**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:

- Draw any ray AB, such that AB = 7.6 cm.
- Draw Ax ray at point A making acute angle.
- Locate the points A
_{1}, A_{2}, A_{3}, ………., A_{13}from A. - Join BA
_{13}. Draw BA_{13}|| A_{5}C.

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:

- ∆ABC is constructed, having AB = 4 cm, BC = 5 cm, CA = 6 cm.
- Draw Bx ray such that it forms acute angle to BC which is adjacent to vertex A.
- Locate 3 points on Bx such that BB
_{1}= B_{1}B_{2}= B_{2}B_{3}. - Join B
_{3}C. Draw a parallel line to B_{3}C which intersect BC at C. - 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 :

- First construct a triangle ABC having sides AB = 5cm, BC = 6 cm, CA = 7 cm.
- Draw a ray Bx such that it makes an acute angle at ‘B’.
- Locate Points B
_{1}, B2, B_{3}, B_{4}, B_{5}, B_{6}, By points on Bx. - Join B
_{5}C. Draw B_{5}C || B_{7}C’, it intersects at ‘C’ which is produced BC line. - Draw AC || C’A’, it meets produced line BC at A’.
- 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:

- Construct ∆ABC having base AB = 8 cm, altitude AD = 4 cm.
- ∠BAx acute angle is constructed at A. From ‘A’ locate A
_{1}, A_{2}, A_{3}. - Join A
_{2}B. Draw A_{2}B || A_{3}B’. B is marked on line AB produced. - Draw AB || B’C’. C’ is marked on AC produced and draw AC’.
- 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:

- Construct a AABC having BC = 6 cm, AB = 5 cm, and ∠ABC = 60°.
- Draw a ray Bx to form acute angle at B and mark BB
_{1}, B_{2}, B_{3}, at equal intervals of distance. - Join B
_{4}C. Draw B_{4}C || B_{3}C’. - Draw AC || A’C’.
- 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 :

- Construct a triangle ABC having BC = 7 cm, ∠A = 45° and ∠B = 105°.
- Draw Bx ray at B to form acute angle.
- Mark points B
_{1}, B_{2}, B_{3}, B_{4}from B. Draw B_{4}C’. - Draw B
_{4}C || B_{3}C’. Draw AC || A’C’. - 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:

- Construct Right angled triangle ABC having sides BC = 4 cm, BA = 3 cm, and ∠B = 90°.
- Draw a ray Bx to form acute angle at B.
- Mark points B
_{1}, B_{2}, B_{3}, B_{4}from B at equal intervals. - Join B
_{4}C. Draw B_{4}C || B_{3}C’. - Draw AC ‘|| C’A’, then ∆A’BC’ is constructed.
- 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.