Constructions — Karnataka (SSLC) Class 9 Mathematics Solutions (Free)

Step 1: Draw line segment AB = 7 cm using straightedge and compass. Step 2: Set compass to more than half the length of AB (e.g., 4 cm). Step 3: From point A, draw an arc above and below the line AB. Step 4: From point B with the same compass radius, draw arcs above and below the line AB, intersecting the first two arcs at points P and Q. Step 5: Join P and Q using straightedge. Line PQ is the perpendicular bisector of AB. Verification: Any point on PQ is equidistant from A and B. The perpendicular bisector meets AB at its midpoint. Final Answer: Line PQ is perpendicular bisector of AB, meeting AB at right angles at its midpoint.

Step 1: Draw an angle of 60° with vertex at O. (Use compass: draw arc of any radius from O intersecting both rays at A and B. From A and B with same radius, draw intersecting arc at C. Angle AOC = 60°.) Step 2: From vertex O, set compass to any convenient radius and draw an arc intersecting both rays of the angle at points P and Q. Step 3: From point P, set compass to more than half of PQ and draw an arc inside the angle. Step 4: From point Q with the same compass radius, draw an arc inside the angle intersecting the first arc at point R. Step 5: Join O and R using straightedge. Ray OR bisects the 60° angle into two 30° angles. Verification: Angle POR = Angle QOR = 30°. Final Answer: Ray OR is the angle bisector, dividing the 60° angle into two equal 30° angles.

Step 1: Draw base BC = 6 cm using straightedge and compass. Step 2: From B, set compass to 5 cm radius and draw an arc above BC. Step 3: From C, set compass to 5 cm radius and draw an arc above BC intersecting the first arc at point A. Step 4: Join AB and AC. Triangle ABC is isosceles with AB = AC = 5 cm and BC = 6 cm. Step 5: To construct altitude from A (vertex angle) to BC: the altitude is perpendicular from A to line BC. Step 6: From A, set compass to more than the perpendicular distance to BC. Draw two arcs intersecting BC extended at points on either side. Step 7: Using compass, from these two intersection points draw equal radius arcs to find point D on BC such that AD ⊥ BC. Step 8: Join AD. This is the altitude from vertex A. Verification: In isosceles triangle, altitude from vertex angle bisects the base, so BD = DC = 3 cm. Final Answer: Altitude AD from A perpendicular to BC at point D, where BD = DC = 3 cm.

Step 1: Draw line segment PQ = 9 cm using straightedge and compass. Step 2: Draw a ray at acute angle from P, different from PQ. Call it PR. Step 3: Using compass with any radius, mark 3 equal divisions (since ratio is 2:1, total parts = 3) on ray PR from P. Mark points P₁, P₂, P₃ where PP₁ = P₁P₂ = P₂P₃. Step 4: Join P₃Q using straightedge. Step 5: From P₂, construct a line parallel to P₃Q using compass and straightedge. Step 6: This parallel line intersects PQ at point M. Point M divides PQ in ratio 2:1. Final Answer: Point M on PQ such that PM:MQ = 2:1, where PM = 6 cm and MQ = 3 cm.

Step 1: Draw base BC = 7 cm using straightedge and compass. Step 2: At point B, construct a 45° angle. (Use compass: first construct 90° by perpendicular bisector method at B. Then bisect the 90° angle using angle bisector to get 45°.) Step 3: On the 45° ray from B, measure 6 cm using compass and mark point A. Step 4: Join A and C using straightedge to complete triangle ABC. Verification: AB = 6 cm, BC = 7 cm, angle ABC = 45°. Final Answer: Triangle ABC with sides AB = 6 cm, BC = 7 cm and included angle ABC = 45°.

Step 1: Draw a line AB using straightedge. Step 2: Mark a point P above line AB (external point). Step 3: From P, set compass to any radius greater than the perpendicular distance from P to line AB. Draw two arcs intersecting line AB at points C and D on either side of the foot of perpendicular. Step 4: From C and D, set compass to same radius (more than half CD) and draw arcs below line AB intersecting at point Q. Step 5: Join P and Q using straightedge. Line PQ intersects AB at point M, and PM ⊥ AB. Verification: Angle PMB = 90°. Final Answer: Line PM perpendicular to AB, where M is foