# 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 :
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.

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 $$\frac{2}{3}$$] X1, X2, X3 on BX1 such that BXj = X1X2 = X2X3.

IV. Join X3C.
V. Draw a line through X2 such that it is parallel to X3C 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 X3C || X2C’

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.

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 $$\frac{2}{3}$$] X1, X2 and X3 such that BX1 = X1X2 = X2X3
VII. Join X2C.
VIII. Draw a line through X3 parallel to X2C 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:

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

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 $$\frac{4}{3}$$ ] X1, X2, X3 and X4 such that BX1 = X1X2 = X2X3 = X3X4.
IV. Join X3C.
V. Draw a line through X4 parallel to X3C 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:

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.

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