Construct a trapezium ABCD in which AB = 6 cm, BC = 4 cm, CD = 3.2 cm, $∠ B = 75^{\circ}$ and DC || AB.

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Solution

Steps of Construction :

(i) Draw a line segment AB = 6 cm.

(ii) At B, draw a ray BX making an angle of $75^{\circ}$ and cut off BC = 4 cm.

(iii) At C, draw a ray CY making an angle of $180^{\circ} - 75^{\circ} = 105^{\circ}$ .

So that CY may be parallel to AB.

(iv) From CY, Cut off CD = 3.2 cm.

(v) Join DA.

ABCD is the required trapezium.

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Similar questions

A B = 6 c m, A D = 4 c m, C D = 3 c m

∠ D A B = 90^{o}

A B ∥ C D

Construct a quadrliateral ABCD; if:

(i) AB = 4.3 cm, BC = 5.4, CD = 5 cm, DA = 4.8 cm and $∠$ ABC = $75^{\circ}$ .

(ii) AB = 6 cm, CD = 4.5 cm, BC = AD = 5 cm and $∠ BCD = 60^{\circ}$

(iii) AB = 8 cm, BC = 5.4 cm, AD = 6 cm, $∠ A = 60^{\circ}$ and $∠ B = 75^{\circ}$ .

(iv) AB = 5 cm, BC = 6.5 cm, CD = 4.8 cm, $∠ B = 75^{\circ}$ and $∠ C = 120^{\circ}$

(v) AB = 6 cm = AC, BC = 4 cm, CD = 5 cm and AD = 4.5 cm.

(vi) AB = AD = 5 cm, BD = 7 cm and BC = DC = 5.5 cm

A B = 6 c m, A D = 4 c m, C D = 3 c m

∠ D A B = 90^{o}

A B ∥ C D

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