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प्रश्न
Construct a trapezium ABCD in which AB || DC, AB = 4.4 cm, ∠ABC = 90°, ∠DAB = 120° and the distance between parallel sides is 3.5 cm.
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उत्तर
Given:
AB || DC, AB = 4.4 cm
∠ABC = 90°
∠DAB = 120°
Distance between the parallel sides (height) = 3.5 cm
Step-wise calculation and construction:
1. Place AB horizontally:
Draw AB = 4.4 cm.
Label A (left) and B (right).
2. Draw the top parallel line through points at distance 3.5 cm from AB:
From any point on AB, construct a line parallel to AB at a perpendicular distance of 3.5 cm. Use a compass to transfer the 3.5 cm distance and draw a line parallel to AB. This is the line that will contain DC.
3. Locate C:
Through B construct BC ⊥ AB because ∠ABC = 90°.
The intersection of this perpendicular with the top parallel is C.
Hence, BC = 3.5 cm.
4. Locate D:
At A construct ∠DAB = 120° measure 120° counterclockwise from AB. Draw the ray from A at 120°.
The intersection of this ray with the top parallel is D.
5. Join AD and CD to complete trapezium ABCD.
