OP Malhotra Class 10 Maths Solutions Chapter 14 Circle Ex 14(g)

Interactive S Chand Class 10 Maths Solutions ICSE Chapter 14 Circle Ex 14(g) engage students in active learning and exploration.

S Chand Class 10 ICSE Maths Solutions Chapter 14 Circle Ex 14(g)

Question 1.
Using ruler and compasses only construct the tangents to the given circle from the point P. Measure the length of each one of them.

Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Take of point P outside the circle.
(iii) Join OP and take its midpoint M.
(iv) With centre M and diameter OP, draw a circle which intersects the given circle at T and S.
(v) Join PT and PS.
PT and PS are required tangents to the circle on measuring PT = PS = 5.5 cm. (approx)

Question 2.
Draw a tangent to a circle which may be perpendicular to a given line.
Solution:
Steps of construction :
(i) Draw a circle with centre O and with a suitable radius.
(ii) Take a point P on it and join OP.
(iii) At P, draw a perpendicular to OP which meet a given line l at S.
Then ST is the required tangent.

Question 3.
Using ruler and compasses only, draw tangents to a circle of radius 4 cm from a point 5 cm from the centre. What is the length of each of them ?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 4 cm.
(ii) Take a point P such that OP = 5 cm.
(iii) Draw its bisector which bisects OP at M.
(iv) With centre M and radius MP draw a circle intersecting the given circle at T and S.
(v) Join PT and PS.
PT and PS are the required tangents to the circle on measuring each of them PT = PS = 3 cm.

Question 4.
Draw a circle of radius 2 cm and construct a tangent to it from an external point without using the centre.
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 2 cm.
(ii) Take a point P outside the circle.
(iii) From P draw a straight line which intersects the circle at A and B.
(iv) With BP as diameter draw a semicircle.
(v) At A, draw a perpendicular which meets the semicircle at C.
(vi) With centre P and radius PC, draw an arc which intersects the given circle at T and S.
(vii) Join PT.
PT is the required tangent.

Question 5.
Through a given point on the circumference of a circle, draw a tangent to the circle without using its centre.
Solution:
Steps of construction :
(i) Draw a circle with centre O and some suitable radius.
(ii) Take a point P on it.
(iii) Take two more points Q and R on the remaining part of the circle and joined PQ, QR and RP.
(iv) Draw an angle ∠QPT equal to ∠R and produce the line TP to S.
Then SPT is the required tangent to the circle.

Question 6.
Draw two tangents, inclined at an angle of 60° to each other, to a circle with radius of 3 cm. Measure the length of one of the tangents and verify by calculation.
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw a radius OS and on PS, draw an angle ∠SOT of 180°- 60°= 120°.
(iii) At S and T, draw lines making 90° each. Which intersect each other at P.
Then PT and SP are the required tangents making an angle of 60° with each other at P on measuring them each one of them is 4.5 cm.

Calculation :
Here radius of the circle (r) = 3 cm
and distance of OP = 5.3 cm
In right △OPT,
PT² = OP² – OT² = (5.3)² – (3)²
= 28.09 – 9 = 19.09 = (4.37)²
∴ PT = 4.37 cm

Question 7.
Using ruler and compasses only construct a triangle ABC having given c = 6 cm, b = 7 cm and ∠C = 30°. Measure side a. Draw carefully the circumcircle of the triangle. Measure its radius. (Two triangles are possible).
Solution:
Steps of construction :
(i) Draw a line segment AC = 7 cm (b = 7 cm)
(ii) At C, draw a ray CX making an angle of 30°.
(iii) With centre A and radius 6 cm (c = 6 cm), draw arc which intersects CX at B and B’.
(iv) JoinAB andAB’.
Then two triangles are possible, △ABC and △AB’C in which a = 1.3 cm or 11.3 cm.
(v) Now draw the perpendicular bisectors of AC and BC which intersect each other at O.
(vi) With centre O and radius equal to OB, draw a circle which passes through A, B and C. Then this is the required circumcircle of the △ABC.
On measuring its radius = 6.5 cm (approx).

Question 8.
Using ruler and compasses only draw a triangle ABC such that AB = 4 cm, BC = 6 cm and ∠B = 90°. Draw the circum-scribed circle of the triangle and state its radius.
Solution:
Solution—
Steps of construction :
(i) Draw a line segment BC = 6 cm.
(ii) At B, draw a ray BX making an angle of 90° and cut off BA = 4 cm.
(iii) Join AC.
(iv) Now draw the perpendicular bisectors of AB and BC intersecting each other at O.
(v) With centre O and radius OA, draw a circle which will pass through A, B and C.
This is required circumcircle of △ABC whose radius is \(\frac { 1 }{ 2 }\)AC = 3.6 cm.

Question 9.
Using ruler and compasses only construct a triangle ABC in which BC = 4 cm. ∠ACB = 45° and the perp. from A on BC is 2.5 cm. Draw a circle circumscribing a triangle ABC and measure its radius.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4 cm.
(ii) At C, draw a ray CX making an angle of 45° and CY making an angle of 90°.
(iii) Cut off CQ = 2.5 cm.
(iv) From Q, draw a line PQ parallel to BC. Which meets CX at A.
(v) JoinAB.
(vi) Draw the perpendicular bisector of AB and BC which intersect each other at O.
(vii) With centre O and radius OA, draw a circle which will pass through A, B and C.
This is the required circumcircle of △ABC whose radius OA = 2.1 cm (approx).

Question 10.
Answer true or false:
The centre of the circumcircle of a right-angled triangle is the mid-point of its hypotenuse.
Solution:
It is true.

