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

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 A1, A2, A3, … , A13 on AX so that AA1 = A1A2 = A2A3 = … = A12A13
• Step 3. Join BA13
• Step 4. Through the point A5, draw a line parallel to A13B(by making angle equal to ∠AA13B) at A5 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 A5C is parallel to A13B
$\cfrac { AA_{ 5 } }{ A_{ 5 }A_{ 13 } }$ = $\cfrac { AC }{ CB }$ (By the Basic Proportionality Theorem)

By construction, $\cfrac { AA_{ 5 } }{ A_{ 5 }A_{ 13 } }$ = $\cfrac { 5 }{ 8 }$
$\cfrac { AC }{ CB }$ = $\cfrac { 5 }{ 8 }$

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 $\cfrac { 2 }{ 3 }$ 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 $\overrightarrow { AX }$, making an acute angle with AB on opposite side of C
• Step 6 – Locate 3 points A1, A2 and A3 on $\overrightarrow { AX }$, such that AA1 = A1A2 = A2A3
• Step 7 – Join BA3
• Step 8 – Through A2, draw a line parallel to BA3, intersecting AB at B’
• Step 9 – Through B’, draw a line parallel to BC, intersecting AC at C’

The sides of △AB’C’ are $\cfrac { 2 }{ 3 }$ times that of △ABC.

Justification

In △AB’A2 and △ABA3
∠A2AB’ = ∠A3AB  (Common)
∠AA2B’ = ∠AA3B   (Corresponding angles)
∴ △AB’A2 ∼ △ABA3 (AA similarity criteria)
$\cfrac { A{ B }^{ \prime } }{ AB }$ = $\cfrac { AA_{ 2 } }{ AA_{ 3 } }$ = $\cfrac { 2 }{ 3 }$     …(i)
In △AB’C’ and △ABC
∠C’AB’ = ∠CAB  (Common)
∠AB’C’ = ∠ABC  (Corresponding angles)
∴ △AB’C’ ∼ △ABC
$\cfrac { AB' }{ AB }$ = $\cfrac { B'C' }{ BC }$ = $\cfrac { C'A }{ CA }$  …(ii)
From equation (i) and (ii)
$\cfrac { AB' }{ AB }$ = $\cfrac { B'C' }{ BC }$ = $\cfrac { C'A }{ CA }$ = $\cfrac { 2 }{ 3 }$
Therefore, sides of △AB’C’ are $\cfrac { 2 }{ 3 }$ times that of △ABC.

3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are $\cfrac { 7 }{ 5 }$ 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 $\overrightarrow { AX }$, making an acute angle with AB on opposite side of C
• Step 6 – Locate 7 points A1, A2,… , A7 on $\overrightarrow { AX }$, such that AA1 = A1A2 = … = A­6A7
• Step 7 – Join BA5
• Step 8 – Through A7, draw a line parallel to BA5, 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 $\cfrac { 7 }{ 5 }$ times that of △ABC.

Justification

In △AB’A7 and △ABA5
∠A7AB’ = ∠A5AB  (Common)
∠AA7B’ = ∠AA5B   (Corresponding angles)
∴ △AB’A7 ∼ △ABA5 (AA similarity criteria)
$\cfrac { AB' }{ AB }$ = $\cfrac { AA_{ 7 } }{ AA_{ 5 } }$ = $\cfrac { 7 }{ 5 }$    …(i)

In △AB’C’ and △ABC
∠C’AB’ = ∠CAB  (Common)
∠AB’C’ = ∠ABC  (Corresponding angles)
∴ △AB’C’ ∼ △ABC
$\cfrac { AB' }{ AB }$ = $\cfrac { B'C' }{ BC }$ = $\cfrac { C'A }{ CA }$…(ii)

From equation (i) and (ii)
$\cfrac { AB' }{ AB }$ = $\cfrac { B'C' }{ BC }$ = $\cfrac { C'A }{ CA }$ = $\cfrac { 7 }{ 5 }$
Therefore, sides of △AB’C’ are $\cfrac { 7 }{ 5 }$ 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 $1\cfrac { 1 }{ 2 }$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 $\overrightarrow { AX }$, making an acute angle with AB on opposite side of C
• Step 6 – Locate 3 points A1, A2, A3 on $\overrightarrow { AX }$, such that AA1 = A1A2 = A­2A3
• Step 7 – Join BA2
• Step 8 – Through A3, draw a line parallel to BA2, 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 $\cfrac { 3 }{ 2 }$ = $1\cfrac { 1 }{ 2 }$ times that of △ABC.

