10th Maths - Book Back Answers - Chapter 4 Unit Exercise 4.3 - English Medium Guides

 


    SSLC / 10th - Maths - Book Back Answers - Chapter 4 Unit Excercise 4.3 - English Medium

    Tamil Nadu Board 10th Standard Maths - Chapter 4 Unit Exercise 4.3: Book Back Answers and Solutions

        This post covers the book back answers and solutions for Chapter 4 Unit Exercise 4.3 – Maths from the Tamil Nadu State Board 10th Standard textbook. These detailed answers have been carefully prepared by our expert teachers at KalviTips.com.

        We have explained each answer in a simple, easy-to-understand format, highlighting important points step by step under the relevant subtopics. Students are advised to read and memorize these subtopics thoroughly. Once you understand the main concepts, you’ll be able to connect other related points with real-life examples and confidently present them in your tests and exams.

        By going through this material, you’ll gain a strong understanding of Chapter 4 Unit Exercise 4.3 along with the corresponding book back questions and answers (PDF format).

    Question Types Covered:

    • 1 Mark Questions: Choose the correct answer, Fill in the blanks, Identify the correct statement, Match the following 
    • 2 Mark Questions: Answer briefly 
    • 3, 4, and 5 Mark Questions: Answer in detail

    All answers are presented in a clear and student-friendly manner, focusing on key points to help you score full marks.

    All the best, Class 10 students! Prepare well and aim for top scores. Thank you!

    Chapter 4 Algebra Unit Ex 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?
    Answer Key:

    Let the initial position of the man be “O” and his final
    position be “B”.
    By Pythagoras theorem
    In the right ∆ OAB,
    OB2 = OA2 + AB2
    = 182 + 242
    = 324 + 576 = 900
    OB =
    900 = 30
    The distance of his current position is 30 m
     
    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).
    Answer Key:
    Distance between Sarah House and James House using “C street”.
    AC2 = AB2 + BC2
    = 22 + 1.52
    = 4 + 2.25 = 6.25
    AC =
    6.5
    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
     
    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?
    Answer Key:

    In the right ∆ABC,
    By Pythagoras theorem
    AC2= AB2 + BC2 = 342 + 412
    = 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
     
    4. In the rectangle WXYZ, XY + YZ = 17 cm, and XZ + YW = 26 cm.
    Calculate the length and breadth of the rectangle?
    Answer Key:
    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,
    XZ2 = WX2 + WZ2
    (XZ)2 = a2 + b2
     XZ = a2+b2
    Similarly WY = a2+b2 XZ + WY = 26
    2 a2+b2= 26 a2+b2 = 13
    Squaring on both sides
    a2 + b2 = 169
    (a + b)2 – 2ab = 169
    172 – 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
    b2 – 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
     
    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.
    Answer Key:

    Let the shortest side of the right ∆ be x.
    Hypotenuse = 6 + 2x
    Third side = 2x + 6 – 2
    = 2x + 4
    In the right triangle ABC,
    AC2 = AB2 + BC2
    (2x + 6)2 = x2 + (2x + 4)2
    4x2 + 36 + 24x = x2 + 4x2 + 16 + 16x
    0 = x2 – 24x + 16x – 36 + 16
    x2 – 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
     
    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.
    Answer Key:

    “C” is the position of the foot of the ladder “A” is the position of the top of the ladder.
    In the right ∆ABC,
    BC2 = AC2 – AB2 = 52 – 42
    = 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,
    ED2 = AB2 + BE2
    52 = (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
     
    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.
    Answer Key:

    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,
    PQ2 = PS2 + QS2 ……….(4)
    Similarly in ∆ PSR
    PR2 = PS2 + SR2 ………..(5)
    Subtract (4) and (5)
    PQ2 – PR2 = PS2 + QS2 – PS2 – SR2
    = QS2 – SR2
    PQ2 – PR2 = 12 QR2
    2PQ2 – 2PR2 = QR2
    2PQ2 = 2PR2 + QR2
    Hence the 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.
    Answer Key:

    Since the Points D, E trisect BC.
    BD = DE = CE
    Let BD = DE = CE = x
    BE = 2x and BC = 3x
    In the right ∆ABD,
    AD2 = AB2 + BD2
    AD2 = AB2 + x2 ……….(1)
    In the right ∆ABE,
    AE2 = AB2 + 2BE2
    AE2 = AB2 + 4X2 ………..(2) (BE = 2x)
    In the right ∆ABC
    AC2 = AB2 + BC2
    AC2 = AB2 + 9x2 …………… (3) (BC = 3x)
    R.H.S = 3AC2 + 5AD2
    = 3[AB2 + 9x2] + 5 [AB2 + x2] [From (1) and (3)]
    = 3AB2 + 27x2 + 5AB2 + 5x2
    = 8AB2 + 32x2
    = 8 (AB2 + 4 x2)
    = 8AE2 [From (2)]
    = R.H.S.
    8AE2 = 3AC2 + 5AD2

     


     


     

     

     

     






    0 Comments:

    Post a Comment

    Recent Posts

    Total Pageviews

    Code

    Blog Archive