TOPICS

Exercise - 11.1

Constructions

**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. Give the justification of the construction.

Given that : A line segment of length 7.6 cm can be divided in the ratio of 5:8 as follows. Steps of Construction : 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. Justification :- 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/BC = 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

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

Given that : A triangle of sides 4 cm, 5cm and 6cm. Steps of Construction : 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. 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'. ΔAB'C' is the required triangle. Justification :- The construction can be justified by proving that AB' = 2AB/3, B'C' = 2BC/3, AC' = 2AC/3 By construction, we have B’C’ || BC ∠AB'C' = ∠ABC (Corresponding angles) In ΔAB'C' and ΔABC, ∠AB'C' = ∠ABC (Proved above) ∠B'AC' = ∠BAC (Common) ∴ ΔAB'C'∼ΔABC (AA similarity criterion) So, AB'/AB = B'C'/BC = AC'/AC -----(1) In ΔAA₂B' and ΔAA₃B, ∠A₂AB' = ∠A₃AB (Common) ∠AA₂B' = ∠AA₃B (Corresponding angles) ∴ΔAA₂B' ∼ΔAA₃B (AA similarity criterion) So, AB'/AB = AA₂/AA₃ AB'/AB = 2/3 -------(1) From equations (1) and (2), we obtain AB'/AB = B'C'/BC = AC'/AC = 2/3 Therefore, AB' = 2AB/3, B'C' = 2BC/3, AC' = 2AC/3

**Question-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.

Given that : A triangle with sides 5 cm, 6 cm and 7 cm. Steps of Construction : 1. Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 5cm 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. 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'. ΔAB'C' is the required triangle. Justification :- The construction can be justified by proving that AB' = 7AB/5, B'C' = 7BC/5, AC' = 7AC/5 By construction, we have B’C’ || BC ∠AB'C' = ∠ABC (Corresponding angles) In ΔAB'C' and ΔABC, ∠AB'C' = ∠ABC (Proved above) ∠B'AC' = ∠BAC (Common) ∴ ΔAB'C'∼ΔABC (AA similarity criterion) So, AB'/AB = B'C'/BC = AC'/AC or AB/AB' = BC/B'C' = AC/AC' -----(1) In ΔAA₅B and ΔAA₇B', ∠A₅AB = ∠A₇AB' (Common) ∠AA₅B = ∠AA₇B' (Corresponding angles) ∴ΔAA₅B ∼ΔAA₇B' (AA similarity criterion) So, AB/AB' = AA₅/AA₇ AB'/AB = 7/5 -------(2) From equations (1) and (2), we obtain AB'/AB = B'C'/BC = AC'/AC = 7/5 Therefore, AB' = 7AB/5, B'C' = 7BC/5, AC' = 7AC/5

**Question-4 :-** Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose side are 1 by 1/2 times the corresponding sides of the isosceles triangle.
Give the justification of the construction.

Given that : 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 3/2 times of ΔABC can be drawn as follows. Steps of Construction : 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 ΔABC 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'. ΔAB'C' is the required triangle. Justification :- The construction can be justified by proving that AB' = 3AB/2, B'C' = 3BC/2, AC' = 3AC/2 By construction, we have B’C’ || BC ∠AB'C' = ∠ABC (Corresponding angles) In ΔAB'C' and ΔABC, ∠AB'C' = ∠ABC (Proved above) ∠B'AC' = ∠BAC (Common) ∴ ΔAB'C'∼ΔABC (AA similarity criterion) So, AB'/AB = B'C'/BC = AC'/AC or AB/AB' = BC/B'C' = AC/AC' -----(1) In ΔAA₂B and ΔAA₃B', ∠A₂AB = ∠A₃AB' (Common) ∠AA₂B = ∠AA₃B' (Corresponding angles) ∴ ΔAA₂B ∼ΔAA₃B' (AA similarity criterion) So, AB/AB' = AA₂/AA₃ AB/AB' = 2/3 AB'/AB = 3/2 From equations (1) and (2), we obtain AB'/AB = B'C'/BC = AC'/AC = 3/2 Therefore, AB' = 3AB/2, B'C' = 3BC/2, AC' = 3AC/2