Justification

In △AB’A3 and △ABA2
∠A3AB’ = ∠A2AB   (Common)
∠AA3B’ = ∠AA2B    (Corresponding angles)
∴ △AB’A3 ∼ △ABA2 (AA similarity criteria)
$\cfrac { AB' }{ AB }$ = $\cfrac { AA_{ 3 } }{ AA_{ 2 } }$ = $\cfrac { 3 }{ 2 }$     …(i)

In △AB’C’ and △ABC
∠C’AB’ = ∠CAB  (Common)
∠AB’C’ = ∠ABC  (Corresponding angles)
∴ △AB’C’ ∼ △ABC
$\cfrac { AB' }{ AB }$ = $\cfrac { B'C' }{ BC }$ = $\cfrac { C'A }{ CA }$    …(ii)

From equation (i) and (ii)
$\cfrac { AB' }{ AB }$ = $\cfrac { B'C' }{ BC }$ = $\cfrac { C'A }{ CA }$ = $\cfrac { 3 }{ 2 }$ = $1\cfrac { 1 }{ 2 }$
Therefore, sides of △AB’C’ are $1\cfrac { 1 }{ 2 }$  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 $\cfrac { 3 }{ 4 }$ 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 $\overrightarrow { AX }$, making an acute angle with AB on opposite side of C
• Step 5 – Locate 7 points A1, A2,… , A4 on $\overrightarrow { AX }$, such that AA1 = A1A2 = … = A­3A4
• Step 6 – Join BA4
• Step 7 – Through A3, draw a line parallel to BA4, 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 $\cfrac { 3 }{ 4 }$ times that of △ABC.

Justification

In △AB’A3 and △ABA4
∠A3AB’ = ∠A4AB   (Common)
∠AA3B’ = ∠AA4B   (Corresponding angles)
∴ △AB’A3 ∼ △ABA4 (AA similarity criteria)
$\cfrac { AB' }{ AB }$ = $\cfrac { AA_{ 3 } }{ AA_{ 4 } }$ = $\cfrac { 3 }{ 4 }$  …(i)

In △AB’C’ and △ABC
∠C’AB’ = ∠CAB (Common)
∠AB’C’ = ∠ABC (Corresponding angles)
∴ △AB’C’ ∼ △ABC
$\cfrac { AB' }{ AB }$ = $\cfrac { B'C' }{ BC }$ = $\cfrac { C'A }{ CA }$  …(ii)

From equation (i) and (ii)
$\cfrac { AB' }{ AB }$ = $\cfrac { B'C' }{ BC }$ = $\cfrac { C'A }{ CA }$ = $\cfrac { 3 }{ 4 }$
Therefore, sides of △AB’C’ are $\cfrac { 3 }{ 4 }$ 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 $\cfrac { 4 }{ 3 }$ 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 $\overrightarrow { BX }$, making an acute angle with BC on opposite side of A
• Step 5 – Locate 4 points B1, B2,… , B4 on $\overrightarrow { BX }$, such that BB1 = B1B2 = … = B­3B4
• Step 6 – Join CB3
• Step 7 – Through B4, draw a line parallel to CB3, 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 $\cfrac { 4 }{ 3 }$ times that of △ABC.

Justification

In △BC’B4 and △BCB3
∠B4BC’ = ∠B3BC      (Common)
∠BB4C’ = ∠BB3C      (Corresponding angles)
∴ △BC’B4 ∼ △BCB3 (AA similarity criteria)
⇒  $\cfrac { BC' }{ BC }$ = $\cfrac { BB_{ 4 } }{ BB_{ 3 } }$ = $\cfrac { 4 }{ 3 }$ …(i)

In △A’BC’ and △ABC
∠A’BC’ = ∠ABC    (Common)
∠BC’A’ = ∠BCA    (Corresponding angles)
∴ △A’BC’ ∼ △ABC
$\cfrac { AB' }{ AB }$ = $\cfrac { B'C' }{ BC }$ = $\cfrac { C'A }{ CA }$ …(ii)

From equation (i) and (ii)
$\cfrac { AB' }{ AB }$ = $\cfrac { B'C' }{ BC }$ = $\cfrac { C'A }{ CA }$ = $\cfrac { 4 }{ 3 }$
Therefore, sides of △A’BC’ are $\cfrac { 4 }{ 3 }$ 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 $\cfrac { 5 }{ 3 }$ 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 $\overrightarrow { AX }$, making an acute angle with AB on opposite side of C
• Step 5 – Locate 5 points A1, A2,… , A5 on $\overrightarrow { AX }$, such that AA1 = A1A2 = … = A­4A5
• Step 6 – Join BA3
• Step 7 – Through A5, draw a line parallel to BA3, 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 $\cfrac { 5 }{ 3 }$ times that of △ABC.

Justification

In △AB’A5 and △ABA3
∠A5AB’ = ∠A3AB   (Common)
∠AA5B’ = ∠AA3B   (Corresponding angles)
∴ △AB’A5 ∼ △ABA3 (AA similarity criteria)
$\cfrac { AB' }{ AB }$ = $\cfrac { AA_{ 5 } }{ AA_{ 3 } }$ = $\cfrac { 5 }{ 3 }$  …(i)

In △AB’C’ and △ABC
∠C’AB’ = ∠CAB  (Common)
∠AB’C’ = ∠ABC   (Corresponding angles)
∴ △AB’C’ ∼ △ABC
$\cfrac { AB' }{ AB }$ = $\cfrac { B'C' }{ BC }$ = $\cfrac { C'A }{ CA }$   …(ii)

From equation (i) and (ii)
$\cfrac { AB' }{ AB }$ = $\cfrac { B'C' }{ BC }$ = $\cfrac { C'A }{ CA }$ = $\cfrac { 5 }{ 3 }$
Therefore, sides of △AB’C’ are $\cfrac { 5 }{ 3 }$ 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