Question 11.
Draw an equilateral triangle of side 4 cm. Draw the circumcircle of it.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4 cm.
(ii) With centres B and C and radius 4 cm, draw arcs which intersect each other at A.
(iii) Join AB and AC.
Then △ABC is an equilateral triangle.

(iv) Draw the perpendicular bisectors of sides AB and BC which intersect each other at O.
(v) With centre O and radius OA, draw a circle which will pass through A, B, C.

Question 12.
Using ruler and compasses only inscribe a circle in the given triangle and measure its radius.

Solution:
Steps of construction :
(i) Draw the given △ABC.
(ii) Draw the angle bisectors of ∠B and ∠C which intersect each other at I.
(iii) From I, draw a perpendicular ID on BC.
(iv) With centre I and radius ID, draw a circle which touches the sides of the triangle ABC at D, E and F.
On measuring the radius ID = 1.4 cm (approx).

Question 13.
Using ruler and compasses only, draw an equilateral triangle of side 5 cm and draw its inscribed circle. Measure the radius of the circle.
Solution:
Steps of construction :
(i) Draw a line segment BC = 5 cm.
(ii) With centres B and C and radius 5 cm, draw arcs intersecting each other at A.
(iii) Join AB and AC.
△ABC is an equilateral triangle.
(iv) Draw the angle bisectors of ∠B and ∠C which intersect each other at 1.
(v) From I, draw ID ⊥ BC.
(VJ) With centre I and radius ID, draw a circle which will touch the sides of △ABC at D, E and F.
On measuring, the radius ID = 1.5 cm (approx).

Question 14.
(a) Using ruler and compasses only, construct a triangle ABC in which AB = 8 cm, BC = 6 cm and CA = 5 cm.
(b) Find its in-centre and mark it I.
(c) With I as centre, draw a circle which will cut off 2 cm chords from each side of the triangle. What is the length of the radius of this circle ?
Solution:
Steps of construction :
(a) (i) Draw a line segment AB = 8 cm.
(ii) With centre A and radius 5 cm and with centre B and radius 6 cm, draw arcs which intersect each other at C.
(iii) Join AC and BC.
△ABC is the required triangle.
(b) (i) Draw the bisector of ∠A and ∠B which intersect each other at I. I is the incentre of the in circle.
(c) (i) From I, draw IL ⊥ AB.
(ii) From L, cut off LP = LQ = 1 cm so that PQ = 2 cm.
(iii) With centre I and radius IP, draw a circle which intersects BC at R and S and CA at T and U.
Then chords PQ = RS = TU = 2 cm.

Question 15.
Inscribe a circle in a regular hexagon of side 3.2 cm.
Solution:
Steps of construction :
(i) Draw a line segment AB = 3.2 cm.
(ii) Draw rays at A and B making angle of 120° each and cut off AF = BC = 3.2 cm.
(iii) Similarly at F and C, draw rays making angle of 120° each and cut off FE = CD = 3.2 cm.

(iv) Join ED.
ABCDEF is a regular hexagon.
(v) Draw the angle bisector of ∠A and ∠B which intersect each other at O.
(vi) From O, draw OL ⊥ AB.
(vii) With centre O and radius OL draw a circle which will touch the sides of regular hexagon ABCDEF.
This is the required incircle of the hexagon.

Question 16.
Draw a regular hexagon of side 2.8 cm and circumscribe a circle on it.
Solution:
Steps of construction :
(i) Draw a line segment AB = 2.8 cm.
(ii) At A and B draw rays making angle of 120° each and cut off AF = BC = 2.8 cm.
(iii) Similarly at F and C, draw rays making angle of 120° and cut off FE = CD = 2.8 cm.
(iv) Join ED.
ABCDEF is a regular hexagon.
(v) Draw the perpendicular bisector of AB and AF which intersect each other at O.
(vi) With centre O and radius OA, draw a circle which passes through the vertices of hexagon ABCDEF. This is the required circumcircle of regular hexagon.

Self Evaluation And Revision (LATEST ICSE QUESTIONS)

Question 1.
(a) In the figure, chords AB and CD when extended meet at X. Given, AB = 4 cm, BX = 6 cm, XD = 5 cm. Calculate the length of CD.

(b) In the given figure, AB is a common tangent to two circles intersecting at C and D. Write down the measure of (∠ACB + ∠ADB). Justify your answer.

Solution:
(a) In a circle two chords AB and CD meet at X outside the circle. AB = 4 cm, BX = 6 cm, then AX = AB + BX = 4 + 6 = 10 cm and XD = 5 cm
∴ XA × XB = XC × XD
⇒ 10 × 6 = XC × 5 ⇒ XC = \(\frac{10 \times 6}{5}\) = 12 cm
∴ CD = CX – DX = 12 – 5 = 7 cm

(b) Two circles intersect each other at C and D, AB is their common tangents AD, BD, AC, BC are joined
Join CD, AB is tangent and AC is chord at A
∴ ∠BAC = ∠ADC …(i)
Similarly AB is tangent and $B C$ is chord

∴ ∠ABC = ∠CDB ….(ii)
Adding (i) and (ii)
∠BAC + ∠ABC = ADC + ∠CDB = ∠ADB …(iii)

But in △ABC,
∠ACB =180°- (∠BAC + ∠ABC)
∠BAC + ∠ABC =180° – ∠ACB ….(iv) From (iii) and (iv)
180° – ∠ACB = ∠ADB
⇒ ∠ACB + ∠ADB = 180°
Hence ∠ACB + ∠ADB = 180°

2. (a) A, B and C are the points on a circle. The tangent at C meets BA produced at T. Given that ∠ATC = 36° and ∠ACT = 48°, calculate the angle subtended by AB at the centre of the circle.
(b) In the given figure, find TP is AT = 16 cm and AB = 12 cm.

