Please solve q 19?

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1 Answer
CW
May 9, 2018

DM=3/2 units

Explanation:

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Given angleDAB=60^@, angleADC=120^@,
=> DC // AB
Let E be the midpoint of AD,
=> AE=ED=sqrt3/2
given M is the midpoint of BC, => EM // DC // AB
=> angleDEM=60^@,
and EM=(AB+DC)/2=(2*AD)/2=(2sqrt3)/2=sqrt3
In DeltaEDM, as EM=2ED, and angleDEM=60^@,
=> angleEMD=30^@, => angleEDM=90^@,
=> DeltaEDM is right angled at D,
=> DM=EMsin60=sqrt3*sqrt3/2=3/2 units

Or if you are not sure if DeltaEDM is a right triangle, you can also use the law of cosines to find the length DM,
DM^2=ED^2+EM^2-2*ED*EM*cos60,
=> DM^2=(sqrt3/2)^2+sqrt3^2-2*sqrt3/2*sqrt3*1/2
=> DM^2=3/4+3-3/2=9/4
=> DM=sqrt(9/4)=3/2 units

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