**Download 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.

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

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