Brilliant’s Composite Mathematics Class 8 Solutions Chapter 14

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Brilliant’s Composite Mathematics Class 8 Solutions Chapter 14 Tangents to a Circle

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Brilliant’s Composite Mathematics Class 8 Math Book, Chapter 14, Tangents to a Circle. Here students can easily find step by step solutions of all the problems for Tangents to a Circle, Exercise 14.1, 14.2, 14.3 and 14.4 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.

Tangents to a Circle Exercise 14.2 Solution

Question no – (2)

Solution :

Required Figure,

First, we draw OP Now, in △ADP, ∠OAP = 90°, ∠AOP = 60° [1/2 ∠AOB]

∴ ∠APO = 180° – (∠OAP + ∠AOP)

= 180° – (90° + 60°)

= 30°

Similarly, ∠BPO = 30°

Thus, ∠APB = 30° + 30° = 60°

Since, ∠OAP + ∠OBP = 180°

and, ∠APB + ∠AOB = 180°

Therefore, the quadrilateral AOBA is cyclic

Question no – (3)

Solution :

Required Figure,

In △PQR, ∠PRQ = 90° [angle in a semicircle]

and, ∠RPQ = 90° – 40° = 50°

∴ ∠PQR = 180° – (∠RPQ + ∠PRQ)

= 180° – (90° + 50°)

= 40°

Tangents to a Circle Exercise 14.3 Solution

Question no – (1)

Solution :

Required Figure,

(i) OP = 5 cm, OA = 3 cm.

∴ AP = √(5)² – (3)² = √25 – 9

= √16 = 4 cm.

(ii) √(13)² – (5)²

= √169 – 25

= √144

= 12 cm.

(iii) √(15)² – (9)²

= √225 – 81

= √144

= 12 cm.

(iv) √(25)² – (7)²

= √625 – 49

= √576

= 24 cm.

Question no – (2)

Solution :

Required Figure,

We have, ∠AOB + ∠APB = 180°, Since the quadrilateral AOBP is cyclic

∴ ∠APB = 180° – ∠AOB

= 180° – 130°

= 50°

Now, OP bisects ∠AOB and ∠APB

∴ ∠AOP = 130°/2 = 65°,  ∠APO = 50°/2 = 25°

∴ In △AOB ∠OAP

= 180° – (65° + 25°)

= 180° – 90° = 90°

Now, ∠OAB = 180° – 130°/2

= 50°/2

= 25°

∴ ∠PAB =  ∠PAO –  ∠OAB

= 90° – 25°

= 65°

Question no – (3)

Solution :

Required Figure,

Lets draw PC ⊥ AB, which bisects AB

Comparing △ACP and △ABC, we have

(i) AC = CB,

(ii) ∠ACP = ∠BCA

(iii) AC is common side so, △ACP ≅ △BCP

∴ ∠CAP = ∠CBP  [Similar angles of congurengt triangle]

Question no – (4)

Solution :

Required Figure,

Since PA and PB are two tangents to the circle from an external point P, PA = PB

= PL + LA = PM + MB ……(i)

Also, LA and LN are two tangents to the circle from a point L

∴ LA = LN ……(ii)

Similarly, MB = MN ……(iii)

With the help of (ii) and (iii), (i) reduces to PL + LN = PM + MN

Question no – (5)

Solution :

Required Figure,

ABCD is a quadrilateral, whose sides touch the circle with centre O.

We have, AF = AE …….(i)

Similarly, BF = BG ……..(ii)

CG = CA …….(iii) and DH = DE …….(iv)

From (i), (ii), (iii), (iv) we have,

AB + CD = AF + FB + CH + HD

= AE + BG + CG + DE

= (AE + DE) + (BG + CG)

= AD + BC (proved)

Question no – (6)

Solution :

Required Figure,

Given, AB = AC and AD = AE

∴ AB – AD = AC – AE

Or, DB = EC ……(i)

Now, DB = BF ……(ii)

Similarly, EC = FC ……. (iii)

Combining (i), (ii), (iii) we have, DB = BF = FC = CE i.e. BF = FC (proved)

Question no – (7)

Solution :

Required Figure,

Here we have AC = CH = 3 cm. and BD = DE 1.5 cm.

∴ CD = CE + ED

= 3 + 15

= 4.5 cm.

Question no – (8)

Solution :

Required Figure,

In △OAP, ∠A = 90°

∴ AO = OQ = AO = radius ∠AOP = 60° = ∠APO = 30°

∴ ∠APB = 60° But AP = BP

∴  ∠PAB =  ∠PBA = 60°

Hence, △PAB is an equilateral triangle

Question no – (10)

Solution :

Required Figure,

Since, ∠DOE = m DE = 140°,

Therefore ∠DPE = 180° – 140° = 40°

Now, in △DEO, ∠OPE

= 180° – 140°/2

= 40°/2

= 20°

∴ ∠PDE = ∠PDO – ∠ODE

= 90° – 20°

= 70°

Question no – (11)

Solution : 

Required Figure,

First we connect O and P Now, comparing △POD and △BOP, we have,

(i) ∠PD = PB

(ii) ∠PDO = ∠PBO

(iii) OP common side

∴ △PDO and △BOP are congruent.

Now, ∠DPO = 180° – (90° + 60°) = 30° = ∠BPO

∴ ∠DPB = 30° + 30° = 60°

Since, ∠BOD + ∠BPD = 120° + 60° = 180° and, ∠ODP + ∠OBP = 90° + 90° = 180°, we can conclude that BODP is a cyclic quadrilateral.

Question no – (12)

Solution :

Required Figure,

We have AQ = AR …….(i)

Now, Perimeter of △ABC

= AB + BC + CA

= AB + (BP + PC) + AC

= AB + BQ + CR + AC

= AQ + AR = 2 AQ

∴ AQ = 1/2 perimeter of △ABC

Question no – (13)

Solution :

Required Figure,

(i) Comparing △OPS and △OQS, we get

(a) SP = SQ

(b) OP = OQ

(c) So is common side

So, △ORS ≅ △OQS.

(ii) ∠OQS = 90°, because SQ is a tangent which touches radius OQ at Q, and we know that whenever a radius and a tangent touch each other, angle between them is 90°

Precious Chapter Solution :  

👉 Chapter 13