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## Kerala State Syllabus 10th Standard Maths Solutions Chapter 7 Tangents

### Tangents Textbook Questions & Answers

Textbook Page No. 163

**Tangents Class 10 Chapter 7 Kerala Syllabus Tangents Class 10 Kerala Syllabus Question 1.**

In each of the two pictures below, a triangle is formed by a tangent to a circle, the radius through the point of contact and a line through the center:

Draw these in your notebook.

Answer:

Draw a circle with radius 2.5cm and center as O. Mark a point P at a distance 5cm from O. Taking OP as the diameter, draw a circle.

Mark A at the point where the two circles intersect each other. Join PA.

Radius of the circle = \(\sqrt{4^{2}-2^{2}}\)

= \(\sqrt{16-4}\) = √12 = 2√3 cm

= 2 × 1.73 = 3.46 cm ~ 3.5 cm

Draw a circle with radius 3.5 cm. OA is the radius. Draw AP = 2 cm, perpendicular to OA. Join O and P, such that PA is the tangent

**Tangents Class 10 Kerala Syllabus Chapter 7 Question 2.**

In the picture, all sides of a rhombus are tangents to a circle.

Draw this picture in your notebook.

Answer:

Draw a circle of radius 4 cm and center O. Sdesof a rhombus are the tangents of the circa Let the angle between the tangents be 40°, then the center angle of arc between them = 180 – 40 = 140°.

Draw a rhombus by given measures.

**Sslc Maths Chapter 7 Kerala Syllabus Question 3.**

Prove that the tangents drawn to a circle at the two ends of a diameter are parallel.

Answer:

A tangent through P is perpendicular to PQ, PQ ⊥RP. A tangent through Q is perpendicular to PQ, PQ ⊥ SQ ∴∠SQP = ZRPQ = 90°. But they are co-interior angles. RP || SQ This means that RP is parallel to SQ. Hence proved.

**Sslc Maths Chapter Tangents Kerala Syllabus Chapter 7 Question 4.**

What sort of a quadrilateral is formed by the tangents at the ends of two perpendicular diameters of a circle?

Answer:

Let centre of the circle be O. Draw 2 mutually perpendicular diameters, PQ and SR.

Tangents drawn from a point outside to circle have equal length.

AP = AS, PD = PR, CQ = CR, QB = BS

AD + BC = AP + PD + BQ + CQ = AS + DR + SB + CR

= AS + SB + DR + CR

= AB + CD

Opposite sides are equal.

Hence ABCD is a square.

Textbook Page No. 166

**Class 10 Maths Tangents Kerala Syllabus Chapter 7 Question 1.**

Draw a circle of radius 2.5 centimeters. Draw a triangle of angles 40°, 60°, 80° with all its sides touching the circle.

Answer:

First, draw a circle with radius 2.5cm. PCOB is a quadrilateral, ∠COB = 360 – (90 + 90 + 40) = 140°. Divide the circle into three as 100°, 120°, 140°. Draw an arc equal distance from A and B find the point R. Similarly, find P and Q. Complete the figure,

**Class 10 Tangents Kerala Syllabus Chapter 7 Question 2.**

In the picture, the small (blue) triangle is equilateral. The sides of the large(red) triangle are tangents to the circumcircle of the small triangle at its vertices.

i. Prove that the large triangle is also equilateral and its sides are double those of the small triangle.

ii. Draw this picture, with sides of the smaller triangle 3 centimeters.

Answer:

i. Let O be the center of the radius

In ΔAOB

OA = OB (Radius)

∴ ∠OAB = ZOBA =30°

∴ ∠AOB = 120° ⇒ ∠APB = 60°

Similarly

∠AOC = 120° ⇒ ∠ARC = 60°

∠BOC = 120° ⇒ ∠CQB = 60°

Angle in the A PQR 60° each.

ΔPQRis an equilateral triangle.

In Δ APB, AP = PB

∠APB = 60°

∠PAB = ∠PBA = 60°

∴ Δ PAB is a equilateral triangle.

PA = PB=AB Similarly

AR = RC = AC and CQ = BQ =BC

Δ ABC is a equilateral triangle. Hence

PR = PA + AR = AB + AC = 2 AC

∴ The large triangle is also equilateral and its sides are double that of the small triangle.

