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Construction 11.1 Important Deleted for CBSE Board 2023 Exams

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Construction 11.3 Important Deleted for CBSE Board 2023 Exams

How to construct Tangents to circle if center of circle is not given? Important Deleted for CBSE Board 2023 Exams

Last updated at Aug. 5, 2021 by Teachoo

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Example 2 Construct a triangle similar to a given triangle ABC with its sides equal its to 5/3 of the corresponding sides of the triangle ABC (scale factor 5/3). Here, we are given Ξ ABC, and scale factor 5/3 β΄ Scale Factor > 1 We need to construct triangle similar to Ξ ABC Letβs 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 5 (the greater of 5 and 3 in 5/3) points π΅_1,π΅_2,π΅_3,π΅_4,π΅_5on BX so that γπ΅π΅γ_1=π΅_1 π΅_2=π΅_2 π΅_3=π΅_3 π΅_4 =π΅_4 π΅_5 Join π΅_3C (3rd point as 3 is smaller in 5/3) and draw a line through π΅_5 parallel to π΅_3 πΆ, to intersect BC extended at Cβ². Draw a line through Cβ² parallel to the line CA to intersect BA extendedat Aβ². Thus, Ξ Aβ²BCβ² is the required triangle Justification Since scale factor is 5/3, we need to prove (π¨^β² π©)/π¨π©=(π¨^β² πͺ^β²)/π¨πͺ=(π©πͺ^β²)/π©πͺ = π/π. By construction, BC^β²/π΅πΆ=(π΅π΅_5)/(π΅π΅_3 )= 5/3. Also, AβCβ is parallel to AC So, they will make the same angle with line BC β΄ β AβCβB = β ACB Justification Since scale factor is 5/3, we need to prove (π¨^β² π©)/π¨π©=(π¨^β² πͺ^β²)/π¨πͺ=(π©πͺ^β²)/π©πͺ = π/π. By construction, BC^β²/π΅πΆ=(π΅π΅_5)/(π΅π΅_3 )= 5/3. Also, AβCβ is parallel to AC So, they will make the same angle with line BC β΄ β AβCβB = β ACB (Corresponding angles) Now, In Ξ AβBCβ and Ξ ABC β B = β B (Common) β AβCβB = β ACB (From (2)) Ξ AβBCβ βΌ Ξ ABC (AA Similarity) Since corresponding sides of similar triangles are in the same ratio (π΄^β² π΅)/π΄π΅=(π΄^β² πΆ^β²)/π΄πΆ=(π΅πΆ^β²)/π΅πΆ So, (π¨^β² π©)/π¨π©=(π¨^β² πͺ^β²)/π¨πͺ=(π©πͺ^β²)/π©πͺ = π/π. (From (1)) Thus, our construction is justified