Geometrical Constructions — Telangana (SSC) Class 9 Mathematics Solutions (Free)

Step 1: Draw a straight line and mark point A on it Step 2: With compass point at A, draw an arc of any radius R cutting the line at B Step 3: With the same radius R and compass point at B, draw an arc cutting the first arc at C Step 4: Draw line AC Step 5: ∠BAC = 60° (since triangle ABC is equilateral when R is same) Final Answer: ∠BAC = 60°

Step 1: Let AB be the given line segment Step 2: With compass point at A, draw arcs above and below AB with radius more than half of AB Step 3: With the same radius and compass point at B, draw arcs intersecting the first arcs at points P and Q Step 4: Draw line PQ Step 5: Line PQ is the perpendicular bisector of AB Step 6: It passes through the midpoint and is perpendicular to AB

Step 1: First construct an angle of 60° (as shown in earlier construction) Step 2: Let the 60° angle be ∠BAC Step 3: Construct the angle bisector of ∠BAC Step 4: To bisect: with compass point at A, draw an arc cutting AB at D and AC at E Step 5: With equal radius and compass points at D and E, draw arcs intersecting at F Step 6: Draw line AF Step 7: ∠BAF = ∠CAF = 30° Final Answer: ∠BAF = 30°

Step 1: Let ∠BAC be the given angle Step 2: With compass point at A, draw an arc cutting AB at P and AC at Q Step 3: With compass point at P and any radius, draw an arc inside the angle Step 4: With the same radius and compass point at Q, draw another arc intersecting the first arc at R Step 5: Draw line AR Step 6: Line AR bisects ∠BAC, so ∠BAR = ∠CAR = ∠BAC/2 Final Answer: AR is the angle bisector of ∠BAC

Step 1: Let AB be the given line and P be the point on it Step 2: With compass point at P, draw equal arcs on both sides cutting the line at Q and R Step 3: With radius more than PQ, draw arcs from Q and R intersecting at S Step 4: Draw line PS Step 5: PS is perpendicular to AB at P, so ∠APM = ∠BPM = 90° Final Answer: PS ⊥ AB at point P