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Maths-

General

Easy

Question

If the sides be a, b, c then $fraction numerator b minus c over denominator r subscript 1 end subscript end fraction$ + $fraction numerator c minus a over denominator r subscript 2 end subscript end fraction$ + $fraction numerator a minus b over denominator r subscript 3 end subscript end fraction$ =

5
4
0
1

The correct answer is: 0

Related Questions to study

If the sides be a, b, c then (r + r1) tan $fraction numerator B minus C over denominator 2 end fraction$ +(r + r2) tan $fraction numerator C minus A over denominator 2 end fraction$ + (r+ r3) tan $fraction numerator A minus B over denominator 2 end fraction$ =

If the sides be a, b, c then (r + r1) tan $fraction numerator B minus C over denominator 2 end fraction$ +(r + r2) tan $fraction numerator C minus A over denominator 2 end fraction$ + (r+ r3) tan $fraction numerator A minus B over denominator 2 end fraction$ =

$open parentheses fraction numerator 1 over denominator r subscript 1 end subscript end fraction plus fraction numerator 1 over denominator r subscript 2 end subscript end fraction close parentheses$ $open parentheses fraction numerator 1 over denominator r subscript 3 end subscript end fraction plus fraction numerator 1 over denominator r subscript 1 end subscript end fraction close parentheses$ =

$open parentheses fraction numerator 1 over denominator r subscript 1 end subscript end fraction plus fraction numerator 1 over denominator r subscript 2 end subscript end fraction close parentheses$ $open parentheses fraction numerator 1 over denominator r subscript 3 end subscript end fraction plus fraction numerator 1 over denominator r subscript 1 end subscript end fraction close parentheses$ =

$I f space fraction numerator left parenthesis x minus 2 right parenthesis left parenthesis 1 minus x right parenthesis left parenthesis x minus 3 right parenthesis cubed left parenthesis x minus 4 right parenthesis squared over denominator left parenthesis x plus 1 right parenthesis end fraction less or equal than 0 space t h e n space x space l i e s space i n$

$I f space fraction numerator left parenthesis x minus 2 right parenthesis left parenthesis 1 minus x right parenthesis left parenthesis x minus 3 right parenthesis cubed left parenthesis x minus 4 right parenthesis squared over denominator left parenthesis x plus 1 right parenthesis end fraction less or equal than 0 space t h e n space x space l i e s space i n$

parallel

(r1 – r) (r2 – r) (r3 – r) =

(r1 – r) (r2 – r) (r3 – r) =

$fraction numerator 1 over denominator r subscript 1 end subscript superscript 2 end superscript end fraction$ + $fraction numerator 1 over denominator r subscript 2 end subscript superscript 2 end superscript end fraction$ + $fraction numerator 1 over denominator r subscript 3 end subscript superscript 2 end superscript end fraction$ + $fraction numerator 1 over denominator r subscript blank superscript 2 end superscript end fraction$ =

$fraction numerator 1 over denominator r subscript 1 end subscript superscript 2 end superscript end fraction$ + $fraction numerator 1 over denominator r subscript 2 end subscript superscript 2 end superscript end fraction$ + $fraction numerator 1 over denominator r subscript 3 end subscript superscript 2 end superscript end fraction$ + $fraction numerator 1 over denominator r subscript blank superscript 2 end superscript end fraction$ =

$I f space fraction numerator left parenthesis 2 x minus 1 right parenthesis left parenthesis x minus 1 right parenthesis squared left parenthesis x minus 2 right parenthesis cubed over denominator left parenthesis x minus 4 right parenthesis to the power of 4 end fraction greater than 0 space t h e n space x space l i e s space i n$

$I f space fraction numerator left parenthesis 2 x minus 1 right parenthesis left parenthesis x minus 1 right parenthesis squared left parenthesis x minus 2 right parenthesis cubed over denominator left parenthesis x minus 4 right parenthesis to the power of 4 end fraction greater than 0 space t h e n space x space l i e s space i n$

parallel

$i f space fraction numerator open parentheses x squared plus 3 close parentheses open parentheses x minus 1 close parentheses to the power of 4 over denominator open parentheses x plus 1 close parentheses end fraction greater than 0 space t h e n space x element of$

$i f space fraction numerator open parentheses x squared plus 3 close parentheses open parentheses x minus 1 close parentheses to the power of 4 over denominator open parentheses x plus 1 close parentheses end fraction greater than 0 space t h e n space x element of$

If (a – b) (s – c) = (b – c) (s – a) then r1, r2, r3 are in -

If (a – b) (s – c) = (b – c) (s – a) then r1, r2, r3 are in -

If in a triangle $open parentheses 1 minus fraction numerator r subscript 1 end subscript over denominator r subscript 3 end subscript end fraction close parentheses$ = 2, then the triangle is-

If in a triangle $open parentheses 1 minus fraction numerator r subscript 1 end subscript over denominator r subscript 3 end subscript end fraction close parentheses$ = 2, then the triangle is-

parallel

The angles of a right-angled triangle are in A.P. The ratio of the in-radius and the perimeter is -

The angles of a right-angled triangle are in A.P. The ratio of the in-radius and the perimeter is -

In a ABC, if $fraction numerator R over denominator r end fraction$  2, then the triangle is -

In a ABC, if $fraction numerator R over denominator r end fraction$  2, then the triangle is -

abc =

abc =

parallel

If the altitudes of a triangle be 3, 4, 6 then its in-radius is -

If the altitudes of a triangle be 3, 4, 6 then its in-radius is -

4Rr cos $fraction numerator A over denominator 2 end fraction$ cos $fraction numerator B over denominator 2 end fraction$ cos $fraction numerator C over denominator 2 end fraction$ =

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If the area of a triangle is 81 square cm and its perimeter is 27cm then its in-radius in centi-metres is -

If the area of a triangle is 81 square cm and its perimeter is 27cm then its in-radius in centi-metres is -

parallel

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