Samacheer Kalvi 10th Maths Solutions Chapter 4 Exercise 4.4 Pdf

Samacheer Kalvi 10th Maths Solutions Chapter 4 Exercise 4.4 Pdf

TN Samacheer Kalvi 10th Maths Solutions Chapter 4 Exercise 4.4 : Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 4 Exercise 4.4 – Geometry on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 4 Exercise 4.4 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 4.4

TN Samacheer Kalvi 10th Maths Solutions Chapter 4 – Geometry

Board TNSCERT Class 10th Maths
Class 10 Class
Subject Maths
Chapter 4 (Exercise 4.4)
Chapter Name Geometry

TNSCERT Class 10th Maths Pdf | all Exercise Solution

 

Exercise – 4.4

 

(1) The length of the tangent to a circle from a point P, which is 25 cm away from the centre is 24 cm. What is the radius of the circle?

Solution:

Now, given that AC = 25cm and BC = 24cm.

Let, the radius of the circle is AB = x cm

Now, ∆ABC, ∠B = 90°

By Pythagoras theorem,

(AC)2 = (AB)2 + (BC)2

(AB)2 = (AC)2 – (BC)2

(x)2 = (25)2 – (24)2

x2 = 625 – 576

x2 = 49

x = 7cm

Therefore, the circle radius is 7cm.

 

(2) LMN is a right angled triangle with ∠L = 90°. A circle is inscribed in it. The lengths of the sides containing the right angle are 6 cm and 8 cm. Find the radius of the circle.

Solution:

Now, LN = 6cm and LM = 8cm

Now, ∆LMN, ∠L = 90°

By Pythagoras theorem,

(MN)2 = (LN)2 + (LM)2

(MN)2 = (6)2 + (8)2

(MN)2 = 36 + 64

(MN)2 = 100

(MN)2 = (10)2

∴ MN = 10cm

Let the circle radius OP = OQ = OR = r cm

Now, we know that,

Area of ∆LMN = 1/2 × LM × LN

= 1/2 × 8 × 6

= 24 cm2

Now, Area of ∆LON = 1/2 × 6 × r

= 3r cm2

Area of ∆OLM = 1/2 × 8 × r = 4r cm2

Area of ∆OMN = 1/2 × 10 × r = 5r cm2

Now, Area of ∆LMN = area of ∆LON + area of ∆OLM + area of ∆OMN

24 = 3r + 4r + 5r

12r = 24

r = 2cm

Therefore the circle radius is 2cm

 

(3) A circle is inscribed in ∆ABC having sides 8 cm, 10 cm and 12 cm as shown in figure, Find AD, BE and CF.

Solution: 

Now, AB = 12cm, BC = 8cm, AC = 10cm

Now, CE = CF [Tangents drawn from Sanu external point are equal]

Similarly, BE = BD and AF = AD

Let CE = CF = x, BE = BD = y, and AF = AD = Z

Now, CE + EB = BC

x + y = 8 —– (i)

Now, AD + DB = AB

Z + y = 12

y + z = 12 —- (ii)

Now, AF + FC = AC

Z + x = 10

x + z = 10 —- (iii)

Now, (i) + (ii) + (iii), we get,

x + y + y + z + x + z = 8 + 12 + 10

2 (x + y + z) = 30

x + y + z = 15 —– (iv)

Now, (iv) – (ii), we get

x + y + z – y – z = 15 – 12

x = 3

Now, (iv) – (i), we get

x + y + z – x – y = 15 – 8

z = 7

Now, (iv) – (iii), we get

x + y + z – x – z = 15 – 10

y = 5

Thus, AD = z = 7 cm

BE = y = 5cm

CF = x = 3cm

 

(4) PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circle such that POR = 120°. Find OPQ.

Solution:

Given that, ∠POR = 120°

Then, ∠POQ = 180° – ∠POR

= 180° – 120°

= 60°

We know that ∠PQO = 90°

Then, ∠OPQ = 180° (∠POQ + ∠PQO)

= 180° – (90° + 60°)

= 180° – (90° + 60°)

= 180°- 150°

= 30°

Thus, ∠OPQ = 30°

 

(5) A tangent ST to a circle touches it at B. AB is a chord such that ABT = 65°. Find AOB, where “O” is the centre of the circle.

Solution:

Given, that ∠ABT = 65°

Now, ∠OBA = 90° – 65° = 25°

Now, OB = OA then, ∠BAO = 25°

Now, ∠AOB =180° – (∠OBA + ∠BAO)

= 180° – (25 + 25) = 130°

Thus, ∠AOB = 130°

 

(6) In figure, O is the centre of the circle with radius 5 cm. T is a point such that OT = 13 cm and OT intersects the circle E, if AB is the tangent to the circle at E, find the length of AB.

