# Samacheer Kalvi 10th Maths Solutions Chapter 4 Unit Exercise 4 Pdf

## Samacheer Kalvi 10th Maths Solutions Chapter 4 Unit Exercise 4 Pdf

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

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

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

TNSCERT Class 10th Maths Pdf | all Unit Exercise Solution

#### Unit Exercise – 4

(1) In the figure, if BDAC an CEAB, prove that

(ii) CA/AB = CE/DB

Solution: (i) ∠ADB = 90° [∵ BD⊥AC] and ∠AEC = 90° [∵ CE⊥AB]

Now, ∠A is common angle. By angle angle properties,

Then, AC/AB = CE/DB

∴ CA/AB = CE/DB [Proved]

(2) In the given figure AB||CD||EF.

If AB = 6cm, CD = x cm, EF = 4 cm, BD = 5cm and DE = y cm. Find x and y.

Solution: Given that, AB = 6cm, CD = x cm, EF = 4cm, BD = 5cm and DE = y

Now, ∠BAE = ∠DEC = 90° and ∠E is common angle.

By angle angle properties,

∆DCE and ∆EAB are similar.

∴ ∆DCE ~ ∆EAB

Now, EC/EA = CD/AB

EC/EA = x/6 —– (i)

Now, ∠A is common angle.

Then by angal angel properties,

∆ACD ~ ∆AEF

Now, AC/AE = CD/EF

AC/AE = x/4 —– (ii)

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

EC/AE + AC/AE = x/6 + x/4

AC+CE/AE = 2x + 3x/12

AE/AE = 5x/12 [∵ AE = AC + CE]

5x/12 = 1

∴ x = 12/5 cm

Now, ∆DCE ~ ∆EAB

∴ DC/AB = ED/BE

x/6 = y/y+4

12/6×5 = y/y+5 [∵ x = 12/5]

30y = 12y + 60

30y – 12y = 60

18y = 60

y = 60/18

y = 10/3 cm

Thus the value of x = 12/5 cm and y = 10/3 cm.

(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: Now, ∆AOB, OD is the bisector of ∠AOB.

By Angle bisector theorem,

Now, ∆BOC, OE is the bisector of ∠BOC by angel bisector theorem,

OB/OC = BE/EC —- (ii)

Now, ∆AOC, OF is the bisector of ∠AOC.

By Angle bisector theorem,

OC/OA = CF/FA —– (iii)

Now, (i) × (ii) × (iii), we get,

OA/OB × OB/OC × OC/OA = AD/DB × BE/EC × CF/FA

∴ AD × BE × CF = DB × EC × FA [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 a same circle.

Solution: Also, ∠ADE + ∠BDE = ∠AED + ∠DEC = 180°

∴ ∠BDE = ∠DEC [∵∠ADE = ∠AED]

Now, ∆ABC,

AB = AC

∴ ∠ABC = ∠ACB

∴ ∠B = ∠C [∵ AB = AC]

Now, rectangle — DECB

∠BDE + ∠B + ∠C + ∠CED = 360°

∠BDE + ∠C + ∠C + ∠BDE = 360° [∵∠BDE = ∠DEC ∠B = ∠C]

2 ∠BDE + 2∠C = 360°

∠BOE + ∠C = 180°

Similarly, ∠DEC + ∠B = 180°

Thus, The points B, C, E and D lie on a circle [Proved]

(5) Two trains leave a railway station at the same time. The first train travels due west and the second train 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: Now, After 2 hours, first train travels = 20×2 = 40km

Second train travels = 30×2 = 60km

∴ AB = 60 km and AB = 40 km,

∆AOC, ∠A = 90°

By Pythagoras theorem,

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

(BC)2 = (60)2 + (40)2 = 3600 + 1600

(BC)2 = 5200 = 13×400 = (20√13)2

∴ BC = 20√13 cm

Thus, the two trains distance is 20√13 cm

(6) D is the midpoint of side BC and AEBC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, Prove that

(i) b2 = p2 + ax + a2/4

(ii) c2 = p2 – ax + a2/4

(iii) b2 + c2 = 2p2 + a2/2

Solution: ∆AED, ∠E = 90°

By Pythagoras theorem,

p2 = h2 + x2 —– (i)

