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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 its to 5/3 of the corresponding sides of the triangle ABC (scale factor 5/3). Given scale factor = 5/3 Steps of Construction: Draw any ray BX making an acute angle with BC on the side opposite to the vertex A. Locate 5 (the greater of 5 and 3 in 5/4 ) points B_1,B_2,B_3,B_4 and B_5 on BX so that 〖BB〗_1=B_1 B_2=B_2 B_3=B_3 B_4=B_4 B_5. Join B_3 (3rd point as 3 is smaller in 5/3) to C and draw a line through B_5 parallel to B_3 C to intersecting extended BC at C′. Draw a line through C′ parallel to line CA to intersect extended BA at A′. Thus, Δ A′BC′ is the required triangle. Justification Here, (BC^′)/BC=(BB_5)/(BB_3 ) = 5/3 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 =5/3. This justifies the construction.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.