Exercise 11.1 - - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

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, A₁, A₂, A₃, A₄ …….. A₁₃, on AX such that AA₁ = A₁A₂ = A₂A₃ and so on.

Step 3 Join BA₁₃.

Step 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.

Justification

The construction can be justified by proving that

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

 … (1)

From the figure, it can be observed that AA₅ and A₅A₁₃ 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 A₁, A₂, A₃ (as 3 is greater between 2 and 3) on line AX such that AA₁ = A₁A₂ = A₂A₃.

Step 4

Join BA₃ and draw a line through A₂ parallel to BA₃ 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 ΔAA₂B’ and ΔAA₃B,

∠A₂AB’ = ∠A₃AB (Common)

∠AA₂B’ = ∠AA₃B (Corresponding angles)

∴ ΔAA₂B’ ∼ ΔAA₃B (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, 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.

Step 4

Join BA₅ and draw a line through A₇ parallel to BA₅ 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 ΔAA₅B and ΔAA₇B’,

∠A₅AB = ∠A₇AB’ (Common)

∠AA₅B = ∠AA₇B’ (Corresponding angles)

∴ ΔAA₅B ∼ ΔAA₇B’ (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) A₁, A₂, and A₃ on AX such that AA₁ = A₁A₂ = A₂A₃.

Step 5

Join BA₂ and draw a line through A₃ parallel to BA₂ 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 ΔAA₂B and ΔAA₃B’,

∠A₂AB = ∠A₃AB’ (Common)

∠AA₂B = ∠AA₃B’ (Corresponding angles)

∴ ΔAA₂B ∼ ΔAA₃B’ (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), B₁, B₂, B₃, B₄, on line segment BX.

Step 4

Join B₄C and draw a line through B₃, parallel to B₄C 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 ΔBB₃C’ and ΔBB₄C,

∠B₃BC’ = ∠B₄BC (Common)

∠BB₃C’ = ∠BB₄C (Corresponding angles)

∴ ΔBB₃C’ ∼ ΔBB₄C (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, B₄, on BX.

Step 4

Join B₃C. Draw a line through B₄ parallel to B₃C 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 ΔBB₃C and ΔBB₄C’,

∠B₃BC = ∠B₄BC’ (Common)

∠BB₃C = ∠BB₄C’ (Corresponding angles)

∴ ΔBB₃C ∼ ΔBB₄C’ (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), A₁, A₂, A₃, A₄, A₅, on line segment AX such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅.

Step 5

Join A₃B. Draw a line through A₅ parallel to A₃B 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 ΔAA₃B and ΔAA₅B’,

∠A₃AB = ∠A₅AB’ (Common)

∠AA₃B = ∠AA₅B’ (Corresponding angles)

∴ ΔAA₃B ∼ ΔAA₅B’ (AA similarity criterion)

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

⇒ 

This justifies the construction.