**Tangent Class 10 Solutions Kerala Syllabus Chapter 7 Question 3.**

The picture shows the tangents at two points on a circle and the radii through the points of contact

i. Prove that the tangents have the same length.

ii. Prove that the line joining the center and the point where the tangents meet bisects the angle between the radii.

iii. Prove that this line is the perpendicular bisector of the chord joining the points of contact.

Answer:

i. Δ OAP, Δ OBP are similar triangles.

In.ΔOAP

OP^{2}=OA^{2}+AP^{2}

AP^{2}=OP^{2 –} OA^{2}

Length of the tangents are equal,

ii. . Consider Δ AOP, ΔBOP

PA= PB

(tangents)

OA = OB (radii)

PO = PO (common side)

∴ ΔAO ~ ΔBOP (Since all the sides are same)

Hence triangles are similar.

∴ ∠OPA = ∠OPB

OP bisects ∠ APB

iii. As ΔOAP ~ ΔBOP

∠POA = ∠POB

Considering

ΔAOM, ΔBOM

OA =OB(radii)

OM = OM (common side)

∠AOM = ∠BOM

ΔAOM ~ ΔBOM (By SAS property).

∴ AM = BM, OP bisects AB

**Sslc Maths Chapter 7 Notes Kerala Syllabus Question 4.**

Prove that the quadrilateral with sides as the tangents at the ends of a pair of perpendicular chords of a circle is cyclic.

What sort of a quadrilateral do we get if one chord is a diameter? And if both chords are diameters?

Answer:

LetAB, AD be tangents

∠PON + ∠PCN = 180°

∠PON = ∠MOQ

∠MAQ + ∠PCN = 180°

(PQ ⊥ MN)

∠MOQ = 90°

∠MAQ = 180° – 90° = 90°,

Similarly

∠NCP = 90°, ∠AMQ || ∠NCP = 180°,

∠A+ ∠ C = 180°

ABCD is a cyclic quadrilateral. If one chord is the diameter then the quadrilateral is a trapezium. If two chords are diameters then the quadrilateral is a square.

Textbook Page No. 172

**Tangents 10th Class Kerala Syllabus Chapter 7 Question 1.**

In the picture, the sides of the large triangle are tangents to the circumcircle of the small triangle, through its vertices.

Calculate the angles of the large triangle

Answer:

In a circle, the angle which a chord makes with the tangents at its ends on any side are equal to the angle which it makes on the part of the circle on the other side.

In Δ ABC, ∠B = 80°, ∠C = 60°

∠R = 180 – (80 + 60) = 40

∠CAR = 80°= ∠ACR

∠QBC = ∠QCB = 40°

∠P = 60°, ∠Q = 100°

∠NCP = 20°

Then the angles of bigger triangle are 60°, 100°, 20°.

**Sslc Tangent Questions And Answers Kerala Syllabus Chapter 7 Question 2.**

In the picture, the sides of the large triangle are tangents of the circumcircle of the smaller triangle, through its vertices.

Calculate the angles of the smaller triangle

Answer:

∠ACB =180-(60+ 40) = 80°

In a circle, the angle which a chord makes with the tangents at its ends on any side are equal to the angle which it makes on the part of the circle on the other side.

AP = AR ∴ ∠APR = ∠ARP = 60°

CR = CQ ∴ ∠CRQ =∠CQR = 50°

PB = BQ ∴ ∠BPQ = ∠BQP = 70°

Angles of the smaller triangle are 50°, 60°, 70°

**Hss Live Maths 10th Kerala Syllabus Chapter 7 Question 3.**

In the picture, PQ, RS, TU are tangents to the circumcircle of ΔABC.

Sort out the equal angles in the picture

Answer:

∠QAB = ∠ACB = ∠ABS

∠RBC = ∠BAC = ∠BCU

∠PAC = ∠ABC = ∠TCA

**10th Maths Tangents Kerala Syllabus Chapter 7 Question 4.**

In the picture, the tangent the circumcle of a regular pentagon through a vertex is shown.

calculate the angle which the tangent makes with the two sides of the pentagon through the point of contact.

Answer:

Sides of the pentagon are equal.

