Page 1 (www.tiwariacademy.net) (Class – X) A Free web support in Education 1 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. Answer 1: 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. Page 2 (www.tiwariacademy.net) (Class – X) A Free web support in Education 1 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. Answer 1: 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. (www.tiwariacademy.net) (Class – X) A Free web support in Education 2 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 2/3of the corresponding sides of the first triangle. Give the justification of the construction. Answer 2: 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 ?ABC is the required triangle. Page 3 (www.tiwariacademy.net) (Class – X) A Free web support in Education 1 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. Answer 1: 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. (www.tiwariacademy.net) (Class – X) A Free web support in Education 2 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 2/3of the corresponding sides of the first triangle. Give the justification of the construction. Answer 2: 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 ?ABC is the required triangle. (www.tiwariacademy.net) (Class – X) A Free web support in Education 3 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'. ?AB'C' is the required triangle. Page 4 (www.tiwariacademy.net) (Class – X) A Free web support in Education 1 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. Answer 1: 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. (www.tiwariacademy.net) (Class – X) A Free web support in Education 2 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 2/3of the corresponding sides of the first triangle. Give the justification of the construction. Answer 2: 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 ?ABC is the required triangle. (www.tiwariacademy.net) (Class – X) A Free web support in Education 3 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'. ?AB'C' is the required triangle. (www.tiwariacademy.net) (Class – X) A Free web support in Education 4 Justification The construction can be justified by proving that By construction, we have B’C’ || BC ? ?A = ?ABC (Corresponding angles) In ?AB'C' and ?ABC, ? = ?ABC (Proved above) ? = ?BAC (Common) ? ? ~ ?ABC (AA similarity criterion) … (1) In ?AA 2 B' and ?AA 3 B, ?A 2 AB' = ?A 3 AB (Common) ?AA 2 B' = ?AA 3 B (Corresponding angles) ? ?AA 2 B' ~ ?AA 3 B (AA similarity criterion) From equations (1) and (2), we obtain This justifies the construction. Page 5 (www.tiwariacademy.net) (Class – X) A Free web support in Education 1 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. Answer 1: 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. (www.tiwariacademy.net) (Class – X) A Free web support in Education 2 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 2/3of the corresponding sides of the first triangle. Give the justification of the construction. Answer 2: 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 ?ABC is the required triangle. (www.tiwariacademy.net) (Class – X) A Free web support in Education 3 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'. ?AB'C' is the required triangle. (www.tiwariacademy.net) (Class – X) A Free web support in Education 4 Justification The construction can be justified by proving that By construction, we have B’C’ || BC ? ?A = ?ABC (Corresponding angles) In ?AB'C' and ?ABC, ? = ?ABC (Proved above) ? = ?BAC (Common) ? ? ~ ?ABC (AA similarity criterion) … (1) In ?AA 2 B' and ?AA 3 B, ?A 2 AB' = ?A 3 AB (Common) ?AA 2 B' = ?AA 3 B (Corresponding angles) ? ?AA 2 B' ~ ?AA 3 B (AA similarity criterion) From equations (1) and (2), we obtain This justifies the construction. (www.tiwariacademy.net) (Class – X) A Free web support in Education 5 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. Answer 3: Step 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. 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'. ?AB'C' is the required triangle.Read More

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