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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:

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.

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.

To round to the nearest tenth, look at the tenth’s place (right after the decimal).

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.

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.

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

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

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.

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.