Textbook Page No. 179

**Tangent Class 10 Kerala Syllabus Chapter 7 Question 1.**

In the picture, a triangle is formed by two mutually perpendicular tangents to a circle and a third tangent. Prove that the perimeter of the triangle is equal to the diameter of the circle.

Answer:

O is the center and OS perpendicular to PS

(Radius is perpendicular to tangent)

OQ perpendicular PQ

(Radius is perpendicular to tangent)

OS = PS = PQ = OQ

(PQRS is a square)

PS= PA + AS

AS = AM (Tangents from A is equal) PS= PB + BQ

Hence BM= BQ (Tangents from B is equal)

PQ = PB + BM

Therefore

PS+ PQ= PA+AM + PB + BM

= PA+PB+AM + BM

= PA+PB+AB = Perimeter of right angled triangle PS + PQ is diameter of the circle.

So, the perimeter of the triangle is equal to the diameter of the circle The picture shows a

**Tangent Questions And Answers Kerala Syllabus Chapter 7 Question 2.**

The picture shows a triangle formed by three tangents to a circle.

Calculate the length of each tangent from the corner of the triangle to the point of contact

Answer:

**Sslc Maths Tangents Model Questions Kerala Syllabus Chapter 7 Question 3.**

In the picture, two circles touch at a point and the common tangent at this point is drawn

i. Prove that this tangent bisects! another common tangent of these circles.

ii. Prove that the points of contact of these two tangents from the vertices of a right triangle.

iii. Draw the picture on the right in your notebook, using convenient lengths. What is special about the quadrilateral formed by joining the points of contact of the circles?

Answer:

i.

PB = PQ and AP = PQ (Since they are tangents from P). PQ is common side.

∴ PA = PB, Therefore, the first tangent bisects the second tangent

ii. AP = QP

=> ∠ PAQ = ∠ PQA = x

BP = QP

=> ∠BQP = ∠PBQ = y

In AAQB

∠QAP + ∠ABQ + ∠AQB = 180°

x + y + x + y = 180°

2(x + y) = 180°

x + y = 90°

∠AQB = 90°

Δ ABQ is a right angled triangle.

iii. It is a rectangle

**Sslc Maths Tangents Notes Kerala Syllabus Chapter 7 Question 4.**

In the picture below, AB is a diameter and p is a point on AB extended. A tangent from P touches the circle at Q.

What is the radius of the circle?

Answer:

AP = 8 cm PQ = 4 cm

QP^{2} = AP × BP

AB = AP – BP

= 8 – 2 = 6

Diameter = 6

∴ Radius = \(\frac { 6 }{ 2 }\) = 3 cm

**Kerala Syllabus 10th Standard Maths Guide Pdf Download Chapter 7 Question 5.**

In the first picture below, the line joining two points on a circle is extended by 4 centimeters and a tangent is drawn from this point. Its length is 6 centimeters, as shown:

The second picture shows the same line extended by 1 centimeter more and a tangent drawn from this point. What is the length of this tangent?

Answer:

In the first circle

PC^{2} = PB × PA

62 = 4 x PA

PA = \(\frac { 36 }{ 4 }\) = 9

AB = AP – PB = 9 – 4 = 5 cm In the second

PC^{2} = PA × PB

PC^{2}= 10 × 5 = 50

PC= √50 = 5 √2cm

= 7.05 cm

Textbook Rage No. 185

**Kerala Syllabus 10th Standard Maths Guide Malayalam Medium Chapter 7 Question 1.**

Draw a triangle of sides 4 centimetres, 5 centimetres, 6 centimetres and draw its incircle. Calculate its radius.

Answer:

Draw a line BC = 6 cm. Draw an arc with a radius of 4 cm with the B as center. Also, draw an arc with a radius of 5 cm taken C as center. Let the bisector of both arcs drawn and they meet at ‘A’. Complete Δ ABC.

Let the bisectors of ∠A and∠B be drawn and they meet at ‘O’. Draw an incircle to the triangle with center O.

**10th Class Maths Malayalam Medium Kerala Syllabus Chapter 7 Question 2.**

Draw a rhombus of sides 5 centimeters and one angle 50° and draw its incircle.

Answer:

Draw AB with length 5 cm. Draw a ray from A making angle 50° with AB. Mark D at 5 cm from A. Draw a perpendicular bisector to ∠A. That is meet BD atO.

