Geometrical Constructions

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Step 1: Draw line segment AB = 6 cm. Step 2: Take A as center, draw arc with radius > 3 cm on both sides of AB. Step 3: Take B as center with same radius, draw arcs intersecting previous arcs at P and

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Proof: In triangles PAQ and PBQ: AP = BP (equal radii), AQ = BQ (equal radii), PQ = PQ (common). Therefore △PAQ ≅ △PBQ (SSS rule). So ∠APO = ∠BPO (CPCT). In △APO and △BPO: AP = BP, ∠APO = ∠BPO, OP = O

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Step 1: Draw ray AB with initial point A. Step 2: Take A as center with any radius, draw arc intersecting AB at D. Step 3: Take D as center with same radius, draw arc intersecting previous arc at E. S

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Step 1: First construct 60° angle using standard method. Step 2: Bisect the 60° angle. Method: Take any radius from vertex, mark arcs on both rays. From these points, draw equal arcs intersecting insi

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Step 1: Draw BC = 5 cm. Step 2: At B, construct ∠CBX = 60°. Step 3: From B on ray BX, mark point D such that BD = 8 cm. Step 4: Join CD. Step 5: Draw perpendicular bisector of CD meeting BD at A. Step

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Conceptual explanation: When we mark D on BX such that BD = AB + AC, and draw perpendicular bisector of CD, point A lies equidistant from C and D. Therefore AC = AD. Since A lies on BD, we have BD = B

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Step 1: Draw QR = 6 cm. Step 2: At Q, construct ∠RQX = 45°. Step 3: On QX, mark point S such that QS = 2 cm (= PQ - PR). Step 4: Join RS. Step 5: Draw perpendicular bisector of RS meeting QX at P. Ste

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Case (i): When AB > AC, use AB - AC. Mark point D on AB extended such that BD = AB - AC. Case (ii): When AB < AC, use AC - AB. Mark point D on AB produced such that BD = AC - AB. Key difference: In ca