# Constructions – Exercise 11.1 – Class 10

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

Solution:

A line segment of length 7.6 cm can be divided in the ratio of 5:8 as follows.

1. Draw line segment AB of 7.6 cm and draw a ray AX making an acute angle with line segment AB.
2. Locate 13 (= 5 + 8) points, A1, A2, A3, A4 …….. A13, on AX such that AA1 = A1A2 = A2A3 and so on.
3. Join BA13.
4. Through the point A5, draw a line parallel to BA13 (by making an angle equal to ∠AA13B) at A5 intersecting AB at point C.

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

The lengths of AC and CB can be measured. It comes out to 2.9 cm and 4.7 cm respectively.

The construction can be justified by proving that

AC/CB = 5/8

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

AC/CB = AA5/A5A13 ——-(1)

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

AA5/A5A13 = 5/8 ———–(2)

On comparing equations (1) and (2), we obtain

AC/CB = 5/8

2.  Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar to it whose sides are 2/3of the corresponding sides of the first triangle. Give the justification of the construction.

Solution:

1: Draw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5 cm radius. Similarly, taking point B as its centre, draw an arc of 6 cm radius. These arcs will intersect each other at point C. Now, AC = 5 cm and BC = 6 cm and ∆ABC is the required triangle.

2: Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.

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

4: Join BA3 and draw a line through A2 parallel to BA3 to intersect AB at point B’.

5: Draw a line through B’ parallel to the line BC to intersect AC at C’. ∆AB’C’ is the required triangle.

The construction can be justified by proving that

AB’ = 2/3 AB, B’C’ = 2/3 BC,AC’ = 2/3 AC

By construction, we have B’C’ || BC

∴ ∠A = ∠ABC (Corresponding angles)

In ∆AB’C’ and ∆ABC,

∠AB’C’ = ∠ABC (Proved)

∠B’AC’ = ∠BAC (common)

∆AB’C’ ~ ∆ABC (AA similarity criteirion)

That implies,

AB’/AB = B’C’/BC = AC’/AC ——–(1)

In ∆AA2B’ and ∆AA3B,

∠A2AB’ = ∠A3AB (Common)

∠AA2B’ = ∠AA3B (Corresponding angles)

∴ ∆AA2B’ ∼ ∆AA3B (AA similarity criterion)

AB’/AB = (AA2/AA3)

AB’ /AB = 2/­3 ——–(2)

From equations (1) and (2), we obtain

AB’/AB = B’C’/BC = AC’/AC = 2/3

⇒ AB’ =2/3 AB

B’C’ =2/3 BC

AC’ =2/3 AC

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. Give the justification of the construction.

Solution:

1. Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 5 cm radius respectively. Let these arcs intersect each other at point C. ∆ABC is the required triangle having length of sides as 5 cm, 6 cm, and 7 cm respectively.
2. Draw a ray AX making acute angle with line AB on the opposite side of vertex C.
3. Locate 7 points, A1, A2, A3, A4 A5, A6, A7 (as 7 is greater between 5and 7), on line AX such that AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7.
4. Join BA5 and draw a line through A7 parallel to BA5 to intersect extended line segment AB at point B’.
5. Draw a line through B’ parallel to BC intersecting the extended line segment AC at C’. ∆AB’C’ is the required triangle.

The construction can be justified by proving that

AB’ = 7/5 AB, B’C’ = 7/5 BC,AC’ = 7/5 AC

In ∆ABC and ∆AB’C’,

∠ABC = ∠AB’C’ (Corresponding angles)

∠BAC = ∠B’AC’ (common)

∆ABC ~ ∆AB’C’ (AA similarity criteirion)

That implies,

AB/AB’ = BC/B’C’ = AC/AC’ ——–(1)

In ∆AA5B and ∆AA7B’,

∠AA5 B = ∠AA7 B’ (Common)

∠AA5 B = ∠AA7 B’ (Corresponding angles)

∴ ∆AA5B ∼ ∆AA7B’ (AA similarity criterion)

AB/AB’ = (AA5/AA7)

AB /AB’ = 5/­7 ——–(2)

From equations (1) and (2), we obtain

AB/AB’ = BC/B’C’ = AC/AC’ =5/7

⇒ AB’ =7/5 AB

B’C’ =7/5 BC

AC’ =7/5 AC

4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose side are 1 ½ times the corresponding sides of the isosceles triangle. Give the justification of the construction.

Solution:

Let us assume that ∆ABC is an isosceles triangle having CA and CB of equal lengths, base AB of 8 cm, and AD is the altitude of 4 cm. A ∆AB’C’ whose sides are 3/2 times of ∆ABC can be drawn as follows.

1. Draw a line segment AB of 8 cm. Draw arcs of same radius on both sides of the line segment while taking point A and B as its centre. Let these arcs intersect each other at O and O’. Join OO’. Let OO’ intersect AB at D.
2. Taking D as centre, draw an arc of 4 cm radius which cuts the extended line segment OO’ at point C. An isosceles ∆ABC is formed, having CD (altitude) as 4 cm and AB (base) as 8 cm.
3. Draw a ray AX making an acute angle with line segment AB on the opposite side of vertex C.
4. Locate 3 points (as 3 is greater between 3 and 2) A1, A2, and A3 on AX such that AA1= A1A2 = A2A3.
5.  Join BA2 and draw a line through A3 parallel to BA2 to intersect extended line segment AB at point B’.
6. Draw a line through B’ parallel to BC intersecting the extended line segment AC at C’. ∆AB’C’ is the required triangle.