Extend line AO up to C such that AO = OC. Join BC and DC. Draw a perpendicular OP to AB through O. Draw a circle with centre as O and OP as radius.

Question 3.

Draw an equilateral triangle and a semi-circle touching its two sides, as in the picture.

Answer:

Draw an equilateral triangle with side 4cm.

The bisector of ∠A is passing through the midpoint of BC. Take the midpoint of AB as P. Join OP. Let draw a circle with OP as radius, we get the required one.

Question 4.

What is the radius of the incircle of a right triangle having perpendicular sides of length 5 centimeters and 12 centimeters?

Answer:

Question 5.

Prove that if the hypotenuse of a right triangle is h and the radius of its incircle is r, then its area is r(h+r).

Answer:

Let the perpendicular sides of right-angled triangle be a, d and hypotenuse be h then

Question 6.

Prove the radius of the incircle of an equilateral triangle is half the radius of its circumcircle.

Answer:

The center of the circumcircle and the incircle are the same.

r = \(\frac { r }{ R }\), Hence the radius of the incircle of an equilateral triangle is half the radius of its circumcircle.

### Tangents Orukkam Questions & Answers

Question 1.

ΔABC is an equilateral triangle. A circumcircle is drawn to it Prove that the triangle formed by the tangents to the circle at the vertices of ABC is another equilateral triangle.

Answer:

∴ ∠AOB = 120° ∠BOC = 120°

∠A = ∠B = ∠c = 60° (v Equilateral triangle)

∴ ∠AQC = 180 – 120° = 60°

(Sum of opposite sides of a quadrilateral is 180°)

∠BPC = 180 – ∠BOC = 60°

∠ARB = 180 –∠AOB = 60°

Three angles of the ΔPQR is 60° each.

So ΔPQR is an equilateral triangle.

Question 2.

Δ lf the perimeter of ΔABC is 10cm, calculate the perimeter of triangle PQR?

Answer:

Perimeter of ΔPQR = 2 × 10 = 20

One side =20/3

Area of ΔPQR =

Question 3.

What is the relation between the perimeters of ΔABC and ΔPQR?

Answer:

Area of Δ ABC =

Area of ΔPQR is four times of the area of ΔABC.

Question 4.

Draw the figure. Mark the circumcenter of AABC.

Answer:

Question 5.

Join OA, OB, OC. Note the cyclic quadrilaterals in it.

Answer:

OBPC, OAQC, OARB are the cyclic quadrilaterals by joining tangents and radius.

Question 6.

Since ∠B = 60°. ∠AOC will be 120° Write is ∠AOC?

Answer:

∠B = 60° hence ∠AOC = 120°

Angle made by an arc on the center which is twice the angle made by it on the alternate arc.

Question 7.

Find ∠P, ∠R . Write conclusions

Answer:

∠P = 60°, ∠R = 60°, both are equal

Question 8.

See three parallelograms like ABCQ. Find the perimeter of the outer triangle by the equality of opposite sides.

Answer:

ABCQ, CABP, CARB are three parallelograms.

In parallelogram. ABCQ , AB = CQ and BC = AQ . AB = BC

QA = QC (tangents from a outside point) = AB + BC = 2BC

PC = PB (tangents from a outside point) = 2 BC

RA = RB (tangents from a outside point) = 2 AC

∴ PQ + QR + PR =QC + PC + QA + RA + RB + PB

= 2 AB + 2 BC + 2 AC = 2(AB + BC + AC) = 2 × 10 (Perimeter of ΔABC is 10) = 20

Perimeter of ΔPQR =20

Question 9.

AC is the diagonal which divides the parallelogram by two equal triangles.

Answer:

The diagonal AC divides parallelogram ABCQ into two equal triangles.

Area of ΔACQ = Area of ΔABC

Similarly, area of ΔBPC

= Area of ΔABC

Area of ΔPQR

= Area (ΔACQ + ΔBPC + ΔABR + ΔABC)

= (4 × Area ΔABC)

∴ The perimeter of the outer triangle is twice of the perimeter of inner triangle.

Question 10.

Draw a circle and mark a point on it. Construct tangent to the circle at this point without using center.

Draw the circle and mark the point(P) DrawachordAB and joinAP and BP.

