__Access Answers of Maths NCERT Chapter 11 – Constructions__

__Access Answers of Maths NCERT Chapter 11 – Constructions__

__Exercise 11.1 Page: 220__

__Exercise 11.1 Page: 220__

**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**.

Construction Procedure:

A line segment with a measure of 7.6 cm length is divided in the ratio of 5:8 as follows.

1. Draw line segment AB with the length measure of 7.6 cm

2. Draw a ray AX that makes an acute angle with line segment AB.

3. Locate the points i.e.,13 (= 5+8) points, such as A1, A2, A3, A4 …….. A13, on the ray AX such that it becomes AA1 = A1A2 = A2A3 and so on.

4. Join the line segment and the ray, BA13.

5. Through the point A5, draw a line parallel to BA13 which makes an angle equal to ∠AA13B

6. The point A5 which intersects the line AB at point C.

7. C is the point divides line segment AB of 7.6 cm in the required ratio of 5:8.

8. Now, measure the lengths of the line AC and CB. It comes out to the measure of 2.9 cm and 4.7 cm respectively.

Justification:

The construction of the given problem can be justified by proving that

AC/CB = 5/ 8

By construction, we have A5C || A13B. From Basic proportionality theorem for the triangle AA13B, we get

AC/CB =AA_{5}/A_{5}A_{13}….. (1)

From the figure constructed, it is observed that AA5 and A5A13 contain 5 and 8 equal divisions of line segments respectively.

Therefore, it becomes

AA_{5}/A_{5}A_{13}=5/8… (2)

Compare the equations (1) and (2), we obtain

AC/CB = 5/ 8

Hence, Justified.

**2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of**

**the corresponding sides of the first triangle.**

Construction Procedure:

1. Draw a line segment AB which measures 4 cm, i.e., AB = 4 cm.

2. Take the point A as centre, and draw an arc of radius 5 cm.

3. Similarly, take the point B as its centre, and draw an arc of radius 6 cm.

4. The arcs drawn will intersect each other at point C.

5. Now, we obtained AC = 5 cm and BC = 6 cm and therefore ΔABC is the required triangle.

6. Draw a ray AX which makes an acute angle with the line segment AB on the opposite side of vertex C.

7. Locate 3 points such as A1, A2, A3 (as 3 is greater between 2 and 3) on line AX such that it becomes AA1= A1A2 = A2A3.

8. Join the point BA3 and draw a line through A2which is parallel to the line BA3 that intersect AB at point B’.

9. Through the point B’, draw a line parallel to the line BC that intersect the line AC at C’.

10. Therefore, ΔAB’C’ is the required triangle.

Justification:

The construction of the given problem can be justified by proving that

AB’ = (2/3)AB

B’C’ = (2/3)BC

AC’= (2/3)AC

From the construction, we get B’C’ || BC

∴ ∠AB’C’ = ∠ABC (Corresponding angles)

In ΔAB’C’ and ΔABC,

∠ABC = ∠AB’C (Proved above)

∠BAC = ∠B’AC’ (Common)

∴ ΔAB’C’ ∼ ΔABC (From AA similarity criterion)

Therefore, AB’/AB = B’C’/BC= AC’/AC …. (1)

In ΔAAB’ and ΔAAB,

∠A_{2}AB’ =∠A_{3}AB (Common)

From the corresponding angles, we get,

∠AA_{2}B’ =∠AA_{3}B

Therefore, from the AA similarity criterion, we obtain

ΔAA_{2}B’ and AA_{3}B

So, AB’/AB = AA_{2}/AA_{3}

Therefore, AB’/AB = 2/3 ……. (2)

From the equations (1) and (2), we get

AB’/AB=B’C’/BC = AC’/ AC = 2/3

This can be written as

AB’ = (2/3)AB

B’C’ = (2/3)BC

AC’= (2/3)AC

Hence, justified.

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

Construction Procedure:

1. Draw a line segment AB =5 cm.

2. Take A and B as centre, and draw the arcs of radius 6 cm and 5 cm respectively.

3. These arcs will intersect each other at point C and therefore ΔABC is the required triangle with the length of sides as 5 cm, 6 cm, and 7 cm respectively.

