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

Step 1: Draw base AB = 5 cm Step 2: With compass point at A and radius 7 cm, draw an arc Step 3: With compass point at B and radius 6 cm, draw another arc Step 4: The arcs intersect at C Step 5: Join AC and BC Step 6: Triangle ABC has sides 5 cm, 6 cm, 7 cm Final Answer: Triangle ABC constructed with given sides

Step 1: Draw a line segment AB = 4 cm Step 2: At A, construct a perpendicular using compass (as shown in earlier method) Step 3: On this perpendicular, mark point D such that AD = 4 cm Step 4: With compass point at D, draw an arc of radius 4 cm Step 5: With compass point at B, draw an arc of radius 4 cm Step 6: These arcs intersect at C Step 7: Join BC, CD, and DA Step 8: ABCD is a square with side 4 cm Final Answer: Square ABCD constructed with side 4 cm

Step 1: Draw base AB = 6 cm Step 2: Construct a perpendicular at a point (say at distance from A) measuring 4 cm height Step 3: Mark the top point C Step 4: Alternatively, at A or B, construct 45° angle Step 5: From C, draw lines to make appropriate angles Step 6: The intersection gives the required triangle Note: Construction depends on exact position of 45° angle specified Final Answer: Triangle constructed with given constraints