Question

follow my above account okk

in a Trapezium ABCD AD=DC=BC=20cm angle A =60°
find
1. length of AB
2. DISTANCE BETWEEN AB and DC
​​

Answer 1

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.

Explanation:

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Answer 2

Answer:

follow my above account okk

in a Trapezium ABCD AD=DC=BC=20cm angle A =60°

find

1. length of AB

2. DISTANCE BETWEEN AB and DC

your answer is

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.