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Draw and describe the locus of a point in rhombus ABCD which is equidistant from AB and AD.

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Answer

Let ABCD be a rhombus.

Draw and describe the locus of a point in rhombus ABCD which is equidistant from AB and AD. Locus, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

AC is diagonal of rhombus which will bisect angle A. As rhombus bisects the vertices angle.

∴ AC bisects ∠A

Let there be a point E and F on the sides AD and AB respectively. Draw a line perpendicular to AC from both the points.

Considering △AEG and △AFG

∠AGE = ∠AFG (Both are equal to 90°)
∠EAG = ∠FAG (Both are equal to half of ∠DAB.)

Hence, by AA axiom △AEG ~ △AFG.

Since triangles are similar so the ratio of their corresponding sides will be similar.

EGFG=AGAGEGFG=1EG=FG.\therefore \dfrac{EG}{FG} = \dfrac{AG}{AG} \\[1em] \Rightarrow \dfrac{EG}{FG} = 1 \\[1em] \Rightarrow EG = FG.

Since, distance from AD and AB is similar,

Hence, the locus of point is the diagonal AC of the rhombus ABCD.

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