**Samacheer Kalvi 10th Maths Solutions Chapter 4 Exercise 4.5** **Pdf**

TN Samacheer Kalvi 10th Maths Solutions Chapter 4 Exercise 4.5 : Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 4 Exercise 4.5 – Geometry on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 4 Exercise 4.5 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 4.5

**TN ****Samacheer Kalvi 10th Maths Solutions Chapter 4 – Geometry**

Board |
TNSCERT Class 10th Maths |

Class |
10 Class |

Subject |
Maths |

Chapter |
4 (Exercise 4.5) |

Chapter Name |
Geometry |

**TNSCERT Class 10th Maths Pdf | all Exercise Solution**

__Exercise – 4.5__

__Exercise – 4.5__

**(1)** **If in triangles ABC and EDF, AB/DE = BC/FD then they will be similar, when **

**(A) ****∠B = ∠E **

**(B) ∠A = ∠D **

**(C) ∠B = ∠D **

**(D) ∠A = ∠F **

**Solution: **

Now, ∠B = ∠D

Then, ∆ABC and ∆EDF similar

∴ (c) ∠B = ∠D

**(2) In ****∆LMN, ****∠L = 60****°, ****∠M = 50****°. If ****∆LMN ****~ ****∆PQR then the value of ****∠R is – **

**(A) 40****°**

**(B) 70° **

**(C) 30° **

**(D) 110° **

**Solution: **

Now, ∆LMN ~ ∆PQR

∠L = 60°, ∠M = 50°

∴ ∠N = 188° – (60+50)

∠N = 70°

Now, ∠N = ∠R

∴ ∠R = 70° (B)

**(3) If ****∆ABC is an isosceles triangle with ****∠C = 90****° and AC = 5cm, then AB is **

**(A) 2.5cm **

**(B) 5 cm **

**(C) 10 cm **

**(D) 5√2 cm **

**Solution:**

Now, AC = BC = 5 (Given)

∆ABC, ∠C = 90°

By Pythagoras theorem,

(AB)^{2} = (AC)^{2} + (BC)^{2}

(AB)^{2} = (5)^{2} + (5)

(AB)^{2} = 25 + 25

(AB)^{2} = (5√2)^{2}

AB = 5√2 (D)

** **

**(4) In a given figure ST****||QR, PS = 2 cm and SQ = 3 cm. Then the ratio of the area of ****∆PQR to the area of ∆PST is **

**(A) 25:4 **

**(B) 25:7 **

**(C) 25:****11 **

**(D) 25:13 **

**Solution:**

Now, ∆PQR ~ ∆PST then,

Area of ∆PQR/area of ∆PST = (PQ)^{2}/(PS)^{2} = (5)^{2}/(2)^{2} = 25/4

∴ required ratio is 25:4 (A)

** **

**(5) The perimeters of two similar triangles ****∆ABC and ∆PQR are 36cm and 24cm respectively. If PQ = 10 cm, then length of AB is **

**(A) 6 2/3 cm **

**(B) 10√6/3 cm **

**(C) 66 2/3 cm **

**(D) 15cm **

**Solution: **

Now, ∆ABC ~ ∆PQR

∴ AB/PQ = AB+BC+CA/PQ+QR+PR

AB/PQ = 36/24

AB = 36×10/24

AB = 15cm (D)

** **

**(6) If in ****∆****ABC, DE****||****BC.AB = 3.6 cm, AC = 2.4 ****cm and AD = 2.1 cm then the length of AE is**

**(A) 1.4 cm **

**(B) 1.8 cm **

**(C) 1.2 cm **

**(D) 1.05 cm **

**Solution: **

Now, ∠ADE = ∠ABC and ∠A is common.

∴ ∆ABC ~ ∆ADE

Then, by Thales theorem,

AB/AD = AC/AE

3.6/2.1 = 2.4/AE

AE = 2.4×2.1/3.6

AE = 1.4 cm (A)

** **

**(7) In a ****∆ABC, AD is the bisector of ****∠BAC. If AB = 8cm, BD = 6 cm and DC = 3cm. The length of the side AC is **

**(A) 6cm **

**(B) 4cm **

**(C) 3cm **

**(D) 8cm **

**Solution: **

Thus, AB/BC = BD/DC = AC = AB×DC/BD = 8×3/6 = 4cm (B)

