**Samacheer Kalvi 10th Maths Solutions Chapter 4 Exercise 4.3** **Pdf**

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**TN ****Samacheer Kalvi 10th Maths Solutions Chapter 4 – Geometry**

Board |
TNSCERT Class 10th Maths |

Class |
10 Class |

Subject |
Maths |

Chapter |
4 (Exercise 4.3) |

Chapter Name |
Geometry |

**TNSCERT Class 10th Maths Pdf | all Exercise Solution**

__Exercise – 4.3__

__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: **

Now, ∆ABC, ∠A = 90° by Pythagoras theorem,

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

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

(BC)^{2} = 324 + 576

(BC)^{2} = 900

(BC)^{2} = (30)^{2}

BC = ±30

BC = 30

BC ≠ -30 [Distance is not negative]

∴ Thus the distance of his current position from the starting point is 30m.

** **

**(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 B street and then A street. How much shorter is the direct path along C street? (Using figure).**

**Solution: **

Now, given figure,

AB = 1.5 miles

BC = 2 miles

∆ABC, ∠B = 40°

By Pythagoras theorem,

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

AC^{2} = (1.5)^{2} + (2)^{ 2}

AC^{2} = 2.25 + 4

AC^{2} = 6.25

AC = 2.5

Now, Sarah’s house to James house distance = 2.5 miles

Now, B-street and A – street to Sarah’s house to jame’s house distance = (1.5+2) miles = 3.5 miles

Now, B-street and A-street to C-street difference = (3.5-2.5) miles = 1 miles

Therefore, 1miles shorter is the direct Path along street.

** **

**(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 ∆ABC, ∠B = 90°

By Pythagoras theorem

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

AC^{2} = (34)^{2 }+ (41)^{2}

AC^{2} = 1156 + 1681

AC^{2} = 2837

AC^{2} = 53.26m

Now, (41 + 34) m = 75m

Required save in distance = (75 – 53.26) m = 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: **

Let, rectangle length = p cm

Let, rectangle breadth = q cm

Now, XY + YZ = 17

P + q = 17 — (i)

Now, ∆XYZ, ∠Y = 90°

By Pythagoras theorem,

(XZ)^{2} = (XY)^{2 }+ (YZ)^{2}

(XZ)^{2} = q^{2} + p^{2}

XZ = √P^{2} + q^{2} —– (ii)

Now, YX = p and WZ = q

∆WYZ, ∠W = 90°

By Pythagoras theorem,

(WY)^{2} = (ZY)^{2} + (WZ)^{2}

(WY)^{2} = P^{2} + q^{2}

WY = √P^{2} + q^{2} — (iii)

Now, given that

XZ + YW = 26

√P^{2} + q^{2} + √p^{2} + q^{2} = 26 [By Pythagoras (ii) and (iii)]

= 2√p^{2} + q^{2} = 26

√p^{2} + q^{2} = 13

p^{2} + q^{2} = 169 [Both side square]

(p + q)^{2} – 2p.q = 169

(17)^{2} – 169 = 2pq

2pq = 289 – 169 = 120

∴ pq = 60 —- (iv)

∴ pq = 60

p = 60/q

From (i) putting p = 60/q, we get

p+q = 17

60/q + q = 17

q^{2} + 60 = 17

q^{2} – 17q + 60 = 0

q^{2} – 12q – 5q + 60 = 0

q (q – 12) – 5 (q – 12) = 0

(q – 5) (q – 12) = 0

q – 5 = 0

q = 5

q – 12 = 0

q = 12

q ≠ 12 [∵ p>q]

From (i) putting q = 5, we get

p + q = 17

p + 5 = 17

p = 12

∴ Therefore, the rectangle length = 12cm and breadth = 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, shortest side = x m

