Maths Crash Course - Live lectures + all videos + Real time Doubt solving!

Examples

Example 1
Important
Deleted for CBSE Board 2023 Exams

Construction 11.1 Deleted for CBSE Board 2023 Exams

Construction 11.2 Important Deleted for CBSE Board 2023 Exams

Construction 11.3 Deleted for CBSE Board 2023 Exams

Construction 11.4 Deleted for CBSE Board 2023 Exams

Construction 11.5 Important Deleted for CBSE Board 2023 Exams

Construction 11.6 Important Deleted for CBSE Board 2023 Exams You are here

Last updated at Aug. 25, 2021 by Teachoo

Maths Crash Course - Live lectures + all videos + Real time Doubt solving!

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 Similarly, Point C lies in perpendicular bisector of AX ∴ CY = CA Steps of Construction : Draw a line segment, say XY equal to BC + CA + AB. Make angles LXY equal to ∠B and MYX equal to ∠C. Bisect ∠LXY and ∠MYX. Let these bisectors intersect at a point A . For proof of BX = BA, see Construction 11.4 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 AXB 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) ∠ 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