06 Apr 2022

Samacheer kalvi 10th Maths – Geometry Ex 4.5

Samacheer Kalvi 10th Maths Book – Unit 4 Exercise 4.5: Book Back Question and Answers

TN Samacheer Kalvi 10th Maths Book uploaded as PDF and its book back question & answers solutions Guide PDF given below. Class 10th Standard New Syllabus Maths Books consist of 8 units. Each chapter is well designed and organized by the TN textbook corporation. Complete solutions for all 8 units of 10th std maths book back questions uploaded and available on our site. For Samacheer kalvi 10th Maths – Geometry Ex 4.5 book back question and answers solutions are given below:

Complete Samacheer Kalvi 10th Maths Guide available in PDF for free download. Check Samacheer Kalvi’s 10th Maths Unit 4 Chapter 5 Book Back Questions and Answers below:





TN Samacheer Kalvi 10th Maths Chapter 4 – Geometry Book Back Exercise is given below.

Chapter 4

Exercise 4.5 Geometry

1. If in triangles ABC and EDF, ABDE=BCFD then they will be similar, when
(1) ∠B = ∠E
(2) ∠A = ∠D
(3) ∠B = ∠D
(4) ∠A = ∠F
Solution:
(1) ∠B = ∠E
Hint:

10th Maths Samacheer Kalvi Book Back Answers

2.In, ∆LMN, ∠L = 60°, ∠M =50° . If ∆LMN ~ ∆PQR then the value of ∠R is
(1) 40°
(2) 70°
(3) 30°
(4) 110°
Solution:
(2) 70°

Samacheer Kalvi 10th Maths Book Back Answers

∆LMN ~ ∆PQR, ∠R = 70°.

3.If ∆ABC is an isosceles triangle with ∠C = 90° and AC = 5 cm, then AB is
(1) 2.5 cm
(2) 5 cm
(3) 10 cm
(4) 52–√ cm
Solution:
(4) 52–√cm
Hint:

Samacheer Kalvi 10th Maths Book Back Answers

 

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

10th Maths Book Back Answers

(1) 25 : 4
(2) 25 : 7
(3) 25 : 11
(4) 25 : 13
Solution:
(1) 25 : 4
Hint:
Ratio of the area of similar triangles is equal to the ratio of the square of their corresponding sides.
∴ 52 : 22 = 25 : 4

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

Samacheer Kalvi 10th Maths Book Back Answers

Solution:
(4) 15 cm
Hint:

Samacheer Kalvi 10th Maths Book Back Answers



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
(1) 1.4 cm
(2) 1.8 cm
(3) 1.2 cm
(4) 1.05 cm
Solution:
(1) 1.4 cm

Samacheer Kalvi 10th Maths Book Back Answers

3⋅62⋅1=2⋅4A⋅E
(3.6) (AE) = 2.1 × 2.4
AE = 1.4 cm

 

7.In a ∆ABC , AD is the bisector of ∠BAC . If AB = 8 cm, BD = 6 cm and DC = 3 cm. The length of the side AC is
(1) 6 cm
(2) 4 cm
(3) 3 cm
(4) 8 cm
Solution:
(2) 4 cm
Hint:

Samacheer Kalvi 10th Maths Book Back Answers

8. In the adjacent figure ∠BAC = 90° and AD ⊥ BC then

Samacheer Kalvi 10th Maths Book Back Answers

(1) BD.CD = BC2
(2) AB.AC = BC2
(3) BD.CD = AD2
(4) AB.AC = AD2
Solution:
(3) BD.CD = AD2

Samacheer Kalvi 10th Maths Book Back Answers

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?
(1) 13 cm
(2) 14 m
(3.) 15 m
(4) 12.8 m
Solution:
(1) 13 cm
Hint:

Samacheer Kalvi 10th Maths Book Back Answers

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

Samacheer Kalvi 10th Maths Book Back Answers

(1) 80°
(2) 85°
(3) 75°
(4) 90°
Solution:
(4) 90°
Hint:
PR = 26
QR = 24
∠PAQ = 90°
PQ = 10
PQ = 262−242−−−−−−−−√=100−−−√ = 10
∴ ∠PAQ = 90°

11.A tangent is perpendicular to the radius at the …………..
(1) centre
(2) point of contact
(3) infinity
(4) chord
Answer:
(2) point of contact