4. Draw a ray AX which makes an acute angle with the line segment AB on the opposite side of vertex C.

5. Locate the 7 points such as A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7} (as 7 is greater between 5 and 7), on line AX such that it becomes AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5} = A_{5}A_{6} = A_{6}A_{7}

6. Join the points BA_{5} and draw a line from A_{7} to BA_{5} which is parallel to the line BA_{5} where it intersects the extended line segment AB at point B’.

7. Now, draw a line from B’ the extended line segment AC at C’ which is parallel to the line BC and it intersects to make a triangle.

8. Therefore, ΔAB’C’ is the required triangle.

Justification:

The construction of the given problem can be justified by proving that

AB’ = (7/5)AB

B’C’ = (7/5)BC

AC’= (7/5)AC

From the construction, we get B’C’ || BC

∴ ∠AB’C’ = ∠ABC (Corresponding angles)

In ΔAB’C’ and ΔABC,

∠ABC = ∠AB’C (Proved above)

∠BAC = ∠B’AC’ (Common)

∴ ΔAB’C’ ∼ ΔABC (From AA similarity criterion)

Therefore, AB’/AB = B’C’/BC= AC’/AC …. (1)

In ΔAA_{7}B’ and ΔAA_{5}B,

∠A_{7}AB’=∠A_{5}AB (Common)

From the corresponding angles, we get,

∠A A_{7}B’=∠A A_{5}B

Therefore, from the AA similarity criterion, we obtain

ΔA A_{2}B’ and A A_{3}B

So, AB’/AB = AA_{5}/AA_{7}

Therefore, AB /AB’ = 5/7 ……. (2)

From the equations (1) and (2), we get

AB’/AB = B’C’/BC = AC’/ AC = 7/5

This can be written as

AB’ = (7/5)AB

B’C’ = (7/5)BC

AC’= (7/5)AC

Hence, justified.

Construction Procedure:

1. Draw a line segment BC with the measure of 8 cm.

2. Now draw the perpendicular bisector of the line segment BC and intersect at the point D

3. Take the point D as centre and draw an arc with the radius of 4 cm which intersect the perpendicular bisector at the point A

4. Now join the lines AB and AC and the triangle is the required triangle.

5. Draw a ray BX which makes an acute angle with the line BC on the side opposite to the vertex A.

6. Locate the 3 points B_{1}, B_{2} and B_{3} on the ray BX such that BB_{1} = B_{1}B_{2} = B_{2}B_{3}

7. Join the points B_{2}C and draw a line from B_{3} which is parallel to the line B_{2}C where it intersects the extended line segment BC at point C’.

8. Now, draw a line from C’ the extended line segment AC at A’ which is parallel to the line AC and it intersects to make a triangle.

9. Therefore, ΔA’BC’ is the required triangle.

Justification:

The construction of the given problem can be justified by proving that

A’B = (3/2)AB

BC’ = (3/2)BC

A’C’= (3/2)AC

From the construction, we get A’C’ || AC

∴ ∠ A’C’B = ∠ACB (Corresponding angles)

In ΔA’BC’ and ΔABC,

∠B = ∠B (common)

∠A’BC’ = ∠ACB

∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

Therefore, A’B/AB = BC’/BC= A’C’/AC

Since the corresponding sides of the similar triangle are in the same ratio, it becomes

A’B/AB = BC’/BC= A’C’/AC = 3/2

Hence, justified.

