In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 11 Constructions Ex 11.1. These solutions are based on new NCERT Syllabus.

## NCERT Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1

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:

Steps of Construction :

I. Draw a line segment AB = 7.6 cm.

II. Draw a ray AX making an acute angle with AB.

III. Mark 13 = (8 + 5) equal points on AX, such that AX_{1} = X_{1}X_{2} = ……….. X_{12}X_{13}.

IV. Join points X_{13} and B.

V. From point X_{5}, draw X_{5}C || X_{13}B, which meets AB at C.

Thus, C divides AB in the ratio 5 : 8 On measuring the two parts, we get AC = 2.9 cm and CB = 4.7 cm.

Justification:

In ∆ABX_{13} and ∆ACX_{5}, we have

CX_{5} || BX_{13}

∴ \(\frac{A C}{C B}=\frac{A X_{5}}{X_{5} X_{13}}=\frac{5}{8}\) [By Thales theorem]

⇒ AC : CB = 5 : 8.

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:

Steps of Construction :

I. Construct a ∆ABC such that AB = 5 cm, BC = 7 cm and AC = 6 cm.

II. Draw a ray BX such that ∠CBX is an acute angle.

III. Mark 7 points of X_{1}, X_{2}, X_{3}, X_{4}, X_{5}, X_{6} and X_{7} on BX such that BX_{1} = X_{1}X_{2} = X_{2}X_{3} = X_{3}X_{4} – X_{4}X_{5} = X_{5}X_{6} = X_{6}X_{7}

IV Join X_{5} to C.

V. Draw a line through X_{7} intersecting BC (produced) at C’ such that X_{5}C || X_{7}C’

VI. Draw a line through C’ parallel to CA to intersect BA (produced) at A’.

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

Justification:

By construction, we have C’A’ || CA

∴ Using AA similarity, ∆ABC ~ ∆A’BC’

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:

Steps of construction :

I. Construct a ∆ABC such that BC = 6 cm, AB = 5 cm and ∠ABC = 60°.

II. Draw a ray BX such that ∠CBX is an acute angle.

Mark four points [greater of 3 and 4 in \(\frac{3}{4}\)] X_{1}, X_{2}, X_{3}, X_{4} on BX such that 4

BX_{1} = X_{1}X_{2} = X_{2}X_{3} = X_{3}X_{4}

IV. Join X_{4}C and draw a line through X_{3} parallel to X_{4}C to intersect BC at C’.

V. Also draw another line through C’ and parallel to CA to intersect BA at A’.

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

Justification:

In ∆BX_{4}C we have

X_{4}C || X_{3}C’ [By construction]

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:

Steps of Construction :

I. Construct the right triangle ABC such that ∠B = 90°, BC = 4 cm and BA = 3 cm.

II. Draw a ray BX such that an acute angle ∠CBX is formed.

III. Mark 5 points X_{1}, X_{2}, X_{3}, X_{4} and X_{5} on BX such that BX_{1} = X_{1}X_{2} = X_{2}X_{3} = X_{3}X_{4} = X_{4}X_{5}.

IV. Join X_{3}C.

V. Draw a line through X_{5} parallel to X_{3}C, intersecting the extended line segment BC at C’.

VI. Draw another 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]