The construction can be justified by proving that

AB’ = 3/2 AB, B’C’ = 3/2 BC,AC’ = 3/2AC

In ∆ABC and ∆AB’C’,

∠ABC = ∠AB’C’ (Corresponding angles)

∠BAC = ∠B’AC’ (common)

∆ABC ~ ∆AB’C’ (AA similarity criteirion)

That implies,

AB/AB’ = BC/B’C’ AC/AC’ ——–(1)

In ∆AA5B and ∆AA7B’,

∠AA2 B = ∠AA3 B’ (Common)

∠AA B = ∠AA3 B’ (Corresponding angles)

∴ ∆AA5B ∼ ∆AA7B’ (AA similarity criterion)

⇒ AB/AB’ = (AA5/AA7)

⇒ AB /AB’ = 2/3 ——–(2)

From equations (1) and (2), we obtain

AB/AB’ = BC/B’C’ AC/AC’ =2/3

⇒ AB’ =3/AB

B’C’ =3/BC

AC’ =3/AC

5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are ¾ of the corresponding sides of the triangle ABC. Give the justification of the construction

Solution:

A ∆A’BC’ whose sides are ¾ of the corresponding sides of ∆ABC can be drawn as follows.

1. Draw a ∆ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
2. Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
3. Locate 4 points (as 4 is greater in 3 and 4), B1, B2, B3, B4, on line segment BX.
4. Join B4C and draw a line through B3, parallel to B4C intersecting BC at C’.
5. Draw a line through C’ parallel to AC intersecting AB at A’. ∆A’BC’ is the required triangle.

The construction can be justified by proving that

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

In ∆A’BC’ and ∆ABC,

∠A’C’B = ∠ACB (Corresponding angles)

∠A’BC’ = ∠BAC (common)

∆A’BC’ ~ ∆ABC (AA similarity criteirion)

That implies,

AB/AB’ = BC/B’C’ AC/AC’ ——–(1)

In ∆BB5C’ and ∆BB4 C,

∠B3BC’ = ∠B4 BC (Common)

∠BB3C’ = ∠BB4 C (Corresponding angles)

∴ ∆BB3C’ ∼ ∆BB4C (AA similarity criterion)

⇒ BC’/BC = (BB3/BB4)

⇒ BC’ /BC = 3/4 ——–(2)

From equations (1) and (2), we obtain

A’B/AB = BC’/BC A’C’/AC =3/4

⇒ A’B =3/AB

BC’ =3/BC

A’C’ =3/AC

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 side of ∆ABC. Give the justification of the construction

Solution:

∠B = 45°, ∠A = 105°

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

∠A + ∠B + ∠C = 180°

105° + 45° + ∠C = 180°

∠C = 180° − 150°

∠C = 30°

The required triangle can be drawn as follows.

1. Draw a ∆ABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°.
2. Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
3. Locate 4 points (as 4 is greater in 4 and 3), B1, B2, B3, B4, on BX.
4. Join B3 Draw a line through B4 parallel to B3C intersecting extended BC at C’.
5. Through C’, draw a line parallel to AC intersecting extended line segment at C’. ∆A’BC’ is the required triangle.

The construction can be justified by proving that

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

In ∆ABC and ∆A’BC’,

∠ABC = ∠A’BC’ (Common angles)

∠ACB = ∠A’C’B (corresponding angles)

∆ABC ~ ∆A’BC’ (AA similarity criteirion)

That implies,

AB/A’B = BC/BC’ AC/A’C’ ——–(1)

In ∆B3BC and ∆B4B C’,

∠B3BC = ∠B4 BC’ (Common)

∠BB3C = ∠BB4 C’ (Corresponding angles)

∴ ∆BB3C ∼ ∆BB4C’ (AA similarity criterion)

⇒ BC/BC’ = (BB3/BB4)

⇒ BC/BC’ = 3/4 ——–(2)

From equations (1) and (2), we obtain

AB/A’B = BC/BC’ AC/A’C’ =3/4

⇒ A’B =4/AB

BC’ =4/BC

A’C’ =4/AC

7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. the construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle. Give the justification of the construction.

Solution:

It is given that sides other than hypotenuse are of lengths 4 cm and 3 cm. Clearly, these will be perpendicular to each other. The required triangle can be drawn as follows.

1. Draw a line segment AB = 4 cm. Draw a ray SA making 90° with it.
2. Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C. Join BC. ∆ABC is the required triangle.
3. Draw a ray AX making an acute angle with AB, opposite to vertex C.
4. Locate 5 points (as 5 is greater in 5 and 3), A1, A2, A3, A4, A5, on line segment AX such that AA1 = A1A2 = A2A3 = A3A4 = A4A5.
5. Join A3 Draw a line through A5 parallel to A3B intersecting extended line segment AB at B’.
6. Through B’, draw a line parallel to BC intersecting extended line segment AC at C’. ∆AB’C’ is the required triangle

The construction can be justified by proving that

AB’ = 5/3 AB, B’C’ = 5/3BC, AC’ = 5/3 AC

In ∆ABC and ∆AB’C’,

∠ABC = ∠AB’C’ (Corresponding angles)

∠BAC = ∠B’AC’ (common angles)

∆ABC ~ ∆AB’C’ (AA similarity criteirion)

That implies,

AB/AB’ = BC/B’C’ AC/AC’ ——–(1)

In ∆AA3B and ∆AA5B’,

∠A3AB = ∠A5 AB’ (Common)

∠AA3B = ∠AA5 B’ (Corresponding angles)

∴ ∆AA3B ∼ ∆AA5B’ (AA similarity criterion)

⇒ AB/AB’ = (AA3/AA5)

⇒ AB/AB’ = 3/5 ——–(2)

From equations (1) and (2), we obtain

AB/AB’ = BC/B’C’ AC/AC’ =3/5

⇒ AB’ =5/AB

B’C’ =5/BC

AC’ =5/AC