(i) Now, ∆AEC, ∠E = 90°,

By Pythagoras theorem,

(AC)2 = (AE)2 + (EC)2

(AC)2 = (AE)2 + (ED + DC)2 [∵ ED + DC = EC]

(b)2 = (h)2 + (x + a/2)2 [∵ D is the midpoint BC BC = a]

b2 = h2 + x2 + a2/4 + x × x a/2

b2 = p2 + ax + a2/4 [By equation (i)]

∴ b2 = p2 +ax + a2/4 [Proved]

(ii) ∆ABE, ∠E = 90°

By Pythagoras theorem,

(AB)2 = (AE)2 + (BE)2

(C)2 = h2 + (BD – ED)2

(C)2 = p2 – x2 + (a/2 – x)2 [∵ equation (–) and D is the midpoint BC]

C2 = p2 – x2 + a2/4 + x2 – 2 × a/2 × x

C2 = p2 – ax + a2/4 [Proved]

(iii) We know that,

b2 = p2 + ax + a2/4 —– (i)

and C2 = p2 – ax + a2/ —– (ii)

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

b2 + c2 = p2 + ax + a2/4 + p2 = ax + a2/4

b2 + c2 = 2p2 + 2 × a2/4

b2 + c2 = 2p2 + a2/2 [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 the mirror B, he can see the reflection of the top of the tree. How height is the tree?

Solution: Let, tree height AE = h m.

Now, ∠DCA = ∠FBA = 90° and ∠A is common.

∴ ∆ACD ~ ∆ABF [By AA properties]

Now, ∆ACD ~ ∆ABF

CD/DF = AC/AB

2/BF = 24/20

BF = 20×2/24

∴ BF = 5/3

Now, ∠EAC = ∠FBC = 90° and ∠C is common angle

∴ ∆ACE ~ ∆BCF [By AA properties]

∴ AE/BF = AC/BC

h/(5/3) = 24/4

h = 24×5/4×3

h = 10m

Thus, the tree height is 10m

(8) An Emu which is 8 feet tall is standing at the foot of a pillar which is 30 feet high. 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, the shadow of the Emu EC = x and distance from the Emu to the pillar BC = y. Now, ∠ABE = ∠DCE = 90°

and ∠E is common angles

∴ ∆ABE ~ ∆DCE [By AA theorem]

Now, AB/DC = BE/CE

30/8 = x+y/x [∵ AB = 30 feet, DC = 8 feet, BE = BC + CE = y + x]

30x = 8x + 8y [Both side divides by 2]

15x = 4x + 4y

15x – 4x = 4y

11x = 4y

x = 4/11 y

Thus, required relation,

The length of the shadow = 4/11 × the distance from the Emu to the Pillar.

(9) Two circles intersect at A and B. From a point P on one of the circles 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: Now, ∠P’ PB = ∠PAB [By alternate segment theorem]

Now, ∠PAB + ∠BAC = 180° —- (i)

Now ABDC is a cyclic quadrilateral.

Then, ∠BAC + ∠BDC = 180° —– (ii),

Now from (i) and (ii), we get

∠PAB + ∠BAC = ∠BAC + ∠BDC

∠PAB = ∠BDC

∴ ∠P’ PB = ∠PAB = ∠BDC

∴ PP’ and DC are straight lines.

PD is a transversal alternate angles are equal.

∴ CD is parallel to the tangent P. [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: The —- AE, CD, BF are concurrent,

By Ceva’s theorem,

DE/EC × CF/FA × AD/DB = 1

3/2 × CF/FA × 5/3 = 1

[BE:EC = 3:2 BE/EC = 3/2]

CF/FA = 2/5

∴ CF:FA = 2:5

Let, CF = 2x an FA = 5x

Now, AC = CF + FA

21 = 2x + 5x

7x = 21

x = 3

Now, CF = 2x = 2 × 3 = 6

Thus, the length of the line segment CF = 6 units.

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

Updated: January 8, 2022 — 11:37 am