In trapezium ABCD, AB is parallel to CD and EF IS median. DC=6cm, EF=4cm,then AB is?
AntrikshSharma:
ef is the median to what
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To find CD, we need to get the segments DP and CQ, which is possible by applying Pythagoras theorem:
AD^2 = AP^2 + DP^2
13^2 = 12^2 + DP^2
169 = 144 + DP^2
Hence, DP = 5
Similarly, BC^2 = BQ^2 + CQ^2
15^2 = 12^2 + CQ^2
225 = 144 + CQ^2
Hence, CQ = 9
Required length, CD = DP + PQ + CQ = 5 + 20 + 9 = 34
AD^2 = AP^2 + DP^2
13^2 = 12^2 + DP^2
169 = 144 + DP^2
Hence, DP = 5
Similarly, BC^2 = BQ^2 + CQ^2
15^2 = 12^2 + CQ^2
225 = 144 + CQ^2
Hence, CQ = 9
Required length, CD = DP + PQ + CQ = 5 + 20 + 9 = 34
we have to find AB
sorry not understand
plz mark it as brainliest
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