# 10th Maths Chapter 4 Geometry Exercise 4.3

### 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.