Solution:
(a) Given : In a circle with centre O
A, B, C are points on the circle CT is the tangent at C which meets BA produced at T ∠ATC = 36° and ∠ACT = 48°

In △ACT,
Ext. ∠BAC = ∠ACB + ∠ATC = 48° + 36° = 84°
∵ CT is tangent and CA is chord of the circle
∴ ∠ABC = ∠ACT = 48°
But ∠BCA + ∠ABC + ∠BAC = 180° (Sum of angles of a triangle)
⇒ ∠BCA + 48° + 84° = 180°
⇒ ∠BCA+ 132°= 180°
⇒ ∠BCA = 180° – 132°
⇒ ∠BCA = 48°
Now arc AB subtends ∠AOB at the centre and ∠BCA at the remaining part of the circle
∴ ∠AOB = 2 ∠BCA = 2 × 48° = 96°

(b) In the figure, PT is the tangent and ABT is a secant AT = 16 cm, AB = 12 cm

∴ BT = AT – AB = 16 – 12 = 4 cm
Now PT² = TA × TB = 16 × 4 = 64 = (8)²
∴ PT or TP = 8 cm

Question 3.
(a) In the given figure, AB is the diameter of a circle with centre O, ∠BCD = 120°. Find : (i) ∠DBA (ii) ∠BAD

(b) In the given circle with diameter AB, find the value of x.

Solution:
(a) In the figure, ∠ADB = 90° (Angle in a semicircle)
(i) ∵ ABCD is a cyclic quadrilateral
∴ ∠DBA + ∠BCD = 180° (Sum of opposite angles)
⇒ ∠DBA + 120°= 180°
⇒ ∠DBA = 180°- 120°
∠DBA = 60°

(ii) In △ADB,
∠ADB = 90°
∴ ∠BAD + ∠DBA = 90°
⇒ ∠BAD + 60° = 90°
⇒ ∠BAD = 90°- 60° = 30°
Hence ∠BAD = 30°

(b) In the figure, AB is the diameter of the circle ∠ACD = 30°
∠ABD = ∠ACD = 30° (Angles in the same segment)
Now in △ADB,
∠ADB = 90° (Angle in a semicircle)
∴ ∠BAD + ∠ABD = 90°
⇒ x + 30° = 90O ⇒ x = 90° – 30° = 60°
Hence x = 60°

Question 4.
In the figure, AB is a diameter and AC is a chord of a circle such that ∠BAC = 30°. The tangent at C intersects AB produced at D. Prove that BC = BD.

Solution:
Given : In the figure, AB is the diameter and AC is a chord of the circle such that ∠BAC = 30°
Tangent at C, meets AB produced at D
To prove : BC = BD
Proof : CD is tangent, CB is chord of the circle
∴ ∠BCD = ∠BAC = 30° (Angle in the alternate segment)
In △ABC, ∠ACB = 90° (Angle in a semicircle)
∴ ∠BAC + ∠ABC = 90°
⇒ ∠ABC = 90° – ∠BAC = 90° – 30° = 60°
But Ext. ABC = ∠BCD + ∠BDC
⇒ 60° = 30° + ∠BDC
⇒ ∠BDC = 60°- 30° = 30°
∴ ∠BCD = ∠BDC = 30°
∴ BC = BD (Sides opposite to equal angles)
Hence proved.

Question 5.
In the given figure, O is the centre of the circle an ∠AOC = 160°. Prove that 3∠y – 2∠x = 140°.

Solution:
Given : In the figure, O is the centre of the circle. ARCP is the cyclic quadrilateral in which ∠AOC = 160°, ∠P = y, ∠R = x
To prove : 3 ∠y – 2∠x = 140°
Proof: ∵∠AOC = 160°
∴ Reflex ∠AOC = 360° – 160° = 200°
Now, arc APC subtends ∠AOC at the centre and ∠ARC at the remaining part of the circle
∴ ∠AOC = 2 ∠ARC = 2x
⇒ 2x = 160° = x = \(\frac{160^{\circ}}{2}\) = 80°
Similarly reflex ∠AOC = 2 ∠APC = 2y
⇒ 200° = 2y ⇒ y = \(\frac{200^{\circ}}{2}\) = 100°
Now 3∠y – 2∠x = 3 × 100 – 2 × 80°
= 300 – 160°= 140°
Hence proved

Question 6.
(a) In the figure, PM is a tangent to the circle and PA = AM, prove that:
(i) △PMB is isosceles (ii) PA.PB = MB².

(b) A circle with centre O, diameter AB and a chord AD is drawn. Another circle is drawn with OA as diameter to cut AD at C. Prove that BD = 20C.
Solution:
(a) Given : In the figure, PM is tangent PA = AM
To prove : (i) △PMB is an isosceles
(ii) PA.PB = MB²
Construction : Join BM.

Proof: In △PAM, PA = AM (given)
∴ ∠1 = ∠2 ….(i) (angles opposite to equal sides)
PM is tangent and MA is chord of the circle
∴ ∠PMA = ∠MBA (Angle in the alternate segment)
⇒ ∠2 = ∠3 ….(I)
From (i) and (ii)
∠1 = ∠3
∴ MB = PM
∴ APMB is an isosceles

(b) Given : A circle with centre O and AB is its diameter, AD is chord
Another circle with AO as diameter is drawn which intersects AD at C
OC and BD are joined
To prove : BD = 2 OC
Construction : Draw a common tangent to the circles at A.