12.How many tangents can be drawn to the circle from an exterior point?
(1) one
(2) two
(3) infinite
(4) zero
Solution:
(2) 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
(1) 100°
(2) 110°
(3) 120°
(4) 130°
Solution:
(2) 110°
Hint:

Samacheer Kalvi 10th Maths Book Back Answers

14.In figure CP and CQ are tangents to a circle T 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
(1) 6 cm
(2) 5 cm
(3) 8 cm
(4) 4 cm
Solution:
(4) 4 cm

Samacheer Kalvi 10th Maths Book Back Answers

BQ = BR
CP = CQ = 11
BC = 7, ∴ BQ = CQ – BC
= 11 – 7 = 4
BR = BQ = 4cm

15.In figure if PR is tangent to the circle at P and O is the centre of the circle, then ∠POQ is
(1) 120°
(2) 100°
(3) 110°
(4) 90°
Solution:
(1) 120°
∠POQ = 180° -(30° + 30°)
= 180° – 60°
= 120°

Samacheer Kalvi 10th Maths Book Back Answers




Chapter 4 Geometry Unit Exercise 4

1.In the figure, if BD⊥AC and CE ∠ AB, prove that

Samacheer Kalvi 10th Maths Book Back Answers

Solution:
In the figure’s ∆AEC and ∆ADB.
We have ∠AEC =∠ADB = 90 (∵ CE ∠AB and BD ∠AC)
and ∠EAC =∠DAB
[Each equal to ∠A]
Therefore by AA-criterion of similarity, we have ∆AEC ~ ∆ADB
(ii) We have
∆AEC ~ ∆ADB [As proved above]
⇒ CABA=ECDB⇒CAAB=CEDB
Hence proved.

2.In the given figure AB||CD || EF . If AB = 6 cm, CD = x cm, EF = 4 cm, BD = 5 cm and DE = y cm. Find x and y.

Samacheer Kalvi 10th Maths Book Back Answers

Solution:
In the given figure, ∆AEF, and ∆ACD are similar ∆s.
∠AEF = ∠ACD = 90°
∠A = ∠A (common)∴ ∆AEF ~ ∆ACD (By AA criterion of similarity)

Samacheer Kalvi 10th Maths Book Back Answers

Samacheer Kalvi 10th Maths Book Back Answers

Substituting x = 2.4 cm in (3)

Samacheer Kalvi 10th Maths Book Back Answers

3. O is any point inside a triangle ABC. The bisector of ∠AOB , ∠BOC and ∠COA meet the sides AB, BC and CA in point D, E and F respectively.
Show that AD × BE × CF = DB × EC × FA
Solution:
In ∆AOB, OD is the bisector of ∠AOB.

Samacheer Kalvi 10th Maths Book Back Answers

In ∆BOC, OE is the bisector of ∠BOC
∴ OBOC=BEEC …………. (2)
In ∆COA, OF is the bisector of ∠COA.
∴ OCOA=CFFA …………… (3)
Multiplying the corresponding sides of (1), (2) and (3), we get

Samacheer Kalvi 10th Maths Book Back Answers

⇒ DB × EC × FA = AD × BE × CF
Hence proved.

4. In the figure, ABC is a triangle in which AB = AC. Points D and E are points on the side AB and AC respectively such that AD = AE . Show that the points B, C, E, and D lie on the same circle.

Samacheer Kalvi 10th Maths Book Back Answers

Solution:
In order to prove that the points B, C, E, and D are concyclic, it is sufficient to show that ∠ABC + ∠CED = 180° and ∠ACB + ∠BDE = 180°.

Samacheer Kalvi 10th Maths Book Back Answers

In ∆ABC, we have AB = AC and AD = AE.
⇒ AB – AD = AC – AE
⇒ DB = EC
Thus we have AD = AE and DB = EC. (By the converse of Thale’s theorem)
⇒ ADDB=AEEC ⇒ DE||BC
∠ABC = ∠ADE (corresponding angles)
⇒ ∠ABC + ∠BDE = ∠ADE + ∠BDE (Adding ∠BDE on both sides)
⇒ ∠ABC + ∠BDE = 180°
⇒ ∠ACB + ∠BDE = 180° (∵ AB = AC ∴ ∠ABC = ∠ACB)
Again DE || BC
⇒ ∠ACB = ∠AED
⇒ ∠ACB + ∠CED = ∠AED + ∠CED (Adding ∠CED on both sides).
⇒ ∠ACB + ∠CED = 180° and
⇒ ∠ABC + ∠CED = 180° (∵ ∠ABC = ∠ACB)
Thus BDEC is a quadrilateral such that
⇒ ∠ACB + ∠BDE = 180° and
⇒ ∠ABC + ∠CED = 180°
∴ BDEC is a cyclic quadrilateral. Hence B, C, E, and D are concyclic points.

