10 Dec 2020 2:52 pm

**Tamilnadu, Samacheer, Kalvi, 10th, sslc, Maths, Solutions, Chapter 4, Algebra, 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,

OB^{2} = OA^{2} + AB^{2}

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

= 324 + 576 = 900

OB = 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”.

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

= 2^{2} + 1.5^{2}

= 4 + 2.25 = 6.25

AC = 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

AC^{2}= AB^{2} + BC^{2} = 34^{2} + 41^{2}

= 1156 + 1681 = 2837

AC = 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,

XZ^{2} = WX^{2} + WZ^{2}

(XZ)^{2} = a^{2} + b^{2}

XZ = a2+b2−−−−−−√

Similarly WY = a2+b2−−−−−−√ ⇒ XZ + WY = 26

2 a2+b2−−−−−−√ = 26 ⇒ a2+b2−−−−−−√ = 13

Squaring on both sides

a^{2} + b^{2} = 169

(a + b)^{2} – 2ab = 169

17^{2} – 2ab = 169 ⇒ 289 – 169 = 2 ab

120 = 2 ab ⇒ ∴ ab = 60

a = 60b ….. (2)

Substituting the value of a = 60b in (1)

60b + b = 17

b^{2} – 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,

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

(2x + 6)^{2} = x^{2} + (2x + 4)^{2}

4x^{2} + 36 + 24x = x^{2} + 4x^{2} + 16 + 16x

0 = x^{2} – 24x + 16x – 36 + 16

∴ x^{2} – 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,

BC^{2} = AC^{2} – AB^{2} = 5^{2} – 4^{2}

= 25 – 16 = 9

BC = 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,

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

5^{2} = (4 + h)^{2} + (1.4)^{2}

25 – 1.96= (4 + h)^{2}

∴ 4 + h = 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 2PQ^{2} = 2PR^{2} + QR^{2}.**Solution:**

Given QS = 3SR

QR = QS + SR

= 3SR + SR = 4SR

SR = 14 QR …..(1)

QS = 3SR

SR = QS3 ……..(2)

From (1) and (2) we get

14 QR = QS3

∴ QS = 34 QR ………(3)

In the right ∆ PQS,

PQ^{2} = PS^{2} + QS^{2} ……….(4)

Similarly in ∆ PSR

PR^{2} = PS^{2} + SR^{2} ………..(5)

Subtract (4) and (5)

PQ^{2} – PR^{2} = PS^{2} + QS^{2} – PS^{2} – SR^{2}

= QS^{2} – SR^{2}

PQ^{2} – PR^{2} = 12 QR^{2}

2PQ^{2} – 2PR^{2} = QR^{2}

2PQ^{2} = 2PR^{2} + QR^{2}

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

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

AD^{2 }= AB^{2} + x^{2} ……….(1)

In the right ∆ABE,

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

AE^{2} = AB^{2} + 4X^{2} ………..(2) (BE = 2x)

In the right ∆ABC

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

AC^{2} = AB^{2} + 9x^{2} …………… (3) (BC = 3x)

R.H.S = 3AC^{2} + 5AD^{2}

= 3[AB^{2} + 9x^{2}] + 5 [AB^{2} + x^{2}] [From (1) and (3)]

= 3AB^{2} + 27x^{2} + 5AB^{2} + 5x^{2}

= 8AB^{2} + 32x^{2}

= 8 (AB^{2} + 4 x^{2})

= 8AE^{2} [From (2)]

= R.H.S.

∴ 8AE^{2} = 3AC^{2} + 5AD^{2}