Question

Construct a triangle ABC with AB+BC+CA=12 cm ∠B=45° and ∠C = 60°

Answer 1

Given the perimeter of the triangle ABC be 12 cm i.e., AB + BC + CA = 12 cm and both the base angles be 45° and 60° i.e., ∠B = 45° and ∠C=60°
STEPS:
(i) Draw a line segment PQ = 12 cm
(ii) At P, construct line PR so that ∠RPO = 45° and at Q, construct a line QS so that ∠SQP = 60°
(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 cpct]
Now, PQ = PB + BC + CQ
= AB + BC + AC
= Given perimeter of the ΔABC drawn.
Also, ∠BPA = ∠BAP [As Δ PMB ≅ Δ AMB]
$\therefore$ ∠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
= ∠SQC = Given base angle ACB.
Thus, given perimeter = perimeter of ΔABC.
given one base angle = angle ABC and,
given other base angle = angle ACB.