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 …….. A_13, 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

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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

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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

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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’ =³/₂ AB

B’C’ =³/₂ BC

AC’ =³/₂ AC

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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’ = ³/₄ 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 ∆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 =³/₄ AB

BC’ =³/₂ BC

A’C’ =³/₄ AC

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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 =⁴/₃ AB

BC’ =⁴/₃ BC

A’C’ =⁴/₃ AC

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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’ =⁵/₃ AB

B’C’ =⁵/₃ BC

AC’ =⁵/₃ AC

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