**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.

- Draw line segment AB of 7.6 cm and draw a ray AX making an acute angle with line segment AB.
- Locate 13 (= 5 + 8) points, A
_{1}, A_{2}, A_{3}, A_{4}…….. A_{13,}on AX such that AA_{1}= A_{1}A_{2}= A_{2}A_{3}and so on. - Join BA13.
- 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 A_{5}C || A_{13}B. By applying Basic proportionality theorem for the triangle AA_{13}B, we obtain

AC/CB = AA_{5}/A_{5}A_{13} ——-(1)

From the figure, it can be observed that AA_{5} and A_{5}A_{13} contain 5 and 8 equal divisions of line segments respectively.

AA_{5}/A_{5}A_{13} = 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 A_{1}, A_{2}, A_{3} (as 3 is greater between 2 and 3) on line AX such that AA_{1} = A_{1}A_{2} = A_{2}A_{3}.

4: Join BA_{3} and draw a line through A_{2 }parallel to BA_{3 }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} = (AA_{2}/AA_{3})

⇒ ^{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:

- 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.
- Draw a ray AX making acute angle with line AB on the opposite side of vertex C.
- Locate 7 points, A
_{1}, A_{2}, A_{3}, A_{4 }A_{5}, A_{6}, A_{7}(as 7 is greater between 5and 7), on line AX such that AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7. - Join BA5 and draw a line through A7 parallel to BA5 to intersect extended line segment AB at point B’.
- 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 ∆AA_{5}B and ∆AA_{7}B’,

∠AA_{5} B = ∠AA_{7} B’ (Common)

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

∴ ∆AA_{5}B ∼ ∆AA_{7}B’ (AA similarity criterion)

⇒ ^{AB}/_{AB’} = (AA_{5}/AA_{7})

⇒ ^{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.

- 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.
- 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.
- Draw a ray AX making an acute angle with line segment AB on the opposite side of vertex C.
- Locate 3 points (as 3 is greater between 3 and 2) A
_{1}, A_{2}, and A_{3}on AX such that AA_{1}= A_{1}A_{2}= A_{2}A_{3}. - Join BA2 and draw a line through A3 parallel to BA2 to intersect extended line segment AB at point B’.
- 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}/_{2}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 ∆AA_{5}B and ∆AA_{7}B’,

∠AA_{2} B = ∠AA_{3} B’ (Common)

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

∴ ∆AA_{5}B ∼ ∆AA_{7}B’ (AA similarity criterion)

⇒ ^{AB}/_{AB’} = (AA_{5}/AA_{7})

⇒ ^{AB }/_{AB’} = ^{2}/_{3} ——–(2)

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

^{AB}/_{AB’} = ^{BC}/_{B’C’ }= ^{AC}/_{AC’} =^{2}/_{3}

⇒ AB’ =^{3}/_{2 }AB

B’C’ =^{3}/_{2 }BC

AC’ =^{3}/_{2 }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.

- Draw a ∆ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
- Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
- Locate 4 points (as 4 is greater in 3 and 4), B
_{1}, B_{2}, B_{3}, B_{4}, on line segment BX. - Join B
_{4}C and draw a line through B_{3}, parallel to B_{4}C intersecting BC at C’. - 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}/_{4 }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 ∆BB_{5}C’ and ∆BB_{4} C,

∠B_{3}BC’ = ∠B_{4} BC (Common)

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

∴ ∆BB_{3}C’ ∼ ∆BB_{4}C (AA similarity criterion)

⇒ ^{BC’}/_{BC} = (BB_{3}/BB_{4})

⇒ ^{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}/_{4 }AB

BC’ =^{3}/_{2 }BC

A’C’ =^{3}/_{4 }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.

- Draw a ∆ABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°.
- Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
- Locate 4 points (as 4 is greater in 4 and 3), B
_{1}, B_{2}, B_{3}, B_{4}, on BX. - Join B
_{3}Draw a line through B_{4}parallel to B_{3}C intersecting extended BC at C’. - 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}/_{3}BC, A’C’ = ^{4}/_{3}AC

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 ∆B_{3}BC and ∆B_{4}B C’,

∠B_{3}BC = ∠B_{4} BC’ (Common)

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

∴ ∆BB_{3}C ∼ ∆BB_{4}C’ (AA similarity criterion)

⇒ ^{BC}/_{BC’} = (BB_{3}/BB_{4})

⇒ ^{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}/_{3 }AB

BC’ =^{4}/_{3 }BC

A’C’ =^{4}/_{3 }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.

- Draw a line segment AB = 4 cm. Draw a ray SA making 90° with it.
- 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.
- Draw a ray AX making an acute angle with AB, opposite to vertex C.
- Locate 5 points (as 5 is greater in 5 and 3), A
_{1}, A_{2}, A_{3}, A_{4}, A_{5}, on line segment AX such that AA_{1}= A_{1}A_{2}= A_{2}A_{3}= A_{3}A_{4}= A_{4}A_{5}. - Join A
_{3}Draw a line through A_{5}parallel to A_{3}B intersecting extended line segment AB at B’. - 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}/_{3}BC, 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 ∆AA_{3}B and ∆AA_{5}B’,

∠A_{3}AB = ∠A_{5} AB’ (Common)

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

∴ ∆AA_{3}B ∼ ∆AA_{5}B’ (AA similarity criterion)

⇒ ^{AB}/_{AB’} = (AA_{3}/AA_{5})

⇒ ^{AB}/_{AB’} = ^{3}/_{5} ——–(2)

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

^{AB}/_{AB’} = ^{BC}/_{B’C’ }= ^{AC}/_{AC’} =^{3}/_{5}

⇒ AB’ =^{5}/_{3 }AB

B’C’ =^{5}/_{3 }BC

AC’ =^{5}/_{3 }AC

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