Construct rhombus ABCD with sides of length 5 cm and diagonal AC of length 6 cm. Find the point R on CD such that RA = RB. Then the length of CR is
Construct rhombus ABCD with sides of length 5 cm and diagonal AC of length 6 cm. Find the point R on CD such that RA = RB. Then the length of CR is
The correct option is B 1 cm
Steps of construction:
(i) Draw a line segment AC = 6 cm.
(ii) Draw the perpendicular bisector of AC.
(iii) Draw arcs on this perpendicular from A, at a distance of 5 cm, which cuts it at B and D(as shown above).
(iv) Complete the rhombus ABCD.
(v) Draw the perpendicular bisector of AB, which intersects CD at R, such that RA = RB.
(Given, R must be such that RA = RB. i.e, point R must be equidistant from A and B. We know that the locus of a point which is equidistant from two fixed points is the perpendicular bisector of the line segment joining the two fixed points. So, we draw the perpendicular bisector of AB so as to get the locus of the point which is equidistant from A and B).
From these constructions, we see that, length of AR is 1 cm.
1 cm
Steps of construction:
(i) Draw a line segment AC = 6 cm.
(ii) Draw the perpendicular bisector of AC.
(iii) Draw arcs on this perpendicular from A, at a distance of 5 cm, which cuts it at B and D(as shown above).
(iv) Complete the rhombus ABCD.
(v) Draw the perpendicular bisector of AB, which intersects CD at R, such that RA = RB.
(Given, R must be such that RA = RB. i.e, point R must be equidistant from A and B. We know that the locus of a point which is equidistant from two fixed points is the perpendicular bisector of the line segment joining the two fixed points. So, we draw the perpendicular bisector of AB so as to get the locus of the point which is equidistant from A and B).
From these constructions, we see that, length of AR is 1 cm.
