Download NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.1 Constructions. This Exercise contains 7 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 10 for Maths NCERT solutions for other Chapters, you can click the link at the end of this Note.

### NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.1 Constructions

**NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.1 Constructions**

In each of the following, give the justification of the construction also:

**1. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.**

**Solution:**

A line segment AB of length 7.6 cm can be divided in the ratio 5:8 in following way:

**Steps of construction:**

**Step 1.**Draw any ray AX, making an acute angle with AB**Step 2.**Locate 13(= 5 + 8) points A_{1}, A_{2}, A_{3}, … , A_{13}on AX so that AA_{1}= A_{1}A_{2}= A_{2}A_{3}= … = A_{12}A_{13}**Step 3.**Join BA_{13}**Step 4.**Through the point A_{5}, draw a line parallel to A_{13}B(by making angle equal to ∠AA_{13}B) at A_{5}intersecting AB at the point C

Then, AC : CB = 5 : 8

The length of AB and AC are 2.9 cm and 4.7 cm respectively.

**Justification **Since A

_{5}C is parallel to A

_{13}B

∴ = (By the Basic Proportionality Theorem)

By construction, =

∴ =

This shows that C divides AB in the ratio 5 : 8.

**2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are of the corresponding sides of the first triangle.**

**Solution:**

**Steps of construction**

**Step 1 –**Draw a line segment AB = 4 cm**Step 2 –**With A as centre and radius 5 cm, draw an arc on one side of AB**Step 3 –**With B as centre and radius 6 cm, draw an arc intersecting the previous arc at C**Step 4 –**Join AC and BC**Step 5 –**Draw , making an acute angle with AB on opposite side of C**Step 6 –**Locate 3 points A_{1}, A_{2}and A_{3}on , such that AA_{1}= A_{1}A_{2}= A_{2}A_{3}**Step 7 –**Join BA_{3}**Step 8 –**Through A_{2}, draw a line parallel to BA_{3}, intersecting AB at B’**Step 9 –**Through B’, draw a line parallel to BC, intersecting AC at C’

The sides of △AB’C’ are times that of △ABC.

**Justification**

In △AB’A_{2} and △ABA_{3 }∠A_{2}AB’ = ∠A_{3}AB (Common)

∠AA_{2}B’ = ∠AA_{3}B (Corresponding angles)

∴ △AB’A_{2} ∼ △ABA_{3} (AA similarity criteria)

⇒ = = …(i)

In △AB’C’ and △ABC

∠C’AB’ = ∠CAB (Common)

∠AB’C’ = ∠ABC (Corresponding angles)

∴ △AB’C’ ∼ △ABC

⇒ = = …(ii)

From equation (i) and (ii)

= = =

Therefore, sides of △AB’C’ are times that of △ABC.

**3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are of the corresponding sides of the first triangle.**

**Solution:**

**Steps of construction**

**Step 1 –**Draw a line segment AB = 5 cm**Step 2 –**With A as centre and radius 6 cm, draw an arc on one side of AB**Step 3 –**With B as centre and radius 7 cm, draw an arc intersecting the previous arc at C**Step 4 –**Join AC and BC**Step 5 –**Draw , making an acute angle with AB on opposite side of C**Step 6 –**Locate 7 points A_{1}, A_{2},… , A_{7}on , such that AA_{1}= A_{1}A_{2}= … = A_{6}A_{7}**Step 7 –**Join BA_{5}**Step 8 –**Through A_{7}, draw a line parallel to BA_{5}, intersecting AB produced at B’**Step 9 –**Through B’, draw a line parallel to BC, intersecting AC produced at C’

The sides of △AB’C’ are times that of △ABC.

**Justification**

In △AB’A_{7} and △ABA_{5 }∠A_{7}AB’ = ∠A_{5}AB (Common)

∠AA_{7}B’ = ∠AA_{5}B (Corresponding angles)

∴ △AB’A_{7} ∼ △ABA_{5 }(AA similarity criteria)

⇒ = = …(i)

In △AB’C’ and △ABC

∠C’AB’ = ∠CAB (Common)

∠AB’C’ = ∠ABC (Corresponding angles)

∴ △AB’C’ ∼ △ABC

⇒ = = …(ii)

From equation (i) and (ii)

= = =

Therefore, sides of △AB’C’ are times that of △ABC.

**4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are times the corresponding sides of the isosceles triangle.**

**Solution:**

**Steps of construction**

**Step 1 –**Draw a line segment AB = 8 cm**Step 2 –**Draw a perpendicular bisector of AB**Step 3 –**Mark D as the mid-point of AB and C on the perpendicular bisector at 4 cm from D**Step 4 –**Join AC and BC**Step 5 –**Draw , making an acute angle with AB on opposite side of C**Step 6 –**Locate 3 points A_{1}, A_{2}, A_{3}on , such that AA_{1}= A_{1}A_{2}= A_{2}A_{3}**Step 7 –**Join BA_{2}**Step 8 –**Through A_{3}, draw a line parallel to BA_{2}, intersecting AB produced at B’**Step 9 –**Through B’, draw a line parallel to BC, intersecting AC produced at C’

The sides of △AB’C’ are = times that of △ABC.