Proof: ∵ TS is tangent and ACD is chord
∴ ∠DAS = ∠AOC = ∠ABD (Angles in the alternate segment)
Now in △AOC and △ABD,
∠A = ∠A (common)
∠AOC = ∠ABD (Proved)
∴ △AOC – △ABD (AA axiom)

Question 7.
PQR is a right angled triangle with PQ = 3 cm and QR = 4 cm. A circle which touches all the sides of the triangle is inscribed in a triangle. Calculate the radius of the circle.
Solution:
In △PQR, ∠Q = 90° and a circle is inscribed in it which touches its sides PQ, QR and RP
at A, B and C respectively
PQ = 3 cm and QR = 4 cm

∴ PR² = PQ² + QR² (Pythagoras theorem)
= (3)² + (4)² = 9 + 16 = 25 = (5)²
∴ PR = 5 cm
Join OA = OB
Let OA = OB – r (radii of the circle)
∴ OAQB is a square
∴ QB = QA = r
∴ RB and RC are the tangents to the circle
∴ RB = RC = 4 – r
Similarly PA and PC are the tangents to the circle.
∴ PC = PA = 3 – r
But PC + RC = PR
⇒ 4 – r + 3 — r = 5 ⇒ 7 – 2r = 5
⇒ 2r = 7 – 5 = 2
∴ r = \(\frac { 2 }{ 2 }\) = 1
∴ Radius of the circle = 1 cm.

Question 8.
In the given figure, ∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°. Find :
(i) ∠BCD
(ii) ∠ADB. Hence show that AC is a diameter.

Solution:
In the figure, ABCD is a cyclic quadrilateral AC and BD are its diagonals
∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°

(i) ∵ ABCD is a cyclic quadrilateral
∴ ∠BAD + ∠BCD = 180° (Sum of opposite angles)
⇒ 65° + ∠BCD = 180°
⇒ ∠BCD = 180° – 65° =115°

(ii) In △ADB,
∠ADB + ∠DAB + ∠ABD = 180° (Sum of angles of a triangle)
⇒ ∠ADB + 65° + 70° = 180°
⇒ ∠ADB+ 135°= 180°
⇒ ∠ADB = 180°- 135°
⇒ ∠ADB = 45°

(iii) Now ∠ADC = ∠ADB + ∠BDC
= 45° + 45° = 90°
∴ But it is an angle of a semicircle
∴ AC is diameter.

Question 9.
(a) In the given figure, AB is a diameter. The tangent at C meets AB produced at Q. If ∠CAB = 34°, find
(i) ∠CBA (ii) ∠CQA

(b) In the given figure, PT touches a circle with centre O at R. Diameter SQ when produced meet PT at P. ∠SPR = x° and ∠QRP =f, show that x° + 2y° = 90°.

Solution:
(a) In the figure, AB is the diameter of circle Tangent at C, meets AB produced at Q

(i) ∠CAB = 34°
∵ ∠ACB = 90° (Angle in a semicircle)
∠CAB + ∠CBA = 90°
⇒ ∠CBA = 90°- ∠CAB
⇒ ∠CBA = 90° – 34° = 56°

(ii) CQ is tangent and CB is chord of the circle
∴ ∠BCQ = ∠CAB = 34°
In △BCQ,
Ext. ∠CBA = ∠BCQ + ∠CQA
⇒ 56° = 34° + ∠CQA
⇒ ∠CQA = 56° – 34°
⇒ ∠CQA = 22°

(b) Given : In a circle with centre O, PT touches the circle at R
Diameter SQ when produced meets PT at P
If ∠SPR = x° and ∠QRP = y°

To prove : x° + 2y° = 90
Proof : PT is tangent and QR is chord of the circle
∴ ∠QSR = ∠QRP = y
In △PQR,
Ext. ∠RQS = ∠SPR + ∠QRP = x + y
∠QRS = 90° (Angle in a semicircle)
∴ ∠RQS + ∠RSQ = 90°
⇒ (x° + y°) + = 90°
⇒ x° + 2y° = 90°
Hence x° + 2y° = 90°
Hence proved.

Question 10.
(i) In the given figure, O is the centre of the circle and ∠PBA = 45°.
Calculate the value of ∠PQB.

(ii) In the given figure, if ∠ACE = 43° and ∠CAF = 62°, find the values of a, b and c.

Solution:
(i) In the figure, AB is the diameter ∠PBA = 45° In △APB,
∠APB = 90° (angle in a semicircle)
∠PAB + ∠PBA = 90°
⇒ ∠PAB + 45° = 90°
⇒ ∠PAB = 90°- 45° = 45°
But ∠PAB = ∠PQB (Angles in the same segment)
∴ ∠PQB = 45°

(ii) In the figure AEDB is a cyclic quadrilateral whose sides AE and BD are produced to meet at F and sides AB and ED are produced to meet at C
∠CAF = 62°, ∠ACE = 43°
In △ACE,
∠CAE + ∠ACE + ∠AEC = 180° (Sum of angles of a triangle)
⇒ 62° + b + 43° + ∠AEC = 180°
⇒ 105° + ∠AEC = 180°
⇒ ∠AEC = 180° – 105° = 75°
⇒ ∠AED = 75° But in cyclic quad. AEDB,
∠ABD + ∠AED = 180° (Sum of opposite angles)
⇒ a + 75° = 180° = a – 180° – 75°= 105°
In △ABF,
∠BAF + AFB + ∠ABF = 180°
⇒ 62° + b + a = 180°
⇒ 62° + 105° = 180°
⇒ 167° + b = 180°
⇒ b= 180°- 167 = 13°
∴ In cyclic quad. AEDB,
Ext. ∠EDF = Int. opp. ∠BAF
⇒ c = 62°
Hence a = 105°, b = 13°, c = 62°

