Ex 11.1, 6 - Draw a triangle ABC with side BC = 7 cm, B = 45, A = 105

Ex 11.1, 6 - Chapter 11 Class 10 Constructions - Part 2
Ex 11.1, 6 - Chapter 11 Class 10 Constructions - Part 3
Ex 11.1, 6 - Chapter 11 Class 10 Constructions - Part 4
Ex 11.1, 6 - Chapter 11 Class 10 Constructions - Part 5 Ex 11.1, 6 - Chapter 11 Class 10 Constructions - Part 6 Ex 11.1, 6 - Chapter 11 Class 10 Constructions - Part 7 Ex 11.1, 6 - Chapter 11 Class 10 Constructions - Part 8

 

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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 sides of Δ ABC. Let’s first draw a rough diagram To construct Δ ABC, we first need to find ∠ C Finding ∠ C In Δ ABC ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle is 180°) 105° + 45° + ∠C = 180° 150° + ∠C = 180° ∠C = 180° − 150° ∠C = 30° Steps to draw Δ ABC Draw base BC of side 7 cm Draw ∠ B = 45° Draw ∠ C = 30° 4. Let point A be the point where the two rays intersect ∴ Δ ABC is the required triangle Now, we need to make a triangle which is 4/3 times its size ∴ Scale factor = 4/3 > 1 Steps of construction Draw any ray BX making an acute angle with BC on the side opposite to the vertex A. Mark 4 (the greater of 4 and 3 in 4/3 ) points 𝐵_1, 𝐵_2, 𝐵_3,𝐵_4 on BX so that 〖𝐵𝐵〗_1=𝐵_1 𝐵_2=𝐵_2 𝐵_3=𝐵_3 𝐵_4 Join 𝐵_3 𝐶 (3rd point as 3 is smaller in 4/3) and draw a line through 𝐵_4 parallel to 𝐵_3 𝐶, to intersect BC extended at C′. Draw a line through C′ parallel to the line AC to intersect AB extended at A′. Thus, Δ A′BC′ is the required triangle Justification Since scale factor is 4/3, we need to prove (𝑨^′ 𝑩)/𝑨𝑩=(𝑨^′ 𝑪^′)/𝑨𝑪=(𝑩𝑪^′)/𝑩𝑪 =𝟒/𝟑 By construction, BC^′/𝐵𝐶=(𝐵𝐵_4)/(𝐵𝐵_3 )=4/3 Also, A’C’ is parallel to AC So, they will make the same angle with line BC ∴ ∠ A’C’B = ∠ ACB Now, In Δ A’BC’ and Δ ABC ∠ B = ∠ B ∠ A’C’B = ∠ ACB Δ A’BC’ ∼ Δ ABC Since corresponding sides of similar triangles are in the same ratio (𝐴^′ 𝐵)/𝐴𝐵=(𝐴^′ 𝐶^′)/𝐴𝐶=(𝐵𝐶^′)/𝐵𝐶 So, (𝑨^′ 𝑩)/𝑨𝑩=(𝑨^′ 𝑪^′)/𝑨𝑪=(𝑩𝑪^′)/𝑩𝑪 =𝟒/𝟑 Thus, our construction is justified

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo