Mathematics
Points A, B and C represent position of three towers such that AB = 60 m, BC = 73 m and CA = 52 m. Taking a scale of 10 m to 1 cm, make an accurate drawing of △ABC. Find by drawing, the location of a point which is equidistant from A, B and C, and its actual distance from any of the towers.
Locus
9 Likes
Answer
Construct the triangle ABC with given conditions.
Construct perpendicular bisectors of all the three sides of triangle.
From figure we see,
FG = Perpendicular bisector of AB.
DE = Perpendicular bisector of AC.
HI = Perpendicular bisector of BC.
These bisectors meet each other at point P. Hence, point P is at equal distance from point A,B and C.
By measuring BP = 37 m.
Hence, the distance of each tower is nearly 37 m.
Answered By
4 Likes
Related Questions
Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°.
Hence,
(i) Construct the locus of points equidistant from BA and BC.
(ii) Construct the locus of points equidistant from B and C.
(iii) Mark the point which satisfies the above two loci as P. Measure and write the length of PC.
Draw two intersecting lines to include an angle of 30°. Use ruler and compasses to locate points which are equidistant from these lines and also 2 cm away from their point of intersection. How many such point exist ?
Without using set square or protractor, construct the quadrilateral ABCD in which ∠BAD = 45°, AD = AB = 6 cm, BC = 3.6 cm and CD = 5 cm.
(i) Measure ∠BCD.
(ii) Locate the point P on BD which is equidistant from BC and CD.
Use ruler and compasses only for this question.
(i) Construct △ABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.
(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 record the length of PB.