Question 11.
Using a ruler, construct a triangle ABC with BC = 6.4 cm, C A = 5.8 cm and ZABC = 60°. Draw its incircle. Measure and record the radius of the incircle.
Solution:
(i) Draw a line segment BC = 6.4 cm.
(ii) At B draw a ray BX making an angle of 60°.
(iii) With centre C and radius 5.8 cm, draw an arc which intersect BX at A and A’.
(iv) Join AC and AC’.
Then △ABC and △A’BC are two possible triangle.
(v) Draw the angle bisectors of ∠B and ∠C which intersect each other at I.
(vi) From I, draw IL ⊥ BC.
(vii) With centre I and radius IL, draw a circle which touches the sides BC, CA and AB at L, as and N respectively.
This is the required incircle of AABC.
On measuring its radius IL, it is 1.8 cm (approx).

Question 12.
In this figure, AB is parallel to DC. ∠BCE = 80° and ∠BAC = 25°. Find : (i) ∠CAD (ii) ∠CBD (iii) ∠ADC

Solution:

In the figure, ABCD is a cyclic quadrilateral in which AB || DC, ∠BCE = 80° and ∠BAC = 25°
In cyclic quad. ABCD,
(i) Ext. ∠BCE = Int. opp. ∠BAD
⇒ 80° = ∠CAD + BAC
⇒ 80° = ∠CAD + 25°
⇒ ∠CAD = 80° – 25°
In right angled △AMO
(OA)² = (AM)² + (OM)²
(15)² = (12)² + (y)²
y² = (15)² – (12)²
y² = 225 – 144
y² ⇒ 81 y = 9 cm
In right angled △CON
(OC)² = (ON)² + (CN)²
(15)² = x² + (9)²
⇒ x² = 225 – 81
⇒ x² = 144 => x = 12 cm
Now, MN = MO + ON = y + x
= 9 cm + 12 cm = 21 cm
∠CAD = 55°

(ii) ∠CBD = ∠CAD (Angles in the same segment)
∴ ∠CBD = 55°

(iii) ∵ AB || DC
∴ ∠BAD + ∠ADC = 180° (co-interior angles)
⇒ ∠BAC + ∠CAD + ∠ADC = 180°
⇒ 25° + 55° + ∠ADC = 180°
⇒ 80° + ∠ADC = 180°
⇒ ∠ADC = 180°- 80°= 100°
∴ ∠ADC =100°

Question 13.
In the figure given, here PQ = QR, ∠RQP = 68°. PC and CQ are tangents to the circle with centre O. Calculate the value of
(i) ∠QOP (ii) ∠QCP

Solution:
In the figure, a circle with centre O Chord PQ = QR, ∠RQP = 68°
PC and QC are the tangents meeting each other at C
In △PQR, PQ = QR (given)
∴ ∠QRP = ∠QPR (angles opposite to equal sides)
But ∠QRP + ∠QPR + ∠PQR = 180°
⇒ ∠QRP + ∠QRP + 68° = 180°
⇒ 2 ∠QRP = 180° – 68°
⇒ ∠QRP = \(\frac{180^{\circ}-68^{\circ}}{2}\) = \(\frac{112^{\circ}}{2}\) = 56°

(i) Arc PQ subtends ∠POQ at the centre and ∠QRP at the remaining part of the circle
∴ ∠POQ = 2∠QRP = 2 × 56° = 112°

(ii) ∵ PC and QC are tangents and OP, OQ are radii of the circle
∴ ∠POQ + QCP = 180°
⇒ 112° + ∠QCP = 180°
⇒ ∠QCP = 180°- 112° = 68°

Question 14.
In the given figure, AE and BC intersect each other at point D. If ∠CDE = 90°, AB = 5 cm, BD = 4 cm and CD = 9 cm, find DE.

Solution:
In the figure, in a circle two chords AE and BC intersect each other at D inside the circle AB is joined.

∠CDE = 90°
AB = 5 cm, BD = 4 cm, CD = 9 cm
In right △ABD,
AB² = AD² + BD² ⇒ (5)² = AD² + (4)²
⇒ 25 = AD² + 16 ⇒ AD² = 25 – 16 = 9 = (3)²
∴ AD = 3 cm
∵ AE and BC intersect each other at D
∴ AD × DE = BD × CD
3 × DE = 4 × 9 ⇒ DE = \(\frac{4 \times 9}{3}\) = 12
∴ DE = 12 cm

Question 15.
Using a ruler and a pair of compasses only, construct
(i) a triangle ABC, given AB = 4 cm, BC = 6 cm and ∠ABC = 90°.
(ii) a circle which passes through the points A, B and C and mark its centre O.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm.
(ii) At B, draw a ray BX making an angle of 90° and cut off BA = 4 cm.
(iii) Join AC.
△ABC is the required triangle.
(iv) Draw the perpendicular bisectors of AB and BC which intersect each other at O.
(v) With centre O and radius OA, draw a circle which passes through A, B and C.
This is the required circle.

Question 16.
(a) In the given figure, O is the centre of the circle, ∠BAD = 75° and chord BC = chord CD. Find :
(i) ∠BOC, (ii) ∠OBD, (iii) ∠BCD.

