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  1. Chapter 11 Class 10 Constructions
  2. Serial order wise
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Draw a triangle ABC with side BC = 7 cm, ∠ B = 45°, ∠ A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of Δ ABC. In Δ ABC   ∠B = 45°, ∠A = 105° Sum of all interior angles in a triangle is 180°.   ∠A + ∠B + ∠C = 180° 105° + 45° + ∠C = 180° ∠C = 180° − 150° ∠C = 30° Steps of construction Draw a ΔABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°. Draw a ray BX making an acute angle with BC on the opposite side of vertex A. Locate 4 points (as 4 is greater in 4 and 3), B_1, B_2, B_3, B_4, on BX. Join B_3C. Draw a line through B_4 parallel to B_3C intersecting extended BC at C'. Through C', draw a line parallel to AC intersecting extended line segment at C'. ΔA'BC' is the required triangle. Justification Here,  (BC^′)/BC=(BB_4)/(BB_3 ) = 4/3 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 =4/3. 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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