STEPS OF CONSTRUCTION
(i) Draw a line segment `PQ=12` cm.
(ii) Make `angleQPR=60^(@)` and `anglePQS=70^(@)`.
(iii) Draw the bisectors of `angleQPR` and `anglePQS` to meet at A.
(iv) Draw the perpendicular bisectors of PA and QA to meet PQ at B and C respectively.
(v) Join AB and AC.
Then, `Delta ABC` is the required triangle.
Justification :
Since B lies on the perpendicular bisector of AP, we have `BA=BP`. Since C lies on the perpendicular bisector of AQ, we have `CA=CQ`. Thus, `AB=PB` and `AC=CQ`.
`:. AB+BC+AC=PB+BC+CQ=PQ=12` cm.
Now, `BA=BP implies angleBPA=angleBAP`
`implies angleABC=angleBPA+angleBAP=2 angleBPA`
`=2xx1/2xx60^(@)=60^(@)`.And, `CA=CQ implies angleCQA=angle CAQ`
`implies angleACB=angleCQA+angleCAQ=2angleCQA`
`=2xx1/2xx70^(@)=70^(@)`.
Verification :
On measurement, we find that
`AB+BC+CA=(4.4+3.5+4.1) cm =12` cm.
`angleB=60^(@)` and `angleC=70^(@)`.
