**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 :**

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.

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:

Steps of Construction :

I. Draw a ∆ABC such that BC = 6 cm, AC = 5 cm and AB = 4 cm.

II. Draw a ray BX making an acute angle ∠CBX.

III. Mark three points [greater of 2 and 3 in \(\frac{2}{3}\)] X_{1}, X_{2}, X_{3} on BX_{1} such that BXj = X_{1}X_{2} = X_{2}X_{3}.

IV. Join X_{3}C.

V. Draw a line through X_{2} such that it is parallel to X_{3}C and meets BC at C’.

VI. Draw a line through C parallel to CA which intersect BA at A’.

Thus, ∆A’BC’ is the required similar triangle.

Justification :

By construction, we have X_{3}C || X_{2}C’

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.

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:

Steps of Construction :

I. Draw BC = 8 cm

II. Draw the perpendicular bisector of BC which intersects BC at D.

III. Mark a point A on the above perpendicular such that DA = 4 cm.

IV. Join AB and AC.

Thus, ∆ABC is the required isosceles triangle.

V. Now, draw a ray BX such that ∠CBX is an acute angle.

VI. On BX, mark three points [greater of 2 and 3 in \(\frac{2}{3}\)] X_{1}, X_{2} and X_{3} such that BX_{1} = X_{1}X_{2} = X_{2}X_{3}

VII. Join X_{2}C.

VIII. Draw a line through X_{3} parallel to X_{2}C and intersecting BC (extended) to C’.

IX. Draw a line through C’ parallel to CA intersecting BA (extended) to A’, thus, ∆A’BC’ is the required triangle.

Justification:

We have C’A’ || CA [By construction]

∴ Using AA similarity, ∆A’BC’ ~ ∆ABC

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

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:

Steps of Construction :

I. Construct a AABC such that BC = 7 cm, ∠B = 45°, ∠A = 105° and ∠C = 30°

II. Draw a ray BX making an acute angle ∠CBX with BC.

III. On BX, mark four points [greater of 4 and 3 in \(\frac{4}{3}\) ] X_{1}, X_{2}, X_{3} and X_{4} such that BX_{1} = X_{1}X_{2} = X_{2}X_{3} = X_{3}X_{4}.

IV. Join X_{3}C.

V. Draw a line through X_{4} parallel to X_{3}C intersecting BC(extended) at C’.

VI. Draw a line through C parallel to CA intersecting the extended line segment BA at A’.

Thus, ∆A’BC’ is the required triangle. Justification:

By construction, we have

C’A’ || CA

∴ ∆ABC ~ ∆A’BC’ [AA similarity]

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.