KSEEB Solutions for Class 10 Maths Chapter 6 Constructions Ex 6.1

Students can Download Class 10 Maths Chapter 6 Constructions Ex 6.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

Karnataka State Syllabus Class 10 Maths Chapter 6 Constructions Ex 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.
Answer:
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 = 3 cm, CB = 4.6 cm.

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.

Question 2.
Construct a triangle of sides 4 cm, 5 cm, and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.
Answer:
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 \(\frac{2}{3}\) of the corresponding sides of the first triangle.

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 an 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, ΔA’BC’ is the required triangle.

Question 3.
Construct a triangle with sides 5 cm, 6 cm, and 7 cm and then another triangle whose sides are \(\frac{7}{5}\) of the corresponding sides of the first triangle.
Answer:
Construct an ∆ABC having sides 5 cm, 6 cm, and 7 cm. Then construct another triangle whose sides are \(\frac{7}{5}\) of the corresponding sides of the first triangle.

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 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 \(\frac{7}{5}\) of the corresponding sides of the first triangle.

Question 4.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are \(1 \frac{1}{2}\) times the corresponding sides of the isosceles triangle.
Answer:
Construct an isosceles triangle with base 8 cm and altitude 4 cm and then another triangle whose sides are \(1 \frac{1}{2}\) times the corresponding sides of the isosceles triangle.

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., \(1 \frac{1}{2}=\frac{3}{2}\)
   ∆ABC ||| ∆AB’C’

Question 5.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm, and ∠ABC = 60°. Then construct a triangle whose sides are \(\frac{3}{4}\) of the corresponding sides of the triangle ABC.
Answer:
Construct a triangle ABC with sides BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are \(\frac{3}{4}\) of the corresponding sides of the triangle ABC.

Steps of Construction:

1. Construct a AABC having BC = 6 cm, AB = 5 cm, and ∠ABC = 60°.
2. Draw a ray Bx to form an 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

Question 6.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are \(\frac{4}{3}\) times the corresponding sides of ∆ABC
Answer:
Draw a triangle ABC with side BC = cm, ∠B = 45°, ∠A = 105°. The construct a triangle whose sides are \(\frac{4}{3}\) times the corresponding sides of ∆ABC.

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 an 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

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 \(\frac{5}{3}\) times the corresponding sides of the given triangle.
Answer:
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 \(\frac{5}{3}\) times the corresponding sides of the given triangle.

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.