∴ Hypotenuse = (2x + 6) m

Third side = (2x + 6) m

∆ABC is right triangle, ∠B = 90°

By Pythagoras theorem,

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

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

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

5x^{2} + 16x + 16 – 4x^{2} – 24x – 36 = 0

x^{2} – 8x – 20 = 0

x^{2} – 10x + 2x – 20 = 0

x (x – 10) + 2 (x – 10) = 0

(x – 10) – (x + 2) = 0

x – 10 = 0

x = 10

x + 1 = 0

x = -2

x ≠ -2 [∵ Length is not negative]

∴ AB = 10m

AC = 2x + 6 = 20 + 6 = 26m

BC = 2x + 4 = 2 × 20 + 4

= 24m

Thus, required side 26m, 24m and 10m

** **

**(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: **

Given, figure, AB = 4m AE = 5m

∆ABE, ∠B = 90°

By Pythagoras theorem,

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

(BE)^{2} = (AE)^{2} – (AB)^{2}

(BE)^{2} = (5)^{2} – (4)^{2}

(BE)^{2} = 25 – 16 = 4

BE = 3

Now, BC = (3 – 1.6) m = 1.4 m

Let, AD = h m, AC = 5m

BD = (h + 4) m

∆BCD, ∠B = 90°

By Pythagoras theorem,

(DC)^{2} = (BD)^{2} + (BC)^{2}

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

h^{2} + 8h + 16 + 1.96 – 25 = 0

(h^{2} + 8h + 16) = 23.04

(h + 4)^{2} = (4.8)^{2}

h + 4 = 4.8

h = 4.8 – 4

h = 0.8

This, required distance 0.8m

** **

**(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 that, QS = 3SR = SR = 1/3 QS

QR = QS + SR

QR = QS + 1/3 QS

QR = 4/3 QS

QS = 3/4 QR

SR = QR – QS

SR = QR – 3/4 QR

SR = 1/4 QR

Now, ∆PQS, ∠S = 90° and ∆PSR, ∠S = 90

By Pythagoras,

(PQ)^{2} = (PS)^{2} + (QS)^{2} —– (i)

By Pythagoras theorem,

(PR)^{2} = (PS)^{2} – (SR)^{2} —– (ii)

Now, (i) – (ii), we get

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

(PQ)^{2} – (PR)^{2} = (QS)^{2} – (SR)^{2}

(PQ)^{2} – (PR)^{2} = (3/4 QR)^{2} – (1/4QR)^{2} [∵ QS = 3/4 QR and SR = 1/ QR]

(PQ)^{2} – (PR)^{2} = 9(QR)^{2}/16 – (QR)^{2}/16

(PQ)^{2} – (PR)^{2} = 9(QR)^{2} – (QR)^{2}/16 = 8/16 (QR)^{2}

(PQ)^{2} – (PR)^{2} = 1/2 (QR)^{2}

2 (PQ)^{2} – 2(PR)^{2} = (QR)^{2}

2 PQ^{2 }= 2PR^{2} + QR^{2} [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 8AE ^{2} = 3AC^{2} + 5AD^{2} **

**Solution: **

Given BD = DE = EC = x (Let)

∴ BC = 3x, BE = 2x

∆ABC, ∠B = 90°,

By Pythagoras theorem,

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

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

(AC)^{2} = (AB)^{2} + 9x^{2} —– (i)

∆ABE, ∠B = 90°,

By Pythagoras,

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

(AE)^{2} = (AB)^{2} + (2x)^{2}

(AE)^{2} = (AB)^{2} + 4x^{2} —- (ii)

∆ABD, ∠B = 90°

By Pythagoras theorem,

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

(AD)^{2} = (AB)^{2} + (x)^{2}

(AD)^{2} = (AB)^{2} + x^{2} —— (iii)

Now, 8 (AE)^{2}

= 8 (AB)^{2} + 4x^{2}) [By equation (ii)]

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

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

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

= 3 (AB^{2} + 9x^{2}) + 5 (AB^{2} + x^{2})

= 3AC^{2} + 5AD^{2 }[By Pythagoras (i) and (iii)]

= 3AC^{2} + 5AD^{2}

Now, ∴ 8AE^{2} = 3AC^{2} + 5AD^{2} [Proved]

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