Solution:

Now, OT = 13cm and OP = 5cm

∆OTP, ∠P = 90°

By Pythagoras theorem,

(OT)2 = (PT)2 + (OP)2

(PT)2 = (OT)2 – (OP)2

(PT)2 = (13)2 – (5)2

PT2 = 169 – 25

PT2 = 144

(PT)2 = (12)2

PT = 12cm

Now, AP = AE = x (Let)

∴ AT = (12 – x) cm, TE = (13 – 5) = 8cm

∆AET, ∠E = 90°

By Pythagoras theorem,

(AT)2 = (AE)2 + (ET)2

(12-x)2 = x2 + (8)2

144 – 24x + x2 = x2 + 64

24x = 80

x = 80/24

x = 10/3

∴ AE = 10/3

Now, AE = EB

AB = AE + EB

=> 10/3 + 10/3 cm

=> 2×10/3 = 20/3 cm

Thus, the length AB = 20/3 cm

 

(7) In two concentric circles, a chord of length 16 cm of larger circle becomes a tangent to the smaller circle whose radius is 6 cm. Find the radius of the larger circle.

Solution:

Given that, OC = 6cm,

AC = CB ∴ AC = CB = 8cm [∵ AB = 16cm]

AB = AC + CB

∆ACO, ∠C = 90°

By Pythagoras theorem,

(AO)2 = (OC)2 + (AC)2 = (6)2 + (8)2

(AO)2 = 36 + 64 = 100 = (10)2

∴ AO = 10

∴ Thus the larger circle radius is 10cm.

 

(8) Two circles with centres O and O’ of radii 3 cm and 4 cm, respectively intersect at two points P and Q, such that OP and O’P are tangents to the two circles. Find the length of the common chord PQ.

Solution:

Given that, OP = 3cm, O’P = 4cm

Now, ∆OPO’, ∠P = 90°

By Pythagoras theorem,

(OO’)2 =  (OP)2 + (PO’)2

(OO’)2 = (8)2 + (4)2 = 9+16

(OO’)2 = 25 = (5)2

OO’ = 5cm

Now, PR = RQ = 2 (Let)

and OR = y (Let), ∴ O’R = (5 – y) cm

∆POR, ∠R = 90°

By Pythagoras theorem,

(PR)2 + (OR)2 = (OP)2

x2 + y2 = (3)2 = 9

∴ x2 + y2 = 9 —–  (i)

Now, ∆PRO’, ∠R = 90°

BY Pythagoras theorem,

(O’P)2 = (O’R)2 + (PR)2

(4)2 = (5-y)2 + (x)2

16 = 25 – 10y + y2 + x2

16 = 25 – 10y + 9 [By equation (i)]

10y = 34 – 16

10y = 18

y = 1.8 cm

From (i) Putting y = 1.8, we get

x2 + y2 = 9

(x)2 + (1.8)2 = 9

x2 = 9 – 3.24

x2 = 5.76

(x)2 = (2.4)2

x = 2.4

Now, PR = 2.4 = RQ

Now, PQ = PR + RQ = 2.4 + 2.4

PQ = 4.8 cm

Therefore, the length of the common chord PQ is 4.8 cm

 

(9) Show that the angle bisectors of a triangle are concurrent.

Solution:

Now, AD is bisector of ∠A

BE is bisector of ∠B

CF is bisector of ∠C

Now, ∆ABC, AD is bisector of ∠A

By Angle bisector theorem

AB/AC = BD/DC —— (i)

∆ABC, BE is bisector of ∠B

By Angle bisector theorem

BC/AB = CE/AE —- (ii)

∆ABC, CF is bisector of ∠C.

By angle bisector theorem,

AC/BC = AF/BF —- (iii)

Now, (i) × (ii) × (iii), we get

AB/AC × BC/AB × AC/BC = BD/DC × CE/AE × AF/BF

BD/DC × CE/AE × AF/BF = 1

By Ceva’s theorem

AD, BE, CF are concurrent.

∴ Thus, The angle bisector of a triangle are concurrent. [Proved]

 

(10) An artist has created a triangular stained glass window and has one strip of small length left before completing the window. She needs to figure out the length of left out portion based on the lengths of the other sides as shown in the figure.

Solution:

Given that AE = 3cm, EC = 4cm, CD = 10cm

DB = 3cm, AF = 5cm, BF = x (Let)

Now, AD, BE, CF are concurrent.

By — theorem,

AE/EC × CD/DB × BF/AF = 1

3/4 × 10/3 × x/5 = 1

x/2 = 1

x = 2

Therefore required BF = 2cm.

 

Here we posted the solution of TN Tamilnadu Board Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry. TNSCERT Samacheer Kalvi 10th Maths Chapter 4 Geometry Solved by Expert Teacher.

Updated: January 7, 2022 — 3:31 pm

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