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Class 9-10, JEE & NEET
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A rhombus whose diagonals are 4 cm and 6 cm in lengths.
We know that, all sides of a rhombus are equal and the diagonals of a rhombus are perpendicular bisectors of one another. So, to construct a rhombus whose diagonals are 4 cm and 6 cm use the following steps.
- Draw the diagonal say AC = 4 cm
- Taking A and C as centres and radius more than ½ AC draw arcs on both sides of the line segment AC to intersect each other.
- Cut both arcs intersect each other at P and Q, then join PQ.
- Let PQ intersect AC at the point O. Thus, PQ is perpendicular bisector of AC.
- Cut off 3 cm lengths from OP and OQ, then we get points B and D.
- Now, join AB, BC, CD, and DA .
Thus, ABCD is the required rhombus.
Since, D and B lie on perpendicular bisector of AC.
DA = DC and BA = BC …(i)
[since, every point on perpendicular bisector of line segment is equidistant from end points
of line segment]
Now, ∠DOC = 90°
Also, OD = OB = 3 cm
Thus, AC is perpendicular bisector or BD.
CD = CB …(ii)
AB = BC =CD = DA
from Eq (i)and (ii)
ABCD is a rhombus.