(b) In the figure, AB = 7 cm and BC = 9 cm
(i) Prove △ACD ~ △DCB
(ii) Find the length of CD.

Solution:
(a) (i) ∠BOD = 2 . ∠BAD = 2 × 75° = 150°
∠BOC = ∠COD

∵ BC = CD (given)
∴ ∠BOD = 2 ∠BOC
∴ ∠BOC = \(\frac { 1 }{ 2 }\) ∠BOD = \(\frac { 1 }{ 2 }\) × 150° = 75°

(ii) ∠OBD = \(\frac { 1 }{ 2 }\) (180° – ∠BOD)
= \(\frac { 1 }{ 2 }\)(180°- 150°)= \(\frac { 1 }{ 2 }\) (30°) = 15°

(iii) ∠BCD = 180° – ∠BAD
= 180° – 75° = 105°

(b) In △ACD and △DCB
∠C = ∠C (common)
∠CAD = ∠CDB
[Angle between chord and tangent is equal to angle made by chord in alternate segment.]
∴ △ACD ~ △DCB (AA axiom)
∴ \(\frac{\mathrm{AC}}{\mathrm{DC}}\) = \(\frac{\mathrm{DC}}{\mathrm{BC}}\)
⇒DC² = AC × BC = (AB + BC) × BC
= (7 + 9) × 9 = 16 × 9 = 144 :
⇒ DC = 12 cm

Question 17.
Using ruler and compasses construct :
(i) a triangle ABC in which AB = 5.5 cm, BC = 3.4 cm and CA = 4.9 cm.
(ii) the locus of points equidistant from A and C.
(iii) A circle touching AB at A and passing through C.
Solution:
Steps of construcltion :
(i) Draw BC = 3.4 and mark the arcs of 5.5 and 4.9 cm from B and C, which intersect each other at A Join A, B and C.
ABC is the required triangle.
(ii) Draw ⊥ bisector of AC.
(iii) Draw an angle of 90° at AB at A which intersects ⊥ bisector at O. Draw circle taking O as centre and OA as radius.

Question 18.
In this following figure O is the centre of the circle and AB is a tangent to it at point B. ∠BDC = 65°, Find ∠BAO.

Solution:
Given ∠BDC = 65° and AB is tangent to circle with centre O.
∵ OB is radius
⇒ OB ⊥ AB
In △BDC
∠DBC + ∠BDC + ∠BCD = 180°
90° + 65° + ∠BCD = 180°

⇒ ∠BCD = (180°- 155°) = 25°
∵ OE = OC = radius
⇒ ∠OEC = ∠OCE ⇒ ∠OEC = 25°
Also, ∠BOE = ∠OEC + ∠OCE [Exterior angle = sum of opposite interior angles in a △]
⇒ ∠BOE = 25° + 25°
⇒ ∠BOE = 50°
⇒ ∠BOA = 50°
In △AOB
∠AOB + ∠BAO + ∠OBA = 180°
50° + ∠BAO + 90° = 180° ⇒ ∠BAO = 40°

Question 19.
Construct a regular hexagon of side 4 cm. Construct a circle circumscribing the hexagon.
Solution:
Steps of construction :
(i) Draw a line segment AB = 4cm
(ii) At A and B draw rays making an angle of 120° each and cut off AF = BC = 4cm.
(iii) At F and C, draw rays making angle of 120° each and cut off EF = CD = 4cm.
(iv) join ED.
ABCDEF is the required hexagon.
(v) Draw perpendicular bisectors of sides AB and BC intersecting each other at O.
(vi) With centre O and radius equal OA or OB draw a circle which passes through the vertices of the hexagon.
This is the required circumcircle of hexagon ABCDEF.

Question 20.
In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If ∠ACO = 30°, find
(i) ∠BCO (ii) ∠AOB (iii) ∠APB

Solution:
(i) ∠BCO = ∠ACO = 30°
(∵ C is the intersecting point of tangent AC and BC)

(ii) ∠OAC = ∠OBC = 90°
∴ ∠AOC = ∠BOC =180° – (90° + 30°) = 60° (∵ sum of the three angles of a △ is 180°)
∴ ∠AOB = ∠AOC + ∠BOC = 60° + 60°= 120°

(iii) ∠APB = \(\frac { 1 }{ 2 }\) ∠AOB = \(\frac{120^{\circ}}{2}\) = 60°
[∵ angle subtended at the remaining part of the circle is half of the subtended at the centre]

Question 21.
ABC is triangle with AB = 10 cm, BC = 8 cm and AC = 6 cm (not drawn to scale). Three circles are drawn touching each other with the vertices as their centres. Find the radii of the three circles.

Solution:
AB = 10 cm, BC = 8 cm, AC = 6 cm
BP + PA = 10 cm ….(i)
BC = 8 cm (given)
BR + RC = 8 cm
BP + RC = 8 …(ii)
Again, AC = 6 cm
AO + OC = 6 cm
PA + RC = 6 cm ….(iii)
Adding (i), (ii) and (iii) we get,
2 (BP + PA + RC) = (10 + 8 + 6) = 24 cm
∴ BP + PA + RC = 12 cm ….(iv)
Subtracting (i) from (iv) we get,
RC = 12 – 10 = 2 cm
Similarly, (iv) – (ii) we get,
PA = 12 – 8 = 4 cm
and (iv) – (iii) we get,
BP = 12 – 6 = 6 cm
∴ Radii : BP – 6 cm, PA = 4 cm, RC = 2 cm

Question 22.
Draw a circle of radius 3.5 cm. Mark a point P outside the circle at a distance of 6 cm from the centre, Construct two tangents from P to the given circle. Measure and write down the length of one tangent.
Solution:
Steps of construction :
(i) Draw a line segment OP = 6 cm
(ii) With centre O and radius 3.5 cm, draw a circle
(iii) Draw the mid point M of OP.
(iv) With centre M and diameter OP, draw a circle which intersect the circle at T and S.
(v) Join PT and PS.
PT and PS are the required tangent on measuring the length of PT = PS = 4.8 cm

Question 23.
(a) In the given figure O is the centre of the circle and AB is a tangent at B. If AB = 15 cm and AC = 7.5 cm, calculate the radius of the circle.

