Cost that triangle ABC with a b + BC + AC is equal to 12 12.5 CM angle b to 45 and angle C is equal to 60
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Answer:
Given the perimeter of the triangle ABC be 12cm i.e., AB + BC + CA = 12cm and both the base angles be 450 and 600 i.e., ∠B = 450 and ∠C = 600 STEPS : (i) Draw a line segment PQ = 12cm (ii) At P, construct line PR so that ∠RPO = 450 and at Q, construct a line QS so that ÐSQP = 600 (iii) Draw bisector of angles RPQ and SQP which meet each other at point A. (iv) Draw perpendicular bisector of AP, which meets PQ at point B. (v) Draw perpendicular bisector of AQ, which meets PQ at point C. (vi) Join AB and AC. Thus, ABC is the required triangle. Proof : Since, MB is perpendicular bisector of AP => ΔQNC ≡ ΔANC [By SAS] PB = AC Similarly, NC is perpendicular bisector of AQ. =>ΔQNC ≡ ΔANC [By SAS] => CQ = AC [By cpctc] Now, PQ = PB + BC + CQ = AB + BC + AC = Given perimeter of the ΔABC drawn. Also, ÐBPA = ÐBAP [As ΔPMB ≡ ΔA MB] ∴ ∠ABC = ∠BPA + ∠BAP [Ext. angle of a triangle = sum of two interior opposite angles] ∠ABC =∠BPA + ∠BAP = 2∠BPA = ∠RPB = ∠ACB [Given] ∠ACB = ∠CQA + ∠CQA = 2∠CQA [∵ ΔQNC ≡ ΔANC ∴ ∠CQA = ∠CAQ] = ∠SQC = Given base angle ACB. Thus, given perimeter = perimeter of ΔABC.