Question
ABCD is a trapezium such that AB∥CD, ∠A:∠D=2:1,∠B:∠C=7:5. Find the angles of the trapezium.
ABCD is a trapezium such that AB∥CD, ∠A:∠D=2:1,∠B:∠C=7:5. Find the angles of the trapezium.
Answer:
:
Let ABCD be a trapezium, where AB∥CD
Let the angles A and D be of measures 2x and x, respectively.
Then, 2x+x=180∘ [∵ in trapezium, the angles on either side of the base are supplementary]
⇒3x=180∘⇒x=60∘
∴∠A=2×60∘=120∘,∠D=60∘
Again, let the angles B and C be 7x and 5x respectively. Then, 7x+5x=180∘
⇒12x=180∘⇒x=15∘
Thus, ∠B=7×15=105∘ and ∠C=5×15=75∘
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:
Let ABCD be a trapezium, where AB∥CD
Let the angles A and D be of measures 2x and x, respectively.
Then, 2x+x=180∘ [∵ in trapezium, the angles on either side of the base are supplementary]
⇒3x=180∘⇒x=60∘
∴∠A=2×60∘=120∘,∠D=60∘
Again, let the angles B and C be 7x and 5x respectively. Then, 7x+5x=180∘
⇒12x=180∘⇒x=15∘
Thus, ∠B=7×15=105∘ and ∠C=5×15=75∘
Was this answer helpful ?
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