Question 199
Construct a quadrilateral NEWS in which NE = 7 cm, EW = 6 cm, $∠ N = 60^{\circ}, ∠ E = 110^{\circ}$ and $∠ S = 85^{\circ}$ .

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Solution

Fourth angle $= 360^{\circ} - (60^{\circ} + 110^{\circ} + 85^{\circ}) = 360^{\circ} - 255^{\circ} = 105^{\circ}$
Steps of Construction
Step I Draw NE = 7 cm.
Step II Make $∠ N E X = 110^{\circ}$ .
Step III With E as centre and radius 6 cm, draw an arc, cutting EX at W.
Step IV Make $∠ E W Y = 105^{\circ}$ .
Step V Make $∠ E N Z = 60^{\circ}$ , so that NZ and WY intersect each other at point S.
Thus, NEWS is the required quadrilateral.

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Similar questions

∠ N = 60^{\circ}, ∠ E = 110^{\circ}

∠ S = 85^{\circ}

∠ N = 60^{\circ}, ∠ E = 110^{\circ}

∠ S = 85^{\circ}

M O R E

M O = 6 c m

O R = 4.5 c m

∠ M = 60^{\circ}

∠ O = 105^{\circ}

∠ R = 105^{\circ}

P L A N

P L = 4 c m

L A = 6.5 c m

∠ P = 90^{\circ}

∠ A = 110^{\circ}

∠ N = 85^{\circ}

H E A R

H E = 5 c m

E A = 6 c m

∠ R = 85^{\circ}

O K A Y

O K = 7 c m

K A = 5 c m

Construct a cyclic quadrilateral EFGH with EF = 6 cm, $∠ G E F = 60^{\circ}$ , FG = 7cm and $∠ E G H = 50^{\circ}$ .

Construct a quadrilateral $B E A R$ in which $B E = 6 c m, E A = 7 c m, R B = R E = 5 c m$ and $B A = 9 c m$ . Measure its fourth side.

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