See the chord in the figure that you have drawn. This chord made the angle ∠PAB on one side. An equal angle will be formed on the other side of the chord at P with P B as one arm.(Use compass and scale method)

Answer:

All angles in the arc PB are equal.

Worksheet 6

Question 11.

In the figure, a circle touches the sides of?

ABC at P, Q, R.

If AB =AC then prove that BR = C R

Why is AP=AQ?

Answer:

AP = AQ

(Tangent drawn from a point A to circle have equal length)

Establish BP = C Q.

Establish BR = C R.

If AB = AC

AP + PB = AQ + QC

AP = AQ (Tangent drawn from a point A to circle have equal length)

∴ PB = QC

BP = BR (Tangent drawn from a point B to circle have equal length)

CR = CQ (Tangent drawn from a point C to circle have equal length)

∴ BR = CR

Question 12.

In the figure AP, BQ, P Qare tangents to the circle. The line AP is parallel to BQ. Find ∠POQ

Draw figure, mark OA, OB, OC.

Establish angle OAP, angle OCP equal.

Take ∠AOP, ∠COP as x.

Triangles BOQ, and COQ are equal.

∠BOQ, ∠COQ = Y

2x + 2y = 180. Write x + y and ∠POQ

Answer:

Δ OAP, Δ OCP are equal triangles.

So, ∠AOP = ∠COP = x

ΔBOQ, ΔCOQ are equal triangles.

So,

∠BOQ= ∠COQ = y

2x + 2y = 180°

Angle on semicircle.

So, x + y = 90°, ∠POQ = x + y = 90°

Question 13.

If a circle can be drawn by touching the sides of a parallelogram inside it will be a rhombus. Prove.

Answer:

Draw the figure AP = AS, BP = BQ, DR= DS, CR = CQ.

Using these equations prove the statement given as the 11^{th} point in the basic concepts.

2 × AB = 2 × AD.

Answer:

AP = AS, BP = BQ,

DR = DS, CR = CQ

(Tangent drawn from a point to circle have equal length)

Sum of opposite sides of a quadrilateral formed by joining the tangents on four points of a circle are equal.

So, 2 × AB = 3 × AD.

Therefore ABCD is a rhombus.

Question 14.

If r is the radius of the incircle of a right triangle prove that \(r=\frac{a+c-b}{2}\)

BP = BQ = r

AP = AR = c – r

CR = CQ = b – r

b = c – r + a – r

Answer:

In figure BP = BQ = r

∴ AP = AR = c – r

(Tangent drawn from a poijt A to circle have equal length)

CR = CQ = a – r

(Tangent drawn from a point C to circle have equal length)

AC = b = c – r + a – r

= c + a – 2r,

c + a – 2r = b

∴ \(r=\frac{a+c-b}{2}\)

Question 15.

Find the inradius of an equilateral triangle of side 10cm. User r = \(\frac { A }{ s }\)

Answer:

### Tangents SCERT Questions and Answers

Question 16.

PQ is a tangent to the circle with center O.

a. Find ∠p?

b. If ∠O = 42° ,what is ∠Q? [Score: 3, Time: 5 minutes]

Answer:

∠P = 90°, ∠Q = 90 – 42 = 48°

Question 17.

Draw this figure using the given measurements. [Score: 3, Time: 4 minutes]

Answer:

Draw a circle of radius 4 centimeters. (1)

Draw a radius and a perpendicular to it. (1)

Complete the triangle after marking angle 60° at the center.

Question 18.

In the figure, AC and BC are tangents to the circle from C. Centre of the circle O.

i. Find ∠A

ii. If ∠C is 2 times ∠O, then what is ∠C ? [Score: 3, Time: 4 minutes]

Answer:

i. ∠A= 90° (1)

ii. ∠C + ∠O = 180° (0)

∠C= 60° (0)

Question 19.

The radius of a circle touching all sides of an equilateral triangle is 3centimetres. Draw this triangle. [Score: 3, Time: 4 minutes]

Answer:

Draw a circle of radius 3 cm. (1)

Mark 120° at the center of circle. (2)

Complete the equilateral triangle. (2)

Question 20.

Radius of an incircle to a triangle is 3 centimeters. Two angles of this triangle are 55° and 63°. Draw this triangle. [Score: 5, Time: 8 minutes]

Answer:

Draw a circle of radius 3 cm. (1)

Mark the angles 180 – 55 = 125°, 180 – 63 = 117°, at the centre (1)

Draw the tangents.