**Question-5 :-** Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.
Give the justification of the construction.

Given that : A ΔA'BC' whose sides are 3/4 of the corresponding sides of ΔABC can be drawn as follows. Steps of Construction : 1. Draw a ΔABC 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'. ΔA'BC' is the required triangle. 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 AB' = 3AB/4, B'C' = 3BC/4, AC' = 3AC/4 By construction, we have B’C’ || BC ∠AB'C' = ∠ABC (Corresponding angles) In ΔAB'C' and ΔABC, ∠AB'C' = ∠ABC (Proved above) ∠B'AC' = ∠BAC (Common) ∴ ΔAB'C'∼ΔABC (AA similarity criterion) So, AB'/AB = B'C'/BC = AC'/AC or AB/AB' = BC/B'C' = AC/AC' -----(1) In ΔBB₃C' and ΔBB₄C, ∠B₃BC' = ∠B₄BC (Common) ∠BB₃C' = ∠BB₄C (Corresponding angles) AB/AB' = 4/3 AB'/AB = 3/4 ------(2) From equations (1) and (2), we obtain AB'/AB = B'C'/BC = AC'/AC = 3/4 Therefore, AB' = 3AB/4, B'C' = 3BC/4, AC' = 3AC/4

**Question-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 ΔABC.
Give the justification of the construction.

Given that : ∠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° Steps of Construction : 1. Draw a ΔABC 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 B₃C. 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'. ΔA'BC' is the required triangle. Justification :- The construction can be justified by proving that AB' = 4AB/3, B'C' = 4BC/3, AC' = 4AC/3 By construction, we have B’C’ || BC ∠AB'C' = ∠ABC (Corresponding angles) In ΔAB'C' and ΔABC, ∠AB'C' = ∠ABC (Proved above) ∠B'AC' = ∠BAC (Common) ∴ ΔAB'C'∼ΔABC (AA similarity criterion) So, AB'/AB = B'C'/BC = AC'/AC or AB/AB' = BC/B'C' = AC/AC' -----(1) In ΔBB₃C and ΔBB₄C', ∠B₃BC = ∠B₄BC' (Common) ∠BB₃C = ∠BB₄C' (Corresponding angles) ∴ΔBB₃C ∼ΔBB₄C' (AA similarity criterion) BC/B'C' = BB₃/BB₄ BC/B'C' = 3/4 B'C'/BC = 4/3 ------(2) From equations (1) and (2), we obtain AB'/AB = B'C'/BC = AC'/AC = 4/3 Therefore, AB' = 4AB/3, B'C' = 4BC/3, AC' = 4AC/3

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

Given that : It is given that sides other than hypotenuse are of lengths 4 cm and 3 cm. Clearly, these will be perpendicular to each other. Steps of Construction : 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. ΔABC 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 A₃B. 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'. ΔAB'C' is the required triangle. Justification :- The construction can be justified by proving that AB' = 5AB/3, B'C' = 5BC/3, AC' = 5AC/3 By construction, we have B’C’ || BC ∠AB'C' = ∠ABC (Corresponding angles) In ΔAB'C' and ΔABC, ∠AB'C' = ∠ABC (Proved above) ∠B'AC' = ∠BAC (Common) ∴ ΔAB'C'∼ΔABC (AA similarity criterion) So, AB'/AB = B'C'/BC = AC'/AC or AB/AB' = BC/B'C' = AC/AC' -----(1) In ΔAA₃B and ΔAA₅B', ∠A₃AB = ∠A₅AB' (Common) ∠AA₃B = ∠AA₅B' (Corresponding angles) ∴ΔAA₃B ∼ΔAA₅B' (AA similarity criterion) So, AB/AB' = AA₃/AA₅ AB/AB' = 3/5 AB'/AB = 5/3 ------(2) From equations (1) and (2), we obtain AB'/AB = B'C'/BC = AC'/AC = 5/3 Therefore, AB' = 5AB/3, B'C' = 5BC/3, AC' = 5AC/3

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