Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

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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,

OB² = OA² + AB²
= 18² + 24²
= 324 + 576 = 900
OB = \(\sqrt { 900 }\) = 30
The distance of his current position is 30 m

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).

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

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

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

Through A one must walk (34m + 41m) 75 m to reach C.
The difference in Distance = 75 – 53.26
= 21.74 m

Question 4.
In the rectangle WXYZ, XY + YZ = 17 cm, and XZ + YW = 26 cm.
Calculate the length and breadth of the rectangle?

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

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

If b = 12 ⇒ a = 5
If b = 6 ⇒ a = 12
Lenght = 12 m and breadth = 5 m

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

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

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

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

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

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)

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

PQ² – PR² = \(\frac { 1 }{ 2 } \) QR²
2PQ² – 2PR² = QR²
2PQ² = 2PR² + QR²
Hence the proved.

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

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