Construct a parallelogram ABCD, in which diagonal AC = 3.8 cm, diagonal BD = 4.6 cm and the angle between AC and BD is 60°.

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Solution

We know that the diagonals of a parallelogram bisect each other.

Steps of construction:
Step 1: Draw AC= $3.8 cm$
Step 2: Bisect AC at O.
Step 3: Make $∠ C O X = 60^{\circ}$
Produce XO to Y.
Step 4:
$O B = \frac{1}{2} (4.6) cm O B = 2.3 cm and O D = \frac{1}{2} (4.6) cm O D = 2.3 cm$
Step 5: Join AB, BC, CD and AD.
Thus, ABCD is the required parallelogram.

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A B C D

A C = 6 c m, B D = 4.6 c m

45^{o}

Construct a parallelogram ABCD, if :

(i) AB = 3.6 cm, BC = 4.5 cm and $∠ ABC = 120^{\circ} .$

(ii) BC = 4.5 cm, CD = 5.2 cm and $∠ ADC = 75^{\circ} .$

(iii) AD = 4 cm, DC = 5 cm and diagonal BD = 7 cm.

(iv) AB = 5.8 cm, AD = 4.6 cm and diagonal AC = 7.5 cm.

(v) diagonal AC = 6.4 cm, diagonal BD = 5.6 cm and angle between the diagonals is $75^{\circ}$ .

(vi) lengths of diagonals AC and BD are 6.3 cm and 7.0 cm respectively, and the angle between them is $45^{\circ}$ .

(vii) lengths of diagonals AC and BD are 5.4 cm and 6.7 cm respectively and the angle between then is $60^{\circ}$ .

60^{\circ}

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