Question
Answer:
:
Given, an isosceles trapezium, where AB∥DC, AD = BC and ∠A=60∘.
Then, ∠B=60∘.
Draw a line parallel to BC through D which intersects the line AB at E (say).
Then, DEBC is a parallelogram, where
BE = CD = 20 cm and DE = BC = 10 cm
Now, ∠DEB+∠CBE=180∘
[adjacent angles are supplementary in parallelogram]
⇒∠DEB=180∘−60∘=120∘
∴InΔADE,∠ADE=60∘ [exterior angle]
Also, ∠DEA=60∘ [∵ AD = DE = 10cm and ∠DAE=60∘]
Then, ΔADE is an equilateral triangle.
∴AE=10cm
⇒AB=AE+EB=10+20=30cm
Hence, x = 30 cm.
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:
Given, an isosceles trapezium, where AB∥DC, AD = BC and ∠A=60∘.
Then, ∠B=60∘.
Draw a line parallel to BC through D which intersects the line AB at E (say).
Then, DEBC is a parallelogram, where
BE = CD = 20 cm and DE = BC = 10 cm
Now, ∠DEB+∠CBE=180∘
[adjacent angles are supplementary in parallelogram]
⇒∠DEB=180∘−60∘=120∘
∴InΔADE,∠ADE=60∘ [exterior angle]
Also, ∠DEA=60∘ [∵ AD = DE = 10cm and ∠DAE=60∘]
Then, ΔADE is an equilateral triangle.
∴AE=10cm
⇒AB=AE+EB=10+20=30cm
Hence, x = 30 cm.
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