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## In fig , XY || seg AC.If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC.Activity : 2AX = 3BX  ${AX \over BX } = {.....\over ....}$              ${ AX +BX \over BX}={.....+.......\over.....}$                   ... by componendo.              ${AB \over BX} = {.... \over ..... }$                              .. (I)             $\Delta$ BCA ~ $\Delta$ BYX                            ....... test of similarity.           ${ BA \over BX} ={ AC \over XY}$                            ... corresponding sides of similar triangles.              ${.... \over .... }= {AC \over 9}$              AC = ....                                                 ... from (I)

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Sol:

2AX = 3BX

${AX \over BX } = {3\over 2}$

${ AX +BX \over BX}={3+2\over2}$                   ... by componendo.

${AB \over BX} = {5 \over 2 }$                              .. (I)

$\Delta$ BCA ~ $\Delta$BYX                            ....... test of similarity.

${ BA \over BX} ={ AC \over XY}$                            ... corresponding sides of similar triangles.

${5\over 2 }= {AC \over 9}$

AC = 22.5                                                 ... from (I)

Similarity August 22 , 2018 0 Comments 101 views