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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 Ξ ABC, and scale factor 3/4 β΄ Scale Factor < 1 We need to construct triangle similar to Ξ ABC Letβ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 π΅_1,π΅_2,π΅_3 and π΅_4 on BX so that γπ΅π΅γ_1=π΅_1 π΅_2=π΅_2 π΅_3=π΅_3 π΅_4. Join π΅_4C and draw a line through π΅_3 (the 3rd point, 3 being smaller of 3 and 4 in 3/4) parallel to π΅_4 πΆ, to intersect BC at Cβ². 4. Draw a line through Cβ² parallel to the line CA to intersect BA at Aβ². Thus, Ξ Aβ²BCβ² is the required triangle Justification Since scale factor is 3/4, we need to prove (π¨^β² π©)/π¨π©=(π¨^β² πͺ^β²)/π¨πͺ=(π©πͺ^β²)/π©πͺ =π/π. By construction, BC^β²/π΅πΆ=(π΅π΅_3)/(π΅π΅_4 )=3/4. Also, AβCβ is parallel to AC So, they will make the same angle with line BC β΄ β  AβCβB = β  ACB Now, In Ξ AβBCβ and Ξ ABC β  B = β  B β  AβCβB = β  ACB Ξ AβBCβ βΌ Ξ ABC Since corresponding sides of similar triangles are in the same ratio (π΄^β² π΅)/π΄π΅=(π΄^β² πΆ^β²)/π΄πΆ=(π΅πΆ^β²)/π΅πΆ So, (π¨^β² π©)/π¨π©=(π¨^β² πͺ^β²)/π¨πͺ=(π©πͺ^β²)/π©πͺ =π/π. Thus, our construction is justified