Complete the triangle. (3)

Question 21.

In the figure PA, PB are tangents through A and B of a circle with center O. If the radius of the circle is r, then prove that OP × OQ = r^{2}. [Score: 3, Time: 5 minutes]

Answer:

Δ OQA, ΔOPA are triangle with equal angles.

The ratio of sides opposite to the equal angles. (1)

Question 22.

In the figure, angles formed by the radius segment of the meeting points of the tan-gent to incircle are given. Find all angles of the triangle. [Score: 3, Time: 5 minutes]

Answer:

Angles 180 – 120 = 60° (1)

180-130 = 50° (1)

Third angle 180 – (60 + 50) = 70° (1)

Question 23.

The incircle triangle ABC touches the tri-angle sides at P, Q& R. as shown in the figure. Find all angles of PQR [Score: 3, Time: 5 minutes]

Answer:

The angles at the center of the circles are 180 – 70 = 110°,

180 – 80 = 100 and

360 + ( 110 + 100) = 150° (2)

The angles of triangle PQR are

Question 24.

Find all angles of triangle AOP and OPT. [Score: 4, Time: 7 minutes]

Answer:

The angles of ΔAOP are 32°, 32°, 116° (2)

The angles of ΔOPT are 64°, 26°, 90° (2)

Question 25.

QP is a tangent of the circle with center O. AR is a diameter. Find all angles of triangle PQR. [Score: 4, Time: 6 minutes]

Answer:

∠PQR = ∠QAR = 30° (1)

∠PRQ = 180 – 60 = 120° (1)

∠P = 180 – (120 + 30) = 30°

Question 26.

In the figure, PQ is a diameter and O is the center of the circle. ∠R = ∠T= 90°

1. Prove that ∠PSR – ∠OSQ.

2. Prove that ΔPSR and ΔSQT are similar. [Score: 3, Time: 5 minutes]

Answer:

1. ∠PSR = ∠PQS (1)

Q is a diameter ZPSQ = 90° (1)

∠PSR = 90 – ∠OSP = ∠OSQ

2. ∠PSR = ∠SQT (1)

∴ Triangle are similar (2 angles are same) (1)

Question 27.

Draw a circle of radiu$3 3 cm. Draw a chord AB = 4 cm of this circle. Draw tangents through A and B. [Score: 3, Time: 5 minutes]

Answer:

Question 28.

In the figure, O is the center. C is a point on the semicircle with diameter OA.

BC is a tangent through B, If OB = 1 cm AB = 3 cm then what is BC f Find all angles of triangle OBC ? [Score: 4, Time: 6 minutes]

Answer:

OB × AB = BC^{2} (1)

1 × 3 = BC^{2} = 3 (1)

BC = √3

The angles of ΔOBC 30°, 60°, 90°. (2)

Question 29.

In the figure, radius of the circle centered at O is 9 cm.

OA = 15 cm. Semicircle with diameter O A cuts the circle with center O at D and BC is a tangent through B.

1. What is the length of BC?

2. If the line PD is perpendicular to OA, then what is the length of PD? [Score: 4, Time: 7 minutes]

Answer:

1. BC^{2} = OB × BA = 9 × 6 = 54

BC = √54cm (1)

2. OP × OA = r^{2}

Question 30.

Hypotenuse of a right triangle is 18 cm and its inradius 3 cm. What is its perimeter? What is its area? . [Score: 3, Time: 5 minutes]

Answer:

Perimeter = 3 + x + x +y +y + 3

= 3 + 3 + 18 + 18 = 42 cm (1)

Area = \(\frac { 42 }{ 2 }\) × 3 = 21 × 3 = 63 sq.cm (1)

Question 31.

Area of a right triangle is 60 sq. centimetres and its inradius 3 cm. What is the length of its hypotenuse? [Score: 3, Time: 5 minutes]

Answer:

Area = 60 sq.cm, radius = 3

### Tangents Exam Oriented Questions and Answers

Short Answer Type Questions (Score 2)

Question 32.

From a point P, the length of the tangent to a circle is 24 cm and distance of P from the center is 25cm. Find radius.

