. In trapezium ABCD, as shown, AB // DC,AD = DC = BC = 20 cm and A = 6

1.	. In trapezium ABCD, as shown, AB // DC,AD = DC = BC = 20 cm and A = 600Find:(i) length of AB(ii) distance between AB and DC.600
Answer» assume a perpendicular to AB from D meets at F. we get DF/AD = Sin60° so, DF = 20√3/2 = 10√3. this is the distance between AB and DC. now AF = 10√3/tan60° = 10√3/√3 = 10. so, AB = 2AF + DC = 2*10+20 = 40.

. In trapezium ABCD, as shown, AB // DC,AD = DC = BC = 20 cm and A = 600Find:(i) length of AB(ii) distance between AB and DC.600

Answer»

assume a perpendicular to AB from D meets at F.

we get DF/AD = Sin60° so, DF = 20*√3/2 = 10√3. this is the distance between AB and DC.

now AF = 10√3/tan60° = 10√3/√3 = 10.

so, AB = 2*AF + DC = 2*10+20 = 40.