If triangle ABC is similar to PQR if M is the midpoint of BC and N is the midpoint of QR.If the area of triangle ABC is 100cm and the area of triangle PQR is 144cm and AM is 4cm then PN is equal to?
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ar(∆ABC)/ ar(∆pqr) = AM²/ PN²
√∆ABC / √∆PQR = AM/ PN
√100/√144 = 4/ PN
PN = 12*4/10
PN= 4.8 cm
MitheshShankar:
dude you can't just directly take the medians AM and PN you first have to prove that ABM is similar to PQN
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52
here is the solution
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