Answer:

OP^{2} = OQ^{2} + QP^{2}

25^{2} = OQ^{2} + 24^{2}

OQ^{2} = 25^{2} – 24^{2}

OQ^{2}= 625 – 576 = 49

OQ = 7 cm

Question 33.

Let PQ be a tangent to a circle at A and AB be a chord. Let C be a point on the circle such that ∠BAC = 54° and ∠BAQ= 62°. Find ∠ABC.

Answer:

∠ABC = l80° – ( ∠BAC + ∠ACB )

∠ABC = i8o° – (54° + 62°) = 64°

Question 34.

Draw a circle of radius 3.5cm and construct an equilateral triangle intersecting all sides with this circle. What is the radius of a circumcircle?

Answer:

Radius of circumcircle = 7 cm

Question 35.

In the figure, AB is the tangent at B of the circle centered at O. How much is ∠OBA? How much is ∠AOB?

Answer:

∠OBA = 90°

∠AOB = 55°

Question 36.

In the figure O is the center of the circle and P, Q,R are points on it. Find the angles of the triangle formed by the tangent at P, Q,R.

Answer:

Angle in a triangle and opposite angle in its center is supplementary. One angle is 40°, other angles are 70° each.

Question 37.

ABCD is a quadrilateral such that all of its sides touch a circle. If AB = 6cm, BC = 6.5cm and CD = 7cm, then find the length of AD.

Answer:

We have, AP = AS

BP = BQ

CR = CQ

DR = DS

Hence,

AP + BP+CR + DR = AS + BQ + CQ + DS

AB +CD = AD + BC

AD = AB + CD – BC = 6 + 7 – 6.5 = 6.5 cm

Short Answer Type Questions (Score 3)

Question 38.

Draw a circle of radius 2cm. Also, construct a regular hexagon intersecting all sides with this circle.

Answer:

ABCDEF is a regular hexagon.

Question 39.

AB 4cm and BC = 6cm. Find the length of AD? Draw a line segment of length √12 cm by using the same idea?

Answer:

AB × AC = AD^{2}

AB = 4cm

AC = AB + BC = 4 + 6= 10cm

4 × 10 = AD2

∴ AD = √40

Inorder to draw √12 , draw AC = 6cm by taking BC = 4 cm and AB = 2 cm.

When a tangent is drawn from A, its length is √12 cm.

Question 40.

The radius of a circle with center O is 8cm. P is a point outside the circle. PQ and PR are the tangents drawn from P to the circle. If ∠QOR = 60° then find the lengths of PQ, PR, and OP.

Answer:

In Δ OPQ the angles are 30°, 60°, and 90°.

∴ The sides are in the ratio 1 : √3: 2

Question 41.

In the figure PA, PB and RS are the three tangents to the circle. Prove that the perimeter of ΔPRS is the sum of the lengths PAandPB.

Answer:

PA = PB (Tangents from an outside point are equal)

RA = RT

ST = SB

Perimeter of ΔPRS = PR + RS + PS

= PR + (RT + ST) + PS

= (PR + RA) + (SB + PS)

(Since RT = RA, ST= SB)

= PA + PB

Question 42.

In the figure, tangents PA and PB are drawn to a circle with center O from an external point P. If CD is a tangent to the circle at E and AP=25cm, find the perimeter of ∠PCD,

Answer:

We know that the lengths of the two tangents from an exterior point to a circle are equal.

CA = CE, DB = DE and PA = PB Now,

the perimeter of = PC + CD + DP

= PC + CE + ED + DP

= PC + CA + DB + DP

= PA + PB = 2PA(PB = PA)

Thus, the perimeter of ∠PCD = 2 × 25 = 50cm

Question 43.

In the figure, PQ is the diameter and RS is a tangent to the circle.

If ∠SPQ = 34°,fmd

a) ∠SQP

b) ∠RSQ

c) ∠SRP

Answer:

∠PSQ = 90° (Angle in a semicircle)

∴ ∠ SQP

= 180 – (90 + 34)

= 180 – 124 = 56°

b. ∠RSQ = ∠SPQ = 34° (Angle between chord and tangent = Angle in the segment on the other side)

c. ∠SRP= 180 – (124 + 34) s = 180 – 158 = 22°

Question 44.