**Justification**

In △AB’A_{3} and △ABA_{2 }∠A_{3}AB’ = ∠A_{2}AB (Common)

∠AA_{3}B’ = ∠AA_{2}B (Corresponding angles)

∴ △AB’A_{3} ∼ △ABA_{2} (AA similarity criteria)

⇒ = = …(i)

In △AB’C’ and △ABC

∠C’AB’ = ∠CAB (Common)

∠AB’C’ = ∠ABC (Corresponding angles)

∴ △AB’C’ ∼ △ABC

⇒ = = …(ii)

From equation (i) and (ii)

= = = =

Therefore, sides of △AB’C’ are times that of △ABC.

**5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are of the corresponding sides of the triangle ABC.**

**Solution:**

**Steps of construction**

**Step 1 –**Draw a line segment AB = 5 cm**Step 2 –**Make ∠CBA = 60° such that BC = 6 cm**Step 3 –**Join AC**Step 4 –**Draw , making an acute angle with AB on opposite side of C**Step 5 –**Locate 7 points A_{1}, A_{2},… , A_{4}on , such that AA_{1}= A_{1}A_{2}= … = A_{3}A_{4}**Step 6 –**Join BA_{4}**Step 7 –**Through A_{3}, draw a line parallel to BA_{4}, intersecting AB produced at B’**Step 8 –**Through B’, draw a line parallel to BC, intersecting AC produced at C’

The sides of △AB’C’ are times that of △ABC.

**Justification**

In △AB’A_{3} and △ABA_{4 }∠A_{3}AB’ = ∠A_{4}AB (Common)

∠AA_{3}B’ = ∠AA_{4}B (Corresponding angles)

∴ △AB’A_{3} ∼ △ABA_{4} (AA similarity criteria)

⇒ = = …(i)

In △AB’C’ and △ABC

∠C’AB’ = ∠CAB (Common)

∠AB’C’ = ∠ABC (Corresponding angles)

∴ △AB’C’ ∼ △ABC

⇒ = = …(ii)

From equation (i) and (ii)

= = =

Therefore, sides of △AB’C’ are times that of △ABC.

**6. Draw a triangle ABC with side BC = 7 cm, ∠ B = 45°, ∠ A = 105°. Then, construct a triangle whose sides are times the corresponding sides of △ABC.**

**Solution:**

In △ABC

∠ A + ∠ B + ∠ C = 180°

105° + 45° + ∠ C = 180°

150° + ∠ C = 180°

∠ C = 30°

** ****Steps of construction**

**Step 1 –**Draw a line segment BC = 7 cm**Step 2 –**Make ∠CBN = 45° and ∠BCM = 30°**Step 3 –**Mark the point of intersection of BN and CM as A**Step 4 –**Draw , making an acute angle with BC on opposite side of A**Step 5 –**Locate 4 points B_{1}, B_{2},… , B_{4}on , such that BB_{1}= B_{1}B_{2}= … = B_{3}B_{4}**Step 6 –**Join CB_{3}**Step 7 –**Through B_{4}, draw a line parallel to CB_{3}, intersecting BC produced at C’**Step 8 –**Through C’, draw a line parallel to AC, intersecting BA produced at A’

The sides of △A’BC’ are times that of △ABC.

**Justification**

In △BC’B_{4} and △BCB_{3 }∠B_{4}BC’ = ∠B_{3}BC (Common)

∠BB_{4}C’ = ∠BB_{3}C (Corresponding angles)

∴ △BC’B_{4} ∼ △BCB_{3} (AA similarity criteria)

⇒ = = …(i)

In △A’BC’ and △ABC

∠A’BC’ = ∠ABC (Common)

∠BC’A’ = ∠BCA (Corresponding angles)

∴ △A’BC’ ∼ △ABC

⇒ = = …(ii)

From equation (i) and (ii)

= = =

Therefore, sides of △A’BC’ are times that of △ABC.

**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 times the corresponding sides of the given triangle.**

**Solution:**

**Steps of construction**

**Step 1 –**Draw a line segment AB = 4 cm**Step 2 –**Draw ∠CBA = 90°, such that BC = 3 cm**Step 3 –**Join AC**Step 4 –**Draw , making an acute angle with AB on opposite side of C**Step 5 –**Locate 5 points A_{1}, A_{2},… , A_{5}on , such that AA_{1}= A_{1}A_{2}= … = A_{4}A_{5}**Step 6 –**Join BA_{3}**Step 7 –**Through A_{5}, draw a line parallel to BA_{3}, intersecting AB produced at B’**Step 8 –**Through B’, draw a line parallel to BC, intersecting AC produced at C’

The sides of △AB’C’ are times that of △ABC.

**Justification**

In △AB’A_{5} and △ABA_{3 }∠A_{5}AB’ = ∠A_{3}AB (Common)

∠AA_{5}B’ = ∠AA_{3}B (Corresponding angles)

∴ △AB’A_{5} ∼ △ABA_{3} (AA similarity criteria)

⇒ = = …(i)

∠C’AB’ = ∠CAB (Common)

∠AB’C’ = ∠ABC (Corresponding angles)

∴ △AB’C’ ∼ △ABC

⇒ = = …(ii)

From equation (i) and (ii)

= = =

Therefore, sides of △AB’C’ are times that of △ABC.

**NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1 Constructions, has been designed by the NCERT to test the knowledge of the student on the topic – Division of a Line Segment**

**Download NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.1 Constructions**