construct a quadrilater ABCD ab=6cm bc =4.5cm angle a=60° angle b=105° and c=105°
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Assuming we are familiar with the ways how to draw lines perpendicularly bisecting each other, or duplicate an included angle obtained during the constr. wk.s. Those wk.s are omitted in here. 1). Draw the side AB on a horizontal, cutting across the line with given compass width, 6cm. Mark the left end of segment A, and the right end B. Vertices C and D are constructed above AB. 2). Mark pt.P on AB at a point, a bit nearer to B. From A and P, draw arcs above AP with radius AP crossing each other at Q. From B and P, draw arcs above BP with radius BP crossing each other at R. Draw 2 rays, one from A thru Q, and the other from B thru R. ∠A =∠BAQ =∠ABR =60°. 3). From R, draw a ray upwards that is perpendicular to BR, then an arc from R with radius BR crossing the perpendicular at S above R. Draw a ray from B thru S. From B, draw an arc with radius of given length, BC=4.5cm, crossing the ray BS at C. ∠RBS =45°, so ∠B =ABR+∠RBS =60°+45° =105° 4). Duplicate ∠B to construct ∠C (=105). Extend the upper side of vertex C ( duplicated ) crossing the ray from A at D. ∠D must be 90 deg. CK ∠D: Draw an circle D with radius DQ, crossing the ray from C at each side of D, M and N. QM=QN, so AD⊥CD. ∠D=90° CKD. The quadrilateral ABCD is the ordered one in which AB=6cm, BC=4.5cm, ∠A=60° and ∠B=∠C=105°.