5. Two trains leave a railway station at the same time. The first train travels due west and the second train is due north. The first train travels at a speed of 20 km/hr and the second train travels at 30 km/hr. After 2 hours, what is the distance between them?
Solution:
After 2 hours, let us assume that the first train is at A and the second is at B.

Samacheer Kalvi 10th Maths Book Back Answers

6.D is the mid point of side BC and AE⊥BC. If i BC a, AC = b, AB = c, ED = x, AD = p and AE = h , prove that
(i) b2 = p2 + ax + a24
(ii) c2 = p2 – ax + a24
(iii) b2 + c2 = 2p2 + a22
Solution:
From the figure, D is the mid point of BC.

Samacheer Kalvi 10th Maths Book Back Answers

We have ∠AED = 90°
∴ ∠ADE < 90° and ∠ADC > 90°
i.e. ∠ADE is acute and ∠ADC is obtuse,

(i) In ∆ADC, ∠ADC is an obtuse angle.
AC2 = AD2 + DC2 + 2DC × DE
⇒ AC2 = AD2 + 12 BC2 + 2 . 12 BC . DE
⇒ AC2 = AD2 + 14 BC2 + BC . DE
⇒ AC2 = AD2 + BC . DE + 14 BC2
⇒ b2 = p2 + ax + 14 a2
Hence proved.

(ii) In ∆ABD, ∠ADE is an acute angle.
AB2 = AD2 + BD2 – 2BD . DE
⇒ AB2 = AD2 + (12BC)2 – 2 × 12 BC . DE
⇒ AB2 = AD2 + 14 BC2 – BC . DE
⇒ AB2 = AD2 – BC . DE + 14 BC2
⇒ c2 = p2 – ax + 14 a2
Hence proved.

(iii) From (i) and (ii) we get .
AB2 + AC2 = 2AD2 + 12 BC2
i.e. c2 + b2 = 2p2 + a22
Hence it is proved.

7. A man whose eye level is 2 m above the ground wishes to find the height of a tree. He places a mirror horizontally on the ground 20 m from the tree and finds that if he stands at a point C which is 4 m from mirror B, he can see the reflection of the top of the tree. How height is the tree?
Solution:
From the figure; ∆DAC, ∆FBC are similar triangles and ∆ACE & ∆ABF are similar triangles.

Samacheer Kalvi 10th Maths Book Back Answers

∴ height of the tree h = 6x = 10 m.

8. An emu which is 8 ft tall standing at the foot of a pillar which is 30 ft height. It walks away from the pillar. The shadow of the emu falls beyond emu. What is the relation between the length of the shadow and the distance from the emu to the pillar?
Solution:
Let OA (emu shadow) the x and AB = y.
⇒ pillar’s shadow = OB = OA + AB
⇒ OB = x + y

Samacheer Kalvi 10th Maths Book Back Answers

9. Two circles intersect at A and B. From a point, P on one of the circle’s lines PAC and PBD are drawn intersecting the second circle at C and D. Prove that CD is parallel to the tangent at P.
Solution:
Let XY be the tangent at P.
TPT: CD is || to XY.
Construction: Join AB.
ABCD is a cyclic quadilateral.
∠BAC + ∠BDC= 180° ………… (1)
∠BDC = 180° – ∠BAC …………. (2)
Equating (1) and (2)
we get ∠BDC = ∠PAB
Similarly we get ∠PBA = ∠ACD
as XY is tangent to the circle at ‘P’
∠BPY = ∠PAB (by alternate segment there)

Samacheer Kalvi 10th Maths Book Back Answers

∴ ∠PAB = ∠PDC
∠BPY = ∠PDC
XY is parallel of CD.
Hence proved.

10. Let ABC be a triangle and D, E, F are points on the respective sides AB, BC, AC (or their extensions). Let AD: DB = 5 : 3, BE : EC = 3 : 2 and AC = 21. Find the length of the line segment CF.
Solution:

Samacheer Kalvi 10th Maths Book Back Answers

Other Important Links for 10th Maths Book Back Answers Solutions:

For 10th Maths Book Unit 4 book back question and answers, check the link – Samacheer kalvi 10th Maths Unit 4 Geometry

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