** **

**(8) In the adjacent figure ****∠BAC = 90****° and AD****⊥BC then **

**(A) BD.CD = BC ^{2} **

**(B) AB.AC = BC ^{2} **

**(C) BD.CD = AD ^{2} **

**(D) AB.AC = AD ^{2} **

**Solution: **

∠ADB = ∠ADC = 90°

Then, ∠BAD = ∠DAC = ∠BAC/2 = 45°

Now, ∆ABD ~ ∆ADC, then,

By Thales theorem,

AD/BD = CD/AD

AD^{2} = BD.CD (C)

** **

**(9) Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m, what is the distance between their tops?**

**(A) 13m **

**(B) 14m **

**(C) 15m **

**(D) 12.8m **

**Solution: **

Now, AE = 11cm,

CD = BE then CD = 6cm

AB = AE – BE

= (11 – 6) = 5 cm

CB = DE = 12cm

Now, ∆ABC, ∠B = 90°

By Pythagoras theorem,

(AC)^{2} = (AB)^{2} + (BC)^{2}

(AC)^{2} (5)^{2} + (12)^{2}

(AC)^{2} = 25 + 144 = 169

(AC)^{2} = (13)^{2}

∴ AC = 13 (A)

** **

**(10) In the given figure, PR = 26cm, QR = 24cm, ****∠PAQ = 90****°, PA = 6 cm and QA = 8cm. Find ****∠PQR **

**(A) 80****° **

**(B) 85° **

**(C) 75° **

**(D) 90° **

**Solution: **

Now, ∆APQ, ∠A = 90°

Then by Pythagoras theorem,

(PQ)^{2} = (AP)^{2} + (AQ)^{2}

(PQ)^{2} = (6)^{2} + (8)^{2} [∵ AP = 6, AQ = 8]

(PQ)^{2} = 36 + 64 = 100 = (10)^{2}

PQ = 10

Now, ∆PQR, PQ = 10cm PR = 26cm and RQ = 24cm

∴ ∆PQR, right angle triangle.

∴ Hypotenuse is RP = 26cm

Then, ∠PQR = 90° (D)

** **

**(11) A tangent is perpendicular to the radius at the**

**(A) Centre **

**(B) Point of contact **

**(C) Infinity **

**(D) Chord**

**Solution: **

(B) Point of contact.

** **

**(12) How many tangents can be drawn to the circle from an exterior point?**

**(A) One **

**(B) Two **

**(C) infinite **

**(D) Zero**

**Solution: **

(B) Two.

** **

**(13) The two tangents from an external points P to a circle with centre at O are PA and PB. If ****∠****APB = 70° then the value of ****∠****AOB is **

**(A) 100°**

**(B) 110° **

**(C) 120° **

**(D) 130° **

**Solution: **

Now, ∠APB = 70°

Then, ∠APO = ∠BPO = ∠APO/2 = 35°

∆APO,

∠AOP = 180° – (∠OAP + ∠APO)

= 180° – (90 + 35)

= 180° – 125

= 55°

∆BPO then,

∠BOP = 180° – (∠OPB + ∠OBP)

= 180° – (35 + 90)

= 180° – 125 = 55°

∴ Now, ∠AOB = ∠AOP + ∠BOP

= 55° + 55° = 110° (B)

** **

**(14) In figure CP and CQ are tangents to a circle with centre at O. ARB is another tangent touching the circle at R. If CP = 11 cm and BC = 7 cm, then the length of BR is**

**(A) 6 cm **

**(B) 5 cm**

**(C) 8 cm **

**(D) 4 cm**

**Solution: **

Now, CP = CQ and CA = CB

Then, AP = CP – CA

= (11 – 7) cm

= 4cm

Now, AB = AR + RB

Now, Point A to AP and AR same tangent then, AP = AR

We know that AR = RB

∴ RB = AR = AP = 4cm (D)

** **

**(15) In figure if PR is tangent to the circle at P and O is the centre of the circle, then ****∠****POQ is **

**(A) 120° **

**(B) 100° **

**(C) 110° **

**(D) 90° **

**Solution: **

Now, ∠OPQ = 90-60 = 30° [∵ ∠OPR = 90°]

Now, ∠OPQ = ∠OQP [∵ OP = OQ = radius]

∆POQ, ∴ ∠POQ = 180° – (∠OPQ + ∠OQP) = 180° – 30+30 = 180 – 60 = 120°

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