# 10th Maths Chapter 4 Geometry Exercise 4.5

10th Standard Maths Chapter 4 Geometry Exercise 4.5 Guide. TN SSLC Samacheer Kalvi Guide Chapter 4 Exercise 4.5 Book Back Answers & Solutions. 10th All Subject Guide – Click Here. Class 1 to 12 All Subject Book Back Answers – Click Here.

## 10th Maths Chapter 4 Geometry Exercise 4.5

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

(1) ∠B = ∠E
(2) ∠A = ∠D
(3) ∠B = ∠D
(4) ∠A = ∠F
Solution:
(1) ∠B = ∠E
Hint:

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°

∆LMN ~ ∆PQR, ∠R = 70°.

#### 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

(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

Solution:
(4) 15 cm
Hint:

#### 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 cmSolution: (1) 1.4 cm

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

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

#### 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 mSolution: (1) 13 cm Hint:

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

(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
(2) point of contact

#### 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 cmSolution: (4) 4 cm

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

## CHAPTER 4 GEOMETRY UNIT EXERCISE 4

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

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.

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

Substituting x = 2.4 cm in (3)

#### 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 × FASolution: In ∆AOB, OD is the bisector of ∠AOB.

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

⇒ 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.

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°.

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)
⇒ ∠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.

#### 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 + a22Solution: From the figure, D is the mid point of BC.

We have ∠AED = 90°

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.

∴ height of the tree h = 6x = 10 m.

#### 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)

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