In a right-angled triangle \[ABC,\,\,\angle BCA=90{}^\circ \]. CD is perpendicular from C to AB. By using the concept of area or areas of triangles, which one of the following relationships holds good?

\[\frac{1}{C{{D}^{2}}}=\frac{1}{B{{C}^{2}}}-\frac{1}{C{{A}^{2}}}\]

\[\frac{1}{C{{D}^{2}}}=\frac{1}{A{{B}^{2}}}+\frac{1}{C{{A}^{2}}}\]

\[\frac{1}{C{{D}^{2}}}=\frac{1}{A{{B}^{2}}}-\frac{1}{C{{A}^{2}}}\]

\[\frac{1}{C{{D}^{2}}}=\frac{1}{B{{C}^{2}}}+\frac{1}{C{{A}^{2}}}\]

In a right-angled triangle \[ABC,\,\,\angle BCA=90{}^\circ \]. CD is p

1.	In a right-angled triangle \[ABC,\,\,\angle BCA=90{}^\circ \]. CD is perpendicular from C to AB. By using the concept of area or areas of triangles, which one of the following relationships holds good?
A.	\[\frac{1}{C{{D}^{2}}}=\frac{1}{B{{C}^{2}}}-\frac{1}{C{{A}^{2}}}\]
B.	\[\frac{1}{C{{D}^{2}}}=\frac{1}{A{{B}^{2}}}+\frac{1}{C{{A}^{2}}}\]
C.	\[\frac{1}{C{{D}^{2}}}=\frac{1}{A{{B}^{2}}}-\frac{1}{C{{A}^{2}}}\]
D.	\[\frac{1}{C{{D}^{2}}}=\frac{1}{B{{C}^{2}}}+\frac{1}{C{{A}^{2}}}\]
Answer» E.

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In a right-angled triangle \[ABC,\,\,\angle BCA=90{}^\circ \]. CD is perpendicular from C to AB. By using the concept of area or areas of triangles, which one of the following relationships holds good?

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