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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 angle equal to ∠ B from point X Let the angle be ∠ LXY. Make angle equal to ∠C from point Y Let the angle be ∠ MYX 4. Bisect ∠ LXY and ∠ MYX. Let these bisectors intersect at a point A . 5. Make perpendicular bisector of AX Let it intersect XY at point B 6. Make perpendicular bisector of AY Let it intersect XY at point C 7. Join AB & AC ∴ Δ ABC is the required triangle Justification To justify, we have to prove AB + BC + AC = XY ∠ LXY = ∠ B ∠ MYX = ∠ C Now,, Point B lies in perpendicular bisector of AX ∴ BX = BA For proof of BX = BA, see Construction 11.4 . Similarly, Point C lies in perpendicular bisector of AX ∴ CY = CA For proof of CY = CA, see Construction 11.4 Now, we know that XY = XQ + BC + CY XY = AB + BC + AC So, XY = AB + BC + AC is proved, now let’s prove ∠ LXY = ∠ B & ∠ MYX = ∠ C (From (1) & (2)) In Δ AXQ Since BX = BA ∴ ∠ BAX = ∠ AXB Now, ∠ ABC is the exterior angle of triangle AXB ∠ ABC = ∠ BAX + ∠ AXB ∠ ABC = ∠ AXB + ∠ AXB (Angles opposite to equal sides are equal) (Exterior angle is sum of interior opposite angles) (From (3)) ∠ ABC = 2 ∠ AXB ∠ ABC = ∠ LXY ∴ ∠ B = ∠ LXY Similarly, we can prove ∠ A = ∠ MYX Hence justified As AX is bisector of ∠ LXY, ∴ ∠ AXB = (∠ 𝐿𝑋𝑌)/2 2∠ AXB = ∠ LXY

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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.