**5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ****∠****ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.**

Construction Procedure:

1. Draw a ΔABC with base side BC = 6 cm, and AB = 5 cm and ∠ABC = 60°.

2. Draw a ray BX which makes an acute angle with BC on the opposite side of vertex A.

3. Locate 4 points (as 4 is greater in 3 and 4), such as B1, B2, B3, B4, on line segment BX.

4. Join the points B4C and also draw a line through B3, parallel to B4C intersecting the line segment BC at C’.

5. Draw a line through C’ parallel to the line AC which intersects the line AB at A’.

6. Therefore, ΔA’BC’ is the required triangle.

Justification:

The construction of the given problem can be justified by proving that

Since the scale factor is 3/4 , we need to prove

A’B = (3/4)AB

BC’ = (3/4)BC

A’C’= (3/4)AC

From the construction, we get A’C’ || AC

In ΔA’BC’ and ΔABC,

∴ ∠ A’C’B = ∠ACB (Corresponding angles)

∠B = ∠B (common)

∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

Since the corresponding sides of the similar triangle are in the same ratio, it becomes

Therefore, A’B/AB = BC’/BC= A’C’/AC

So, it becomes A’B/AB = BC’/BC= A’C’/AC = 3/4

Hence, justified.

**6. Draw a triangle ABC with side BC = 7 cm, ****∠**** B = 45°, ****∠**** A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of ∆ ABC.**

To find ∠C:

Given:

∠B = 45°, ∠A = 105°

We know that,

Sum of all interior angles in a triangle is 180°.

∠A+∠B +∠C = 180°

105°+45°+∠C = 180°

∠C = 180° − 150°

∠C = 30°

So, from the property of triangle, we get ∠C = 30°

Construction Procedure:

The required triangle can be drawn as follows.

1. Draw a ΔABC with side measures of base BC = 7 cm, ∠B = 45°, and ∠C = 30°.

2. Draw a ray BX makes an acute angle with BC on the opposite side of vertex A.

3. Locate 4 points (as 4 is greater in 4 and 3), such as B1, B2, B3, B4, on the ray BX.

4. Join the points B3C.

5. Draw a line through B4 parallel to B3C which intersects the extended line BC at C’.

6. Through C’, draw a line parallel to the line AC that intersects the extended line segment at C’.

7. Therefore, ΔA’BC’ is the required triangle.

Justification:

The construction of the given problem can be justified by proving that

Since the scale factor is 4/3, we need to prove

A’B = (4/3)AB

BC’ = (4/3)BC

A’C’= (4/3)AC

From the construction, we get A’C’ || AC

In ΔA’BC’ and ΔABC,

∴ ∠A’C’B = ∠ACB (Corresponding angles)

∠B = ∠B (common)

∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

Since the corresponding sides of the similar triangle are in the same ratio, it becomes

Therefore, A’B/AB = BC’/BC= A’C’/AC

So, it becomes A’B/AB = BC’/BC= A’C’/AC = 4/3

Hence, justified.

**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 5/3 times the corresponding sides of the given triangle.**

Given:

The sides other than hypotenuse are of lengths 4cm and 3cm. It defines that the sides are perpendicular to each other

Construction Procedure:

The required triangle can be drawn as follows.

1. Draw a line segment BC =3 cm.

2. Now measure and draw ∠= 90°

3. Take B as centre and draw an arc with the radius of 4 cm and intersects the ray at the point B.

4. Now, join the lines AC and the triangle ABC is the required triangle.

5. Draw a ray BX makes an acute angle with BC on the opposite side of vertex A.

6. Locate 5 such as B1, B2, B3, B4, on the ray BX such that such that BB_{1} = B_{1}B_{2} = B_{2}B_{3}= B_{3}B_{4} = B_{4}B_{5}

7. Join the points B3C.

8. Draw a line through B5 parallel to B3C which intersects the extended line BC at C’.

9. Through C’, draw a line parallel to the line AC that intersects the extended line AB at A’.

10. Therefore, ΔA’BC’ is the required triangle.

Justification:

The construction of the given problem can be justified by proving that

Since the scale factor is 5/3, we need to prove

A’B = (5/3)AB

BC’ = (5/3)BC

A’C’= (5/3)AC

From the construction, we get A’C’ || AC

In ΔA’BC’ and ΔABC,

∴ ∠ A’C’B = ∠ACB (Corresponding angles)

∠B = ∠B (common)

∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

Since the corresponding sides of the similar triangle are in the same ratio, it becomes

Therefore, A’B/AB = BC’/BC= A’C’/AC

So, it becomes A’B/AB = BC’/BC= A’C’/AC = 5/3

Hence, justified.