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  1. Chapter 11 Class 9 Constructions
  2. Serial order wise
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Construction 11.6 : To construct a triangle, given its perimeter and its two base angles. Given the base angles, ∠B and ∠C and AB + AC + BC, we have to construct ΔABC. Steps of Construction : Draw a line segment XY equal to AB + AC + BC. Make ∠ LXY equal to ∠B and ∠ MYX equal to ∠C. Bisect ∠ LXY and ∠ MYX. Let these bisectors intersect at a point A . Draw perpendicular bisectors of line AX and AY Let PQ intersect XY at B and RS intersect XY at C. Join AB and AC Thus, Δ ABC is a the required triangle Justification To justify, we have to prove AB + BC + AC  = XY ∠ LXY = ∠ B ∠ MYX = ∠ C In Δ AXQ, PQ is the perpendicular bisector of AX, ∴ BX = BA Similarly, we can say CY = CA Now, we know that XY = XQ + BC + CY XY = AB + BC  + AC In Δ AXQ Since AX = BA ∴ ∠ BAX =  ∠ AXB Now, ∠ ABC is the exterior angle of triangle AXQ ∠ ABC = ∠ BAX + ∠ AXB ∠ ABC = ∠ AXB + ∠ AXB ∠ ABC = 2 ∠ AXB ∠ ABC = ∠ LXY Thus, ∠ B = ∠ LXY Similarly, we can prove ∠ A = ∠ MYX Hence justified

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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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