(b) In the given figure, AB is the diameter of a circle with centre O.
∠BCD = 130°. Find :
(i) ∠DAB (ii) ∠DBA

(c) In the △PQR, PQ = 24 cm, QR = 7 cm and ∠PQR = 90°. Find the radius of the inscribed circle.

Solution:
(a) Join OB
∠OBA = 90°
(Radius through the point of contact is perpendicular to the tangent)

In △OBA (By Pythagoras Theorem)
OA² = OB² + AB²
⇒ OB² – OA² – AB²
⇒ r² = (r + 7.5)² – 15²
⇒ r² = r² + 56.25 + 15r – 225
⇒ 15r = 168.75 => r = 11.25
Hence, radius of the circle = 11.25 cm

(b) JoinDB

(i) ∠DAB + ∠DCB= 180°
[Opposite angles of a cyclic quadrilateral are supplementary]
∠DAB = 180° – 130° = 50°

(ii) In △ADB, ∠ADB = 90° [Angle in a semi-circle is 90°]
So, by angle sum property
∠DBA = 180° – (∠DAB + ∠ADB)
= 180°- (50°+ 90°) = 40°

(c) Let the circle touch the sides PQ, QR and PR at A, B and C respectively.
We know that the tangents drawn from an exterior point to a circle are equal.
So, PA = PC …(i)
QR = QB …(ii)
RB = RC …(iii)

Also, ∠OAQ = ∠OBQ = 90°
(Radius through the point of contact is perpendicular to the tangent)
Also, ∠Q = 90°
So, OAQB is a square
In △PQR (By Pythagoras Theorem)
PR² = PQ² + QR² = 24² + 7²
= 576 + 49 = 625 ⇒ PR = 25 cm
Now, PA = PC
⇒ 24 – x = 25 – CR
⇒ 24 – x = 25 – RB
⇒ 24 – x = 25 – 7 + x
⇒ 6 = 2x ⇒ x = 3
Hence, radius of the inscribed circle = 3 cm

Question 24.
(a) In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°
(i) Prove that AC is a diameter of the circle.
(ii) Find ∠ACB.

(b) In the given circle with centre O, ∠ABC = 100°, ∠ACD = 40°
and CT is a tangent to the circle at C. Find ∠ADC and ∠DCT.

Solution:
(a) (i) In △ABD
65° + 70° + ∠ADB = 180°
∠ADB =180°- 65° – 70° = 45°
Also, ∠ADC = ∠ADB + ∠BDC
∴ ∠ADC = 45° + 45° = 90°
⇒ AC is diameter [∵ Angle in semi circle is 90°]

(ii) ∠ACB = ∠ADB = 45° [angle in same segment]
(b) ABCD is a cyclic quadrilateral
∴ ∠ABC + ∠ADC = 180°
100 + ∠ADC = 180°
∠ADC = 180 – 100 = 80°
In △ADC [By angle sum property]
∠ACD + ∠CDA + ∠DAC = 180°
40 + 80° + ∠DAC = 180°
∠DAC = 180 – 80 – 40 = 60°
Now ∠DAC = 60°
⇒ ∠DCT = 60° [angle is alt. segment]

Question 25.
(a) In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate:
(i) ∠BDC
(ii) ∠BEC
(iii) ∠BAC

(b) In the given figure, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find:
(i) AB.
(ii) the length of tangent PT.

Solution:
(a) ∠DBC = 58°
BD is diameter
∴ ∠DCB = 90° (Angle in semi circle)

(i) In △BDC
∠BDC + ∠DCB + ∠CBD = 180° (By angle sum property)
∠BDC = 180° – 90° – 58° = 32°

(ii) ∠BEC = 180°- 32° (opp. angle of cyclic quadrilateral)
= 148°

(iii) ∠BAC = ∠BDC = 32° (Angles in same segment)

(b) PT is tangent and PDC is secant out to the circle

∴ PT² = PC × PD
PT² = (5 + 7.8) × 5 = 12.8 × 5
PT² = 64 ⇒ PT = 8 cm
In △OTP (By Pythagoras Theorem)
PT² + OT² = OP²
8² + x² = (x + 4)² ⇒ 64 + x² = x² + 16 + 8x
64 – 16 = 8x ⇒ 48 = 8x
x = \(\frac { 48 }{ 8 }\) = 6 cm
∴ Radius = 6 cm
AB = 2 × 6 = 12 cm

Question 26.
(a) In the figure given below, O is the centre of the circle and SP is a tangent. If ∠SRT = 65°, find the value of x,y and z.

(b) AB and CD are two chords of a circle intersecting at P. Prove that AP × PB = CP × PD.

