JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

1 Easy

1 of 1

Application +4 | -1

x,y,z

are perpendicular drawn from the vertices of triangle having sides

a,b

and

c

, then the value of

\frac{{bx}}{c} + \frac{{cy}}{a} + \frac{{az}}{b}

will be

View answer and explanation

Correct option: A

\frac{{{a^2} + {b^2} + {c^2}}}{{2R}}

Option A (

\frac{{{a^2} + {b^2} + {c^2}}}{{2R}}

) is correct because it is derived from the area of the triangle and the circumradius.

More Questions from PROPERTIES OF TRIANGLES

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{p_1},{p_2},{p_3}

\Delta \;ABC

A,B,C

\Delta

p_1^{ - 2} + p_2^{ - 2} + p_3^{ - 2}

5k,\;6k,\;5k

k

\Delta \,ABC

a = \left( {b - c} \right)\sec \theta

\frac{{2\,\sqrt {bc} }}{{b - c}}\,\,\sin \,\,\frac{A}{2} =

\Delta \;ABC,\frac{{b + c}}{a}

\Delta = {a^2} - {\left( {b - c} \right)^2}

\Delta

\Delta \;ABC

\tan A

A

2 s