**Question** 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.

**Answer**:

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

**Step 1** Draw line segment AB of 7.6 cm and draw a ray AX making an acute angle with line segment AB.

**Step 2** 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.

**Step 3** Join BA_{13}.

**Step 4 **Through the point A_{5}, draw a line parallel to BA_{13} (by making an angle equal to ∠AA_{13}B) at A_{5} 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.

**Justification**

The construction can be justified by proving that

By construction, we have A_{5}C || A_{13}B. By applying Basic proportionality theorem for the triangle AA_{13}B, we obtain

… (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.

… (2)

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

This justifies the construction.

**Question** 2:

Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar to it whose sides are of the corresponding sides of the first triangle.

Give the justification of the construction.

**Answer**:

**Step 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 ΔABC is the required triangle.

**Step 2**

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

**Step 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}.

**Step 4**

Join BA_{3} and draw a line through A_{2 }parallel to BA_{3} to intersect AB at point B’.

**Step 5**

Draw a line through B’ parallel to the line BC to intersect AC at C’.

ΔAB’C’ is the required triangle.

**Justification**

The construction can be justified by proving that

By construction, we have B’C’ || BC

∴ ∠A = ∠ABC (Corresponding angles)

In ΔAB’C’ and ΔABC,

∠ = ∠ABC (Proved above)

∠ = ∠BAC (Common)

∴ Δ ∼ ΔABC (AA similarity criterion)

… (1)

In ΔAA_{2}B’ and ΔAA_{3}B,

∠A_{2}AB’ = ∠A_{3}AB (Common)

∠AA_{2}B’ = ∠AA_{3}B (Corresponding angles)

∴ ΔAA_{2}B’ ∼ ΔAA_{3}B (AA similarity criterion)

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

This justifies the construction.

**Question** 3:

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

Give the justification of the construction.

**Answer**:

**Step 1**

Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 7 cm radius respectively. Let these arcs intersect each other at point C. ΔABC is the required triangle having length of sides as 5 cm, 6 cm, and 7 cm respectively.

**Step 2**

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

**Step 3**

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 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}.

**Step 4**

Join BA_{5} and draw a line through A_{7} parallel to BA_{5} to intersect extended line segment AB at point B’.

**Step 5**

Draw a line through B’ parallel to BC intersecting the extended line segment AC at C’. ΔAB’C’ is the required triangle.

**Justification**

The construction can be justified by proving that

In ΔABC and ΔAB’C’,

∠ABC = ∠AB’C’ (Corresponding angles)

∠BAC = ∠B’AC’ (Common)

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

… (1)

In ΔAA_{5}B and ΔAA_{7}B’,

∠A_{5}AB = ∠A_{7}AB’ (Common)

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

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

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

⇒

This justifies the construction.

**Question** 4:

Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose side are ** **times the corresponding sides of the isosceles triangle.

Give the justification of the construction.

**Answer**:

Let us assume that ΔABC 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 ΔAB’C’ whose sides are times of ΔABC can be drawn as follows.

**Step 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.

**Step 2**

Taking D as centre, draw an arc of 4 cm radius which cuts the extended line segment OO’ at point C. An isosceles ΔABC is formed, having CD (altitude) as 4 cm and AB (base) as 8 cm.

**Step 3**

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

**Step 4**

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}.

**Step 5**

Join BA_{2} and draw a line through A_{3} parallel to BA_{2} to intersect extended line segment AB at point B’.

**Step 6**

Draw a line through B’ parallel to BC intersecting the extended line segment AC at C’. ΔAB’C’ is the required triangle.

**Justification**

The construction can be justified by proving that

In ΔABC and ΔAB’C’,

∠ABC = ∠AB’C’ (Corresponding angles)

∠BAC = ∠B’AC’ (Common)

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

… (1)

In ΔAA_{2}B and ΔAA_{3}B’,

∠A_{2}AB = ∠A_{3}AB’ (Common)

∠AA_{2}B = ∠AA_{3}B’ (Corresponding angles)

∴ ΔAA_{2}B ∼ ΔAA_{3}B’ (AA similarity criterion)

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

⇒

This justifies the construction.

**Question** 5:

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

Give the justification of the construction.

**Answer**:

A ΔA’BC’ whose sides are of the corresponding sides of ΔABC can be drawn as follows.

**Step 1**

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

**Step 2**

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

**Step 3**

Locate 4 points (as 4 is greater in 3 and 4), B_{1}, B_{2}, B_{3}, B_{4}, on line segment BX.

**Step 4**

Join B_{4}C and draw a line through B_{3}, parallel to B_{4}C intersecting BC at C’.

**Step 5**

Draw a line through C’ parallel to AC intersecting AB at A’. ΔA’BC’ is the required triangle.

**Justification**

The construction can be justified by proving

In ΔA’BC’ and ΔABC,

∠A’C’B = ∠ACB (Corresponding angles)

∠A’BC’ = ∠ABC (Common)

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

… (1)

In ΔBB_{3}C’ and ΔBB_{4}C,

∠B_{3}BC’ = ∠B_{4}BC (Common)

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

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

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

⇒

This justifies the construction.

**Question** 6:

Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are times the corresponding side of ΔABC. Give the justification of the construction.

**Answer**:

∠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.

**Step 1**

Draw a ΔABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°.

**Step 2**

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

**Step 3**

Locate 4 points (as 4 is greater in 4 and 3), B_{1, }B_{2, }B_{3, }B_{4}, on BX.

**Step 4**

Join B_{3}C. Draw a line through B_{4} parallel to B_{3}C intersecting extended BC at C’.

**Step 5**

Through C’, draw a line parallel to AC intersecting extended line segment at C’. ΔA’BC’ is the required triangle.

**Justification**

The construction can be justified by proving that

In ΔABC and ΔA’BC’,

∠ABC = ∠A’BC’ (Common)

∠ACB = ∠A’C’B (Corresponding angles)

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

… (1)

In ΔBB_{3}C and ΔBB_{4}C’,

∠B_{3}BC = ∠B_{4}BC’ (Common)

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

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

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

⇒

This justifies the construction.

**Question** 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 times the corresponding sides of the given triangle. Give the justification of the construction.

**Answer**:

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.

**Step 1**

Draw a line segment AB = 4 cm. Draw a ray SA making 90° with it.

**Step 2**

Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C. Join BC. ΔABC is the required triangle.

**Step 3**

Draw a ray AX making an acute angle with AB, opposite to vertex C.

**Step 4**

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}.

**Step 5**

Join A_{3}B. Draw a line through A_{5} parallel to A_{3}B intersecting extended line segment AB at B’.

**Step 6**

Through B’, draw a line parallel to BC intersecting extended line segment AC at C’. ΔAB’C’ is the required triangle.

**Justification**

The construction can be justified by proving that

In ΔABC and ΔAB’C’,

∠ABC = ∠AB’C’ (Corresponding angles)

∠BAC = ∠B’AC’ (Common)

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

… (1)

In ΔAA_{3}B and ΔAA_{5}B’,

∠A_{3}AB = ∠A_{5}AB’ (Common)

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

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

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

⇒

This justifies the construction.