find the area of triangle ABC in which Ab =8 cm AC =10 CM ANGLE ABC =45 DEGREE
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Answer:40
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insha126:
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In ∆ABC ,AD is median . D. will be the mid point of BC.
BD=CD= BC/2= 14/2=7cms.
AB^2+AC^2 =2(AD^2+BD^2)
8^2+10^2=2(AD^2+7^2)
164/2=AD^2+49
82–49=AD^2
33=AD^2
AD=√33 cms.
BD=CD= BC/2= 14/2=7cms.
AB^2+AC^2 =2(AD^2+BD^2)
8^2+10^2=2(AD^2+7^2)
164/2=AD^2+49
82–49=AD^2
33=AD^2
AD=√33 cms.
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