KSEEB SSLC Class 10 Maths Solutions Chapter 6 Constructions Ex 6.1

In this chapter, we provide KSEEB SSLC Class 10 Maths Solutions Chapter 6 Constructions Ex 6.1 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 10 Maths Solutions Chapter 6 Constructions Ex 6.1 pdf, free KSEEB SSLC Class 10 Maths Solutions Chapter 6 Constructions Ex 6.1 book pdf download. Now you will get step by step solution to each question.

Karnataka State Syllabus Class 10 Maths Solutions Chapter 6 Constructions Ex 6.1

In each of the following, give the justification of the construction also :
Construction Lesson Class 10 Exercise 6.1 Question 1.
Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
Solution:
Divide a line segment of 7.6 cm length in the ratio 5 : 8 and measure.
m : n = 5 : 8
m + n = 5 + 8 = 13
AC : CB = 5 : 8.
by measurement: AC = 3cm, CB = 4.6cm.
Construction Lesson Class 10 Exercise 6.1 KSEEB Solutions
Steps of Construction:

  1. Draw any ray AB, such that AB = 7.6 cm.
  2. Draw Ax ray at point A making acute angle.
  3. Locate the points A1, A2, A3, ………., A13 from A.
  4. Join BA13. Draw BA13 || A5C.
    Now, AC : CB = 5 : 8
    If measured, AC = 3 cm, CB = 4.6 cm.

10th Maths Construction Exercise 6.1 Question 2.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 23 of the corresponding sides of the first triangle.
Solution:
Construct a triangle ABC having sides 4 cm, 5 cm and 6 cm. For this we have to construct a triangle similar to it whose sides are 23 of the corresponding sides of the first triangle.
10th Maths Construction Exercise 6.1 KSEEB Solutions
Steps of Construction:

  1. ∆ABC is constructed, having AB = 4 cm, BC = 5 cm, CA = 6 cm.
  2. Draw Bx ray such that it forms acute angle to BC which is adjacent to vertex A.
  3. Locate 3 points on Bx such that BB1 = B1B2 = B2B3.
  4. Join B3C. Draw a parallel line to B3C which intersect BC at C.
  5. Draw a parallel line through C’ to CA and it intersects BA at A’.
    Now, A’BC’ is required triangle.

Exercise 6.1 Class 10 Construction Question 3.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 75 of the corresponding sides
of the first triangle.
Solution:
Construct an ∆ABC having sides 5 cm, 6 cm and 7 cm. Then construct another triangle whose sides are 75 of the corresponding sides of the first triangle.
Exercise 6.1 Class 10 Construction KSEEB Solutions
Steps of Construction :

  1. First construct a triangle ABC having sides AB = 5cm, BC = 6 cm, CA = 7 cm.
  2. Draw a ray Bx such that it makes an acute angle at ‘B’.
  3. Locate Points B1, B2, B3, B4, B5, B6, By points on Bx.
  4. Join B5C. Draw B5C || B7C’, it intersects at ‘C’ which is produced BC line.
  5. Draw AC || C’A’, it meets produced line BC at A’.
  6. Now required ∆A’BC’ is obtained whose sides are 75 of the corresponding sides of the first triangle.

KSEEB Solutions For Class 10 Maths Construction Question 4.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 112 times the corresponding sides of the isosceles triangle.
Solution:
Construct an isosceles triangle with base 8 cm and altitude 4 cm and then another triangle whose sides are 112 times the corresponding sides of the isosceles triangle.
KSEEB Solutions For Class 10 Maths Construction
Steps of Construction:

  1. Construct ∆ABC having base AB = 8 cm, altitude AD = 4 cm.
  2. ∠BAx acute angle is constructed at A. From ‘A’ locate A1, A2, A3.
  3. Join A2B. Draw A2B || A3B’. B is marked on line AB produced.
  4. Draw AB || B’C’. C’ is marked on AC produced and draw AC’.
  5. Now ∆AB’C’ is constructed which is similar to ∆ABC and corresponding sides,
    i.e., 112=32
    ∆ABC ||| ∆AB’C’

10th Maths Construction Exercise 6.1 Solutions Question 5.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm, and ∠ABC = 60°. Then construct a triangle whose sides are 34 of the corresponding sides of the triangle ABC.
Solution:
Construct a triangle ABC with sides BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 34 of the corresponding sides of the triangle ABC.
10th Maths Construction Exercise 6.1 Solutions KSEEB Solutions
Steps of Construction:

  1. Construct a AABC having BC = 6 cm, AB = 5 cm, and ∠ABC = 60°.
  2. Draw a ray Bx to form acute angle at B and mark BB1, B2, B3, at equal intervals of distance.
  3. Join B4C. Draw B4C || B3C’.
  4. Draw AC || A’C’.
  5. Now ∆ABC’ is equal to ∆ABC.
    ∴ ∆A’BC’ ||| ∆ABC
    3 : 4

Construction Class 10 KSEEB Question 6.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 43 times the corresponding sides of ∆ABC
Solution:
Draw a triangle ABC with side BC = cm, ∠B = 45°, ∠A = 105°. The construct a triangle whose sides are 43 times the corresponding sides of ∆ABC.
10th Maths Construction Exercise 6.1 Solutions KSEEB Solutions
Steps of Construction :

  1. Construct a triangle ABC having BC = 7 cm, ∠A = 45° and ∠B = 105°.
  2. Draw Bx ray at B to form acute angle.
  3. Mark points B1, B2, B3, B4 from B. Draw B4C’.
  4. Draw B4C || B3C’. Draw AC || A’C’.
  5. Now, ∆A’BC’ ||| ∆ABC is constructed.
    ∆A’BC’ : ∆ABC

Constructions Class 10 Exercise 6.1 Question 7.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 53 times the corresponding sides of the given triangle.
Solution:
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 53 times the corresponding sides of the given triangle.
Constructions Class 10 Exercise 6.1 KSEEB Solutions
Steps of Construction:

  1. Construct Right angled triangle ABC having sides BC = 4 cm, BA = 3 cm, and ∠B = 90°.
  2. Draw a ray Bx to form acute angle at B.
  3. Mark points B1, B2, B3, B4 from B at equal intervals.
  4. Join B4C. Draw B4C || B3C’.
  5. Draw AC ‘|| C’A’, then ∆A’BC’ is constructed.
  6. Now, ∆A’BC’ ||| ∆ABC.
    ∴ ∆A’BC’ : ∆ABC = 5 : 3.

All Chapter KSEEB Solutions For Class 10 Maths

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All Subject KSEEB Solutions For Class 10

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