Question
Question 203
Construct a trapezium ABCD, where AB∥CD, AD = BC = 3.2 cm, AB = 6.4 cm and CD = 9.6 cm. Measure ∠B and ∠A.
[Hint Difference of two parallel sides gives an equilateral triangle]
Answer: Option A
:
Steps of Construction
Step I Draw a line segment DC = 9.6 cm.
Step II With D as center, draw an angle measure 60∘. Now, cut-off it with an arc 3.2 cm called point A.
Step III Now, draw a parallel AB to CD.
Step IV Taking C as center, cut an arc B measure 3.2 cm on previous parallel line.
Step V Draw a line segment BE = 3.2 cm from arc B.
Step VI Join B to E and C.
Thus, we have required trapezium ABCD in which ∠A=120∘ and ∠B=∠EBC+∠ABE=60∘+60∘=120∘. (Since, BEC is an equilateral triangle and ABED is a parallelogam)
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:
Steps of Construction
Step I Draw a line segment DC = 9.6 cm.
Step II With D as center, draw an angle measure 60∘. Now, cut-off it with an arc 3.2 cm called point A.
Step III Now, draw a parallel AB to CD.
Step IV Taking C as center, cut an arc B measure 3.2 cm on previous parallel line.
Step V Draw a line segment BE = 3.2 cm from arc B.
Step VI Join B to E and C.
Thus, we have required trapezium ABCD in which ∠A=120∘ and ∠B=∠EBC+∠ABE=60∘+60∘=120∘. (Since, BEC is an equilateral triangle and ABED is a parallelogam)
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