# 10th Maths Chapter 4 Geometry Exercise 4.3

10th Standard Maths Chapter 4 Geometry Exercise 4.3 Guide. TN SSLC Samacheer Kalvi Guide Chapter 4 Exercise 4.3 Book Back Answers & Solutions. 10th All Subject Guide – Click Here. Class 1 to 12 All Subject Book Back Answers – Click Here.

### 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:
Using Pythagoras theorem

AC2 = AB2 + BC2
= (18)2 + (24)2
= 324 + 576
= 900
AC = 900−−−√ = 30 m
∴ The distance from the starting point is 30 m.

### 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:
By using Pythagoras theorem
AC2 = AB2 + BC2
= 22 + (1.5)2
= 4 + 2.25
= 6.25
AC = 2.5 miles.
If one chooses C street the distance from James house to Sarah’s house is 2.5 miles
If one chooses A street and B street he has to go 2 + 1.5 = 3.5 miles.
2.5 < 3.5, 3.5 – 2.5 = 1 Through C street is shorter by 1.0 miles.
∴ The direct path along C street is shorter by 1 mile.

### 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:
By using Pythagoras

AC2 = AB2 + BC2
= 342 + 412
= 1156+ 1681
= 2837
AC = 53.26 m
Through B one must walk 34 + 41 = 75 m walking through a pond one must comes only 53.2 m
∴ The difference is (75 – 53.26) m = 21.74 m
∴ To the nearest, one can save 21.74 m.

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

Solution:
XY + YZ = 17 cm …………. (1)
XZ + YW = 26 cm ………… (2)
(2) ⇒ XZ = 13, YW = 13
(∵ In rectangle diagonals are equal).
(1) ⇒ XY = 5, YZ = 12 XY + YZ = 17
⇒ Using Pythagoras theorem
52 + 122 = 25 + 144 = 169 = 132
∴ In ∆XYZ = 132 = 52 + 122 it is verified
∴ The length is 12 cm and the breadth is 5 cm.

### 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 a is the shortest side.
c is the hypotenuse
b is the third side.

∴ The sides of the triangle are 10m, 24m, 26m.
Verification 262 = 102 + 242
676 = 100 + 576 = 676

### 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:
Let the distance by which the top of the slide moves upwards be assumed as ‘x’.

From the diagram, DB = AB – AD
= 3 – 1.6 ⇒ DB = 1.4 m
also BE = BC + CE
= 4 + x
∴ DBE is a right angled triangle
DB2 + BE2 = DE2 ⇒ (1.4)2 + (4 + x)2= 52
⇒ (4 + x)2 = 25 – 1.96 ⇒ (4 + x)2 = 23.04
⇒ 4 + x = 23.04−−−−√ = 4.8
⇒ x = 4.8 – 4 ⇒ x = 0.8 m

### 7. The perpendicular PS on the base QR of ∆PQR intersects QR at S, such that QS = 3 SR. Prove that 2PQ2 = 2PR2 + QR2.

Solution:

In ∆PQS,
PQ2 = PS2 + QS2 ………… (1)
In ∆PSR,
PR2 = PS2 + SR2 ……….. (2)
(1) – (2) ⇒ PQ2 – PR2 = QS2 – SR2 …………. (3)

Hence it proved.

### 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 D and E are the points of trisection of BC,
therefore BD = DE = CE
Let BD = DE = CE = x
Then BE = 2x and BC = 3x
In right triangles ABD, ABE and ABC, (using Pythagoras theorem)
We have AD2 = AB2 + BD2
⇒ AD2 = AB2 + x2 ……………. (1)
AE2 = AB2 + BE2
⇒ AB2 + (2x)2
⇒ AE2 = AB2 + 4×2 ………… (2)
and AC2 = AB2 + BC2 = AB2 + (3x)2
AC2 = AB2 + 9×2
Now 8 AE2 – 3 AC2 – 5 AD2 = 8 (AB2 + 4×2) – 3 (AB2 + 9×2) – 5 (AB2 + x2)
= 8AB2 + 32×2 – 3AB2 – 27×2 – 5AB2 – 5×2
= 0
∴ 8 AE2 – 3 AC2 – 5 AD2 = 0
8 AE2 = 3 AC2 + 5 AD2.
Hence it is proved.