Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Students can download Maths Chapter 4 Geometry Ex 4.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.3

Question 1.
A man goes 18 m due east and then 24 m due north. Find the distance of his current position from the starting point?
Solution:
Let the initial position of the man be “O” and his final
position be “B”.
By Pythagoras theorem
In the right ∆ OAB,
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 1
OB2 = OA2 + AB2
= 182 + 242
= 324 + 576 = 900
OB = \(\sqrt { 900 }\) = 30
The distance of his current position is 30 m

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Question 2.
There are two paths that one can choose to go from Sarah’s house to James house. One way is to take C street, and the other way requires to take A street and then B street. How much shorter is the direct path along C street? (Using figure).
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 2
Solution:
Distance between Sarah House and James House using “C street”.
AC2 = AB2 + BC2
= 22 + 1.52
= 4 + 2.25 = 6.25
AC = \(\sqrt { 6.25 }\)
AC = 2.5 miles
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 3
Distance covered by using “A Street” and “B Street”
= (2 + 1.5) miles = 3.5 miles
Difference in distance = 3.5 miles – 2.5 miles = 1 mile

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Question 3.
To get from point A to point B you must avoid walking through a pond. You must walk 34 m south and 41 m east. To the nearest meter, how many meters would be saved if it were possible to make a way through the pond?
Solution:
In the right ∆ABC,
By Pythagoras theorem
AC2= AB2 + BC2 = 342 + 412
= 1156 + 1681 = 2837
AC = \(\sqrt { 2837 }\)
= 53.26 m
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 4
Through A one must walk (34m + 41m) 75 m to reach C.
The difference in Distance = 75 – 53.26
= 21.74 m

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Question 4.
In the rectangle WXYZ, XY + YZ = 17 cm, and XZ + YW = 26 cm.
Calculate the length and breadth of the rectangle?
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 5
Solution:
Let the length of the rectangle be “a” and the breadth of the rectangle be “b”.
XY + YZ = 17 cm
b + a = 17 …….. (1)
In the right ∆ WXZ,
XZ2 = WX2 + WZ2
(XZ)2 = a2 + b2
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 6
XZ = \(\sqrt{a^{2}+b^{2}}\)
Similarly WY = \(\sqrt{a^{2}+b^{2}}\) ⇒ XZ + WY = 26
2 \(\sqrt{a^{2}+b^{2}}\) = 26 ⇒ \(\sqrt{a^{2}+b^{2}}\) = 13
Squaring on both sides
a2 + b2 = 169
(a + b)2 – 2ab = 169
172 – 2ab = 169 ⇒ 289 – 169 = 2 ab
120 = 2 ab ⇒ ∴ ab = 60
a = \(\frac { 60 }{ b } \) ….. (2)
Substituting the value of a = \(\frac { 60 }{ b } \) in (1)
\(\frac { 60 }{ b } \) + b = 17
b2 – 17b + 60 = 0
(b – 2) (b – 5) = 0
b = 12 or b = 5
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 7
If b = 12 ⇒ a = 5
If b = 6 ⇒ a = 12
Lenght = 12 m and breadth = 5 m

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Question 5.
The hypotenuse of a right triangle is 6 m more than twice of the shortest side. If the third side is 2 m less than the hypotenuse, find the sides of the triangle.
Solution:
Let the shortest side of the right ∆ be x.
∴ Hypotenuse = 6 + 2x
Third side = 2x + 6 – 2
= 2x + 4
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 8
In the right triangle ABC,
AC2 = AB2 + BC2
(2x + 6)2 = x2 + (2x + 4)2
4x2 + 36 + 24x = x2 + 4x2 + 16 + 16x
0 = x2 – 24x + 16x – 36 + 16
∴ x2 – 8x – 20 = 0
(x – 10) (x + 2) = 0
x – 10 = 0 or x + 2 = 0
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 9
x = 10 or x = -2 (Negative value will be omitted)
The side AB = 10 m
The side BC = 2 (10) + 4 = 24 m
Hypotenuse AC = 2(10) + 6 = 26 m

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Question 6.
5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.
Solution:
“C” is the position of the foot of the ladder “A” is the position of the top of the ladder.
In the right ∆ABC,
BC2 = AC2 – AB2 = 52 – 42
= 25 – 16 = 9
BC = \(\sqrt { 9 }\) = 3m.
When the foot of the ladder moved 1.6 m toward the wall.
The distance between the foot of the ladder to the ground is
BE = 3 – 1.6 m
= 1.4 m
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 10
Let the distance moved upward on the wall be “h” m
The ladder touch the wall at (4 + h) M
In the right triangle BED,
ED2 = AB2 + BE2
52 = (4 + h)2 + (1.4)2
25 – 1.96= (4 + h)2
∴ 4 + h = \(\sqrt { 23.04 }\)
4 + h = 4. 8 m
h = 4.8 – 4
= 0.8 m
Distance moved upward on the wall = 0.8 m

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Question 7.
The perpendicular PS on the base QR of a ∆PQR intersects QR at S, such that QS = 3 SR. Prove that 2PQ2 = 2PR2 + QR2.
Solution:
Given QS = 3SR
QR = QS + SR
= 3SR + SR = 4SR
SR = \(\frac { 1 }{ 4 } \) QR …..(1)
QS = 3SR
SR = \(\frac { QS }{ 3 } \) ……..(2)
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 11
From (1) and (2) we get
\(\frac { 1 }{ 4 } \) QR = \(\frac { QS }{ 3 } \)
∴ QS = \(\frac { 3 }{ 4 } \) QR ………(3)
In the right ∆ PQS,
PQ2 = PS2 + QS2 ……….(4)
Similarly in ∆ PSR
PR2 = PS2 + SR2 ………..(5)
Subtract (4) and (5)
PQ2 – PR2 = PS2 + QS2 – PS2 – SR2
= QS2 – SR2
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 12
PQ2 – PR2 = \(\frac { 1 }{ 2 } \) QR2
2PQ2 – 2PR2 = QR2
2PQ2 = 2PR2 + QR2
Hence the proved.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Question 8.
In the adjacent figure, ABC is a right angled triangle with right angle at B and points D, E trisect BC. Prove that 8AE2 = 3AC2 + 5AD2.
Solution:
Since the Points D, E trisect BC.
BD = DE = CE
Let BD = DE = CE = x
BE = 2x and BC = 3x
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 13
In the right ∆ABD,
AD2 = AB2 + BD2
AD2 = AB2 + x2 ……….(1)
In the right ∆ABE,
AE2 = AB2 + 2BE2
AE2 = AB2 + 4X2 ………..(2) (BE = 2x)
In the right ∆ABC
AC2 = AB2 + BC2
AC2 = AB2 + 9x2 …………… (3) (BC = 3x)
R.H.S = 3AC2 + 5AD2
= 3[AB2 + 9x2] + 5 [AB2 + x2] [From (1) and (3)]
= 3AB2 + 27x2 + 5AB2 + 5x2
= 8AB2 + 32x2
= 8 (AB2 + 4 x2)
= 8AE2 [From (2)]
= R.H.S.
∴ 8AE2 = 3AC2 + 5AD2

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