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
(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,
BC2 = AC2 + AB2
(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,
AC2 = AB + BC2
AC2 = (1.5)2 + (2) 2
AC2 = 2.25 + 4
AC2 = 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
AC2 = AB2 + BC2
AC2 = (34)2 + (41)2
AC2 = 1156 + 1681
AC2 = 2837
AC2 = 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 = q2 + p2
XZ = √P2 + q2 —– (ii)
Now, YX = p and WZ = q
∆WYZ, ∠W = 90°
By Pythagoras theorem,
(WY)2 = (ZY)2 + (WZ)2
(WY)2 = P2 + q2
WY = √P2 + q2 — (iii)
Now, given that
XZ + YW = 26
√P2 + q2 + √p2 + q2 = 26 [By Pythagoras (ii) and (iii)]
= 2√p2 + q2 = 26
√p2 + q2 = 13
p2 + q2 = 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
q2 + 60 = 17
q2 – 17q + 60 = 0
q2 – 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
4x2 + 24x + 36 = x2 + 4x2 + 16x + 16
5x2 + 16x + 16 – 4x2 – 24x – 36 = 0
x2 – 8x – 20 = 0
x2 – 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
h2 + 8h + 16 + 1.96 – 25 = 0
(h2 + 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 2PQ2 = 2PR2 + QR2
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 PQ2 = 2PR2 + QR2 [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:
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 + 9x2 —– (i)
∆ABE, ∠B = 90°,
By Pythagoras,
(AE)2 = (AB)2 + (BE)2
(AE)2 = (AB)2 + (2x)2
(AE)2 = (AB)2 + 4x2 —- (ii)
∆ABD, ∠B = 90°
By Pythagoras theorem,
(AD)2 = (AB)2 + (BD)2
(AD)2 = (AB)2 + (x)2
(AD)2 = (AB)2 + x2 —— (iii)
Now, 8 (AE)2
= 8 (AB)2 + 4x2) [By equation (ii)]
= 8 AB2 + 32x2
= 3AB2 + 5AB2 + 27x2 + 5x2
= 3AB2 + 27x2 + 5AB2 + 5x2
= 3 (AB2 + 9x2) + 5 (AB2 + x2)
= 3AC2 + 5AD2 [By Pythagoras (i) and (iii)]
= 3AC2 + 5AD2
Now, ∴ 8AE2 = 3AC2 + 5AD2 [Proved]
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