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प्रश्न
Construct a ti.PQR, in which PQ=S. 5 cm, QR=3. 2 cm and PR=4.8 cm. Draw the locus of a point which moves so that it is always 2.5 cm from Q.
उत्तर
Steps of oonstruction:
(i) Draw PQ = 5.5 cm
(ii) With P as centre and radius 4.8 cm draw an arc.
(iii) With Q as centre and radius 3.2 cm cut another arc which meets the first arc at R. Join PR and QR. PQR is the required triangle.
(iv) Draw perpendicular bisector of PR.
(v) Q as centre and radius as 2.5 cm, draw an arc which intersects the perpendicular bisector of PR at O.
O is the required point which is at a distance of 2.5 cm from Q.
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संबंधित प्रश्न
Use ruler and compasses only for this question:
I. Construct ABC, where AB = 3.5 cm, BC = 6 cm and ABC = 60o.
II. Construct the locus of points inside the triangle which are equidistant from BA and BC.
III. Construct the locus of points inside the triangle which are equidistant from B and C.
IV. Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and records the length of PB.
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- equidistant from BA and BC.
- 4 cm from M.
- 4 cm from N.
Mark the point P, which is 4 cm from both M and N, and equidistant from BA and BC. Join MP and NP, and describe the figure BMPN.
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