Construct a parallelogram ABCD in which BC = 5 cm, $∠ B C D = 120^{\circ}$ CD = 4.8 cm.

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Solution

Steps of Construction:

(i) Draw a line segment BC = 5 cm.

(ii) At C, draw a ray making an angle of $120^{\circ}$ and cut off CD = 4.8 cm.

(iii) With centre B and radius 4.8 cm. with centre D and radius 5 cm. draw arcs intersecting each other at A.

(iv) Join AB and AD

ABCD is the required parallelogram.

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Similar questions

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 C D

∠ B A D = 45^{o}

A D = A B = 6 c m, B C = 3.6 c m, C D = 5 c m .

∠ B C D .

Is it possible to construct a unique quadrilateral ABCD with AB = 5cm, BC = 6cm, CD= 7cm, $∠ A B C =$ $40^{\circ}$ and $∠ B C D =$ $55^{\circ}$ ?

∠ B = 40^{\circ}

∠ C = 120^{\circ}

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