34+30)\left[a^2-b^2=(a-b)(a+b)\right]$
$=4 \times 64$
$\therefore PM =\sqrt{4 \times 64}$
(i) [Taking square root on both sides]
$=2 \times 8=16 cm$
ii. Now, $P M=\frac{1}{2}(P Q)$ [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
$16=\frac{1}{2}(P Q)[\text { From }(i)]$
$\therefore P Q=16 \times 2$
$=32 cm$
$\therefore$ The length of the chord of the circle is $32\ cm .$

View full question & answer→

Question 93 Marks

Diameter of a circle is 26 cm and length of a chord of the circle is 24 cm. Find the distance of the chord from the centre.
Given: In a circle with centre O,
PO is radius and PQ is its chord,
seg OR ⊥ chord PQ, P-R-Q
PQ = 24 cm, diameter (d) = 26 cm
To Find: Distance of the chord from the centre (OR)

Answer

Radius $( OP )=\frac{d}{2}=\frac{26}{2}=13 cm$......(i)
$\therefore P R=\frac{1}{2} P Q$ [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
$
=\frac{1}{2} \times 24=12 cm
$
ii. In $\triangle ORP , \angle ORP =90^{\circ}$
$\therefore OP ^2= OR ^2+ PR ^2[$ Pythagoras theorem]
$\therefore 13^2=O R^2+12^2[$ From (i) and (ii)]
$\therefore 169=O R^2+144$
$\therefore O R^2=169-144$
$\therefore O R^2=25$
$\therefore O R=\sqrt{ } 25=5 cm$ [Taking square root on both sides]
$\therefore$ The distance of the chord from the centre of the circle is $5 cm$.

View full question & answer→

Question 103 Marks

Distance of chord $A B$ from the centre of a circle is $8\ cm$ . Length of the chord $A B$ is $12\ cm$. Find the diameter of a circle.
Given: In a circle with centre O ,
$O A$ is radius and $A B$ is its chord,
seg $O P \perp$ chord $A B, A-P-B$
$AB=12 cm, OP=8 cm$
To Find: Diameter of the circle

Answer

i. $A P=\frac{1}{2} A B$ [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
$\therefore A P=\frac{1}{2} \times 12=6 cm \ldots \text { (i) }$
ii. In $\triangle OPA , \angle OPA =90^{\circ}$
$\therefore OA ^2= OP ^2+ AP ^2$ [Pythagoras theorem]
$=8^2+6^2[$ From (i) $]$
$=64+36$
$\therefore OA ^2=100$
$\therefore O A=\sqrt{100}$ [Taking square root on both sides]
$=10 cm$

View full question & answer→

Maths 3 Mark Question

Take a timed test