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  1. Chapter 11 Class 10 Constructions
  2. Serial order wise
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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). Given scale factor = 3/4 Steps of Construction: Draw any ray BX making an acute angle with BC on the side opposite to the vertex A. Locate 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′. Draw a line through C′ parallel to the line CA to intersect BA at A′. Thus, Δ A′BC′ is the required triangle Justification Here, BC^′/BC=(BB_3)/(BB_4 )=3/4. Also, A’C’ is parallel to AC So, the will make the same angle with line BC ∴ ∠ A’C’B = ∠ ACB Now, In Δ A’BC’ and 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. This justifies the construction.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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