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Hello guys can someone help me with this, my dead line is too close hehe

please guys no trolling

Hello guys can someone help me with this, my dead line is too close hehe please guys-example-1

1 Answer

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We are given the measurements:

  • AE = 8m-13
  • PX = 5m+20

Now,


\boxed { \mathcal{ \green{AE = PX}}}

Since, the diagonals of a rectangle are equal..

  • Put the values of AE and PX equal to find m


\pink \implies \purple { \sf8m -13 = 5m + 20 }


\pink \implies \purple { \sf8m -13 - 5x = 20 }


\pink \implies \purple { \sf8m - 5m = 20 + 13 }


\pink \implies \purple { \sf3m = 33 }


\boxed{ \mathfrak { \red{m = 11}}}

#4


\blue \longmapsto \orange{ \tt \: AE = 8m - 13}


\blue \longmapsto \orange{ \tt \: AE = 8(11) - 13}


\blue \longmapsto \orange{ \tt \: AE = 88- 13}


\boxed{ \mathfrak{ \red{AE = 75 \: cm}}}

#5


\blue \longmapsto \orange{ \tt \:PX = 5m+ 20}


\blue \longmapsto \orange{ \tt \:PX = 5(11)+ 20}


\blue \longmapsto \orange{ \tt \:PX = 55+ 20}


\boxed{ \mathfrak{ \red{ PX= 75 \: cm}}}

#6


\boxed{ \mathfrak { \red{m = 11}}}

answered
User Zeeshan Mehdi
by
8.5k points

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