Examples
Example 2 Important Deleted for CBSE Board 2025 Exams
Construction 11.1 Important Deleted for CBSE Board 2025 Exams
Construction 11.2 Deleted for CBSE Board 2025 Exams
Construction 11.3 Important Deleted for CBSE Board 2025 Exams
How to construct Tangents to circle if center of circle is not given? Important Deleted for CBSE Board 2025 Exams
Last updated at April 16, 2024 by Teachoo
Example 1 Construct a triangle similar to a given triangle ABC with its sides equal to 3/4 of the corresponding sides of the triangle ABC (i.e. of scale factor 3/4). Here, we are given Δ ABC, and scale factor 3/4 ∴ Scale Factor < 1 We need to construct triangle similar to Δ ABC Let’s follow these steps 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 3 and 4 in 3/4 ) points 𝐵_1,𝐵_2,𝐵_3 and 𝐵_4 on BX so that 〖𝐵𝐵〗_1=𝐵_1 𝐵_2=𝐵_2 𝐵_3=𝐵_3 𝐵_4. Join 𝐵_4C and draw a line through 𝐵_3 (the 3rd point, 3 being smaller of 3 and 4 in 3/4) parallel to 𝐵_4 𝐶, to intersect BC at C′. 4. Draw a line through C′ parallel to the line CA to intersect BA at A′. Thus, Δ A′BC′ is the required triangle Justification Since scale factor is 3/4, we need to prove (𝑨^′ 𝑩)/𝑨𝑩=(𝑨^′ 𝑪^′)/𝑨𝑪=(𝑩𝑪^′)/𝑩𝑪 =𝟑/𝟒. By construction, BC^′/𝐵𝐶=(𝐵𝐵_3)/(𝐵𝐵_4 )=3/4. 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 =𝟑/𝟒. Thus, our construction is justified