(c) Construct a regular hexagon of side 5 cm. Construct a circle circumscribing the hexagon. All traces of construction must be clearly shown.
Solution:
(a) Consider the following figure:
TS ⊥ SP,
∠TSR = ∠OSP = 90°
In △TSR (By angle sum property)
∠TSR + ∠TRS + ∠RTS = 180°
⇒ 90° + 65° + x = 180°
⇒ x= 180° – 90°- 65°
⇒ x = 25°
⇒ y = 2x
[Angle subtended at the centre is double that of the angle subtended by the arc at the same centre]
⇒ y = 2 × 25°
y = 50°
In △OSP,
∠OSP + ∠SPO + ∠POS = 180°
⇒ 90° + z + 50° = 180°
⇒ z = 180° – 140°
∴ z = 40°
Hence x = 25°, y = 50° and z = 40°

(b) Construction : Join AD and CB.

In △APD and △CPB
∠A = ∠C (Angles in the same segment)
∠D = ∠B (Angles in the same segment)
⇒ △APD ~ △CPB
⇒ \(\frac{\mathrm{AP}}{\mathrm{CP}}\) = \(\frac{\mathrm{PD}}{\mathrm{PB}}\)
(Corresponding sides of similar triangles are proportinal)
⇒ AP × PB = CP × PD

(c) Each side of the regular hexagon = 5 cm Each interior angle of the regular hexagon

Steps of construction:
(i) Using the given data, construct the regular hexagon ABCDEF with each side equal to 5 cm.
(ii) Draw the perpendicular bisectors of sides AB and BC and make them intersect each other at point O.
(iii) With O as the centre and OA as the radius draw a circle which will pass through all the vertices of the regular hexagon ABCDEF.

Question 27.
(a) In the figure given below, AD is a diameter. O is the centre of the circle. AD is parallel to BC and ∠CBD = 32°. Find :
(i) ∠OBD
(ii) ∠AOB
(iii) ∠BED

(b) In the given figure PQRS is a cyclic quadrilateral PQ and SR produced meet at T.

(ii) Find SP if TP = 18 cm, RQ = 4 cm and TR = 6 cm.
(iii) Find area of quadrilateral PQRS if area of △PTS = 27 cm².
Solution:
(a) (i) AD is parallel to BC, that is, OD is parallel to BC and BD is transversal.
∴ ∠ODB = ∠CBD = 32°(Altemate angles)
In △OBD,
OD = OB (Radii of the same circle)
⇒ ∠ODB = ∠OBD = 32°

(ii) AD is parallel to BC, that is, AO is parallel to BC and OB is transversal.
∴ ∠AOB = ∠OBC (Alternate angles)
∠OBC = ∠OBD + ∠DBC
⇒ ∠OBC = 32° + 32°
⇒ ∠OBC = 64°
∴ ∠AOB = 64°

(iii) In △OAB,
OA = OB (Radii of the same circle)
∴ ∠OAB = ∠OBA = x (say)
∠OAB + ∠OBA + ∠AOB = 180°
⇒ x + x + 64° = 180°
⇒ 2x = 180°- 64°
⇒ 2x = 116°
⇒ x = 58°
∴ ∠OAB = 58°
That is ∠DAB = 58°
∴∠DAB = ∠BED = 58° (Angles inscribed in the same arc are equal)

(b) (i) Since PQRS is a cyclic quadrilateral ∠RSP + ∠RQP = 180° (Since sum of the opposite angles of a cyclic quadrilateral is 180°)
⇒ ∠RQP = 180° – ∠RSP …(i)
∠RQT + ∠RQP = 180° (Since angles from a linear pair)
⇒ ∠RQP = 180° – ∠RQT …(ii)
From (i) and (ii),
180° – ∠RSP = 180°- ∠RQT
⇒ ∠RSP = ∠RQT …(iii)
In △TPS and △TRQ,
∠PTS = ∠RTQ (common angle)
∠RSP = ∠RQT [From (iii)]
∴ △TPS ~ △TRQ (AA similarity criterion)

(ii) Since △TPS ~ △TRQ, implies that corresponding sides are proportional that is,
\(\frac{S P}{Q R}\) = \(\frac{TP}{T R}\)
⇒ \(\frac{S P}{4}\) = \(\frac{18}{6}\)
⇒ SP = \(\frac{18 \times 4}{6}\)
⇒ SP = 12 cm

(iii) Since △TPS ~ △TRQ

⇒ ar(△TRQ = \(\frac{27 \times 4}{12}\)
⇒ ar(△TRQ) = 9 cm²
ar(PAQRS) = ar(△TPS) – ar(△TRQ)
⇒ ar(PAQRS) = 27 – 9
ar(PAQRS) = 18 cm²

Question 28.
Draw a line AB = 5 cm. Mark a point C on AB such that AC = 3 cm. Using a ruler and a compasses only, construct:
(i) A circle of radius 2.5 cm, passing through A and C.
(ii) Construct two tangents to the circle from the external point B. Measure and record the length of the tangents.
Solution:
Steps of construction :
(i) Draw AB = 5 cm using a ruler.
(ii) With A as the centre cut an arc of 3 cm on AB to obtain C.
(iii) With A as the centre and radius 2.5 cm, draw an arc above AB.
(iv) With same radius, and C as the centre draw an arc to cut the previous arc and mark the intersection as O.
(v) With O as the centre and radius 2.5 cm, draw a circle so that points A and C lie on the circle formed.
(vi) Join OB.
(vii) Draw the perpendicular bisector of OB to obtain the mid-point of OB, M.
(viii) With M as the centre and radius equal to OM, draw a circle to cut the previous circle at points P and Q.
(ix) Join PB and QB. PB and QB are the required tangents to the given circle from exterior point B.
QB = PB = 3 cm
That is, length of the tangents is 3 cm.