Prove that the angles formed by the tangents from the endpoints of a chord are equal.

Answer:

Each angle between a chord and the tangent at one of its ends is equal to the angle in the segment on the other side of the chord. [Angles in the alternate segments are equal].

∴ ∠P = x ie ∠RBA = x.

ie ∠QAB = ∠RBA

Long Answer Type Questions (Score 4)

Question 45.

In the following figure, PQ is the tangent and PB is the scent. Prove that PQ2 = PA × PB

Answer:

Considering the triangles PAQ and PQB.

∠APQ = ∠BPQ (Common angle)

∠PQA = ∠QBP (Angle between tangent and chord = the angle made by the chord on its . complimentary arc)

ΔPAQ ~ ΔPQB

[A.A Similarly]

[Ratio of the similar sides are equal]

PQ^{2} = PA × PB

Question 46.

AB, BC, and AC are the tangents of the circle. If AR = 5cm, BP = 4cm and BC = 10cm, find the perimeter of the triangle.

Answer:

AR = AQ = 5 cm

BP = BR = 4 cm

BC – BP = PC

∴ PC = 10 – 4 = 6cm,

PC = PQ = 6cm

∴Perimeter = 2 × 5 + 2 × 4 + 2 × 6

= 2 (5 + 4 + 6 ) = 30 cm

Long Answer Type Questions (Score 5)

Question 47.

AB and AC are the tangents of the circle with center O.

Show that the quadrilateral ABOC is a cyclic quadrilateral.

Answer:

AB and AC are tangents

∠B = ∠C = 90°

∠B + ∠C = 180°

The sum of the four angles of a quadrilateral is 360°. ie, ∠A + ∠O = 180°

The opposite angles of quadrilateral ABOC are complementary.

∴ ABOC is a cyclic quadrilateral.

Question 48.

AB, BC, CD and AD are the tangents of the circle. If AP = x, BP = y, CR = z, SD =w then show that the perimeter of the quadrilateral ABCD is 2(x + y + z + w).

Answer:

AP = AS = x;

BP = BQ = y

CR = CQ = z;

SD = DR = w

ie, AB = x + y

BC = y + z;

CD = z + w

AD = x + z

∴ Perimeter = 2x + 2y + 2z + 2w = 2(x + y + z + w)

Question 49.

In the picture PQ, RS, and TU are the tan-gents of the circle with center O. Find the pairs of angles with same measures

Answer:

∠QAB = ∠RBA ; ∠QAB = ∠ACB

∠RBA =∠ACB ; ∠SBC = ∠UCB

∠SBC = ∠BAC ; ∠UCB = ∠BAC

∠ACT = ∠PAC ; ∠ACT = ∠ABC

∠PAC = ∠ABC

### Tangents Memory Map

Tangent to the circle is perpendicular to the radius through the point of tangency.

The tangents from an exterior point to a circle and radii to the points of tangency form a cyclic quadrilateral. In figure PAOB is a cyclic quadrilateral.

Tangents from an exterior point to a circle are equal. If PA, PB are the tangents then PA=PB.

The angle between a chord of a circle and the tangent at one end of the chord is equal to angle formed by the chord in the other side of the circle.

If a circle touches the sides of a quadrilateral, that circle will be the incircle of that quadrilateral. Sum of the opposite sides of such quadrilateral are equal. In the figure, ABCD is a quadrilateral having incircle AB + CD = AD + BD.

If P is an exterior point to a circle, a line from P touches the circle at T on the circle and a line intersects the circle at A and B then PA × PB = PT2.

The center of the circle which touches two lines will be a point on the bisector of the angle between the lines. The bisectors of the angles of a triangle passes through a point. That point will be the incenter of the triangle.

The circle drawn inside a triangle which touches the sides of the triangle is called incircle. The circle drawn outside the triangle which touches the sides of the triangle are excircles.

The radius of the incircle of a triangle is obtained by dividing area of the circle by its semi perimeter.

If a, b, c are the sides of a triangle then the area of the triangle

A= \(\sqrt{\mathrm{S}(\mathrm{S}-\mathrm{a})(\mathrm{S}-\mathrm{b})(\mathrm{S}-\mathrm{c})}, s=\frac{a+b+c}{2}\)

This is popularly known as Hero’s formula.