KSEEB SSLC Class 10 Maths Solutions Chapter 6 Constructions Ex 6.1

KSEEB SSLC Class 10 Maths Solutions Chapter 6 Constructions Ex 6.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 6 Constructions Exercise 6.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 6 Constructions Exercise 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.

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 \(\frac{2}{3}\) 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 \(\frac{2}{3}\) 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 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 \(\frac{7}{5}\) 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 \(\frac{7}{5}\) 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 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 \(\frac{7}{5}\) 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 \(1 \frac{1}{2}\) 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 \(1 \frac{1}{2}\) 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’

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 \(\frac{3}{4}\) 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 \(\frac{3}{4}\) 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 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 \(\frac{4}{3}\) 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 \(\frac{4}{3}\) 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 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 \(\frac{5}{3}\) 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 \(\frac{5}{3}\) 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.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 6 Constructions Ex 6.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 6 Constructions Exercise 6.1, drop a comment below and we will get back to you at the earliest.