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

**10th Maths Exercise 4.3 Samacheer Kalvi 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:

Using Pythagoras theorem

AC^{2} = AB^{2} + BC^{2}

= (18)^{2} + (24)^{2}

= 324 + 576

= 900

AC = \(\sqrt{900}\) = 30 m

∴ The distance from the starting point is 30 m.

**Ex 4.3 Class 10 Samacheer 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:

By using Pythagoras theorem

AC^{2} = AB^{2} + BC^{2}

= 2^{2} + (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.

**10th Maths Exercise 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:

By using Pythagoras

AC^{2} = AB^{2} + BC^{2}

= 34^{2} + 41^{2}

= 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.

**Ex 4.3 Class 10 Question 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

5^{2} + 12^{2} = 25 + 144 = 169 = 13^{2}

∴ In ∆XYZ = 13^{2} = 5^{2} + 12^{2} it is verified

∴ The length is 12 cm and the breadth is 5 cm.

**Class 10th Maths Samacheer Kalvi Solution 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 a is the shortest side.

c is the hypotenuse

b is the third side.

∴ The sides of the triangle are 10m, 24m, 26m.

Verification 26^{2} = 10^{2} + 24^{2}

676 = 100 + 576 = 676

**Exercise 4.3 Class 10 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:

Let the distance by which 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

DB^{2} + BE^{2} = DE^{2} ⇒ (1.4)^{2} + (4 + x)^{2}= 5^{2}

⇒ (4 + x)^{2} = 25 – 1.96 ⇒ (4 + x)^{2} = 23.04

⇒ 4 + x = \(\sqrt{23.04}\) = 4.8

⇒ x = 4.8 – 4 ⇒ x = 0.8 m

**Ex 4.3 Class 10 Maths Question 7.**

The perpendicular PS on the base QR of ∆PQR intersects QR at S, such that QS = 3 SR. Prove that 2PQ^{2} = 2PR^{2} + QR^{2}.

Solution:

In ∆PQS,

PQ^{2} = PS^{2} + QS^{2} ………… (1)

In ∆PSR,

PR^{2} = PS^{2} + SR^{2} ……….. (2)

(1) – (2) ⇒ PQ^{2} – PR^{2} = QS^{2} – SR^{2} …………. (3)

Hence it proved.

**Class 10 Maths 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 8AE^{2} = 3AC^{2} + 5AD^{2}.

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 AD^{2} = AB^{2} + BD^{2}

⇒ AD^{2} = AB^{2} + x^{2} ……………. (1)

AE^{2} = AB^{2} + BE^{2}

⇒ AB^{2} + (2x)^{2}

⇒ AE^{2} = AB^{2} + 4x^{2} ………… (2)

and AC^{2} = AB^{2} + BC^{2} = AB^{2} + (3x)^{2}

AC^{2} = AB^{2} + 9x^{2}

Now 8 AE^{2} – 3 AC^{2} – 5 AD^{2} = 8 (AB^{2} + 4x^{2}) – 3 (AB^{2} + 9x^{2}) – 5 (AB^{2} + x^{2})

= 8AB^{2} + 32x^{2} – 3AB^{2} – 27x^{2} – 5AB^{2} – 5x^{2}

= 0

∴ 8 AE^{2} – 3 AC^{2} – 5 AD^{2} = 0

8 AE^{2} = 3 AC^{2} + 5 AD^{2}.

Hence it is proved.