Given, to construct ΔPQR measurements PQ=5cm,QR=6 and RP=8.2cm
Rough draft :- Ref. image
Main draft :- Ref. image
Construction steps :-
i) Draw a line segment RP=8.2cm as the base.
ii) Now, taking 'R' as centre draw an arc of radius 6cm
iii) Now, taking 'P' as centre cut the previous arc with radius 5cm.
iv) Now join the vertices, a ΔPQR is formed.
v) Taking radius of more than a half draw the perpendicular bisectors on either sides of the triangle.
vi) Now, try to cut the previous perpendicular bisectors
vii) The joining point of all the perpendicular bisectors is named as (S) circum centre.
viii) Now, from (S) draw a circle which touches all the vertices of the $
Delta PQR$.
ix) Thus, circum circle ofΔPQR is formed.
Conclusion :-
Therefore, the required circum circle for △PQR is formed.
![1343482_1261951_ans_e14cd4eabd764d7193d9f56a066bf20e.JPG](https://search-static.byjusweb.com/question-images/toppr_invalid/questions/1343482_1261951_ans_e14cd4eabd764d7193d9f56a066bf20e.JPG)