Step 1: Draw a line EF and mark a point O on it. |
Step 2: Place the pointer of the compass at O and draw an arc of convenient radius which cuts the line EF at point P. |
Step 3: With the pointer at A (as centre) now draw an arc that passes through O. |
Step 4: Let the two arcs intersect at Q. Join OQ. We get \[\angle \text{QOP}\]whose measure is\[\text{60 }\!\!{}^\circ\!\!\text{ }\]. |
(i) Perpendicular bisector of the diameter of a circle passes through the P of the circle. |
(ii) If B is image of A in line l and D is image of C in line l, then AC = Q . |
(iii) Angle bisector is a ray which divides the angle in R equal parts. |
P | Q | R |
Centre | BD | 2 |
P | Q | R |
Centre | AD | 1 |
P | Q | R |
Centre | AB | 1 |
P | Q | R |
Centre | BC | 2 |
1. With A and B as centres and a radius greater than AP construct two arcs, which cut each other at Q. |
2. Join PQ. Then \[\overline{\text{PQ}}\] is perpendicular to l. We write \[\overline{\text{PQ}}\bot l\]. |
3. With P as centre and a convenient radius, construct an arc intersecting the line l at two points A and B. |
4. Given a point P on a line l. |
(i) It is possible to divide a line segment in 5 equal parts by perpendicularly bisecting a given line segment 5 times. |
(ii) With a given centre and a given radius, only one circle can be drawn. |
(iii) If we bisect an angle of a square, we get two angles of \[\text{45 }\!\!{}^\circ\!\!\text{ }\] each. |
(i) | (ii) | (iii) |
F | T | T |
(i) | (ii) | (iii) |
F | T | F |
(i) | (ii) | (iii) |
T | F | T |
(i) | (ii) | (iii) |
T | T | F |
Statement 1: Two lines are said to be perpendicular if they intersect each other at an angle of \[\text{90 }\!\!{}^\circ\!\!\text{ }\]. |
Statement 2: A unique circle can be drawn passing through the given centre. |
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