in an isosceles trapezium abcd, side bc ll side ad. \_a =72°. find \_b please tell fast
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Through B, draw a straight line parallel to AD which meets CD at E.
Now, since AB∥DE and AD∥BE,
∴ABED is a parallelogram.
Thus, ED=AB=18cm
As, BE∥AD and CD is a transversal,
∴∠BED=∠D=60
o
(∵∠ABE=∠D).
Since, in an isosceles trapezium, the base angles are equal, ∠C=∠D=60
o
.
In ΔBEC,∠BEC+∠ECD+∠CBE=180
o
(Angle sum property of a triangle)
⇒60
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+60
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+∠CBE=180
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⇒120
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+∠CBE=180
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⇒∠CBE=180
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−120
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⇒∠CBE=60
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As the measure of ∠BEC=60
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,∠ECB=60
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and ∠CBE=60
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,ΔCBE is an equilateral triangle, Thus
∴CE=BC=12cm(BC=AD=12cm)
Now, CE+ED=12+18
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