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प्रश्न
Construct a triangle ABC, with AB = 5.6 cm, AC = BC = 9.2 cm. Find the points equidistant from AB and AC; and also 2 cm from BC. Measure the distance between the two points obtained.
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उत्तर
Steps of construction:
- Draw a line segment AB = 5.6 cm
- From A and B, as centers and radius 9.2 cm, make two arcs which intersect each other at C.
- Join CA and CB.
- Draw two lines n and m parallel to BC at a distance of 2 cm
- Draw the angle bisector of ∠BAC which intersects m and n at P and Q respectively.
P and Q are the required points which are equidistant from AB and AC.
On measuring the distance between P and Q is 4.3 cm.
संबंधित प्रश्न
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Describe completely the locus of a point in the following case:
Point in a plane equidistant from a given line.
Without using set squares or protractor construct a triangle ABC in which AB = 4 cm, BC = 5 cm and ∠ABC = 120°.
(i) Locate the point P such that ∠BAp = 90° and BP = CP.
(ii) Measure the length of BP.
