TOPICS
Exercise - 11.1

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.

Solution :-
  Given that : 
  A line segment of length 7.6 cm can be divided in the ratio of 5:8 as follows.
construction
  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.

Solution :-
  Given that : 
  A triangle of sides 4 cm, 5cm and 6cm.
construction
  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.

Solution :-
  Given that : 
  A triangle with sides 5 cm, 6 cm and 7 cm.
construction
  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.

Solution :-
  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.
construction
  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.

Solution :-
  Given that : 
  A ΔA'BC' whose sides are 3/4 of the corresponding sides of ΔABC can be drawn as follows.
construction
  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.

Solution :-
  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°
construction
  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.

Solution :-
  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.
construction
  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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