Can anyone help me woth this triangles related q. please provide steps too. Thanks
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since AM=MB i.e. m is the mid point
join CM ( which will form median )
so, ar(∆BMC)=ar[∆AMC)
nD again join cm
therefore ar[∆MDC)= ar(∆EMD) (on the sme bace nd bw sm parallel )
so area of ∆BMC =1/2 ar∆ABC
nd ar∆BEC=1/2 ABC
BEC = 1/2×24
= 12
join CM ( which will form median )
so, ar(∆BMC)=ar[∆AMC)
nD again join cm
therefore ar[∆MDC)= ar(∆EMD) (on the sme bace nd bw sm parallel )
so area of ∆BMC =1/2 ar∆ABC
nd ar∆BEC=1/2 ABC
BEC = 1/2×24
= 12
mohitjoshi560:
I could make out how u got the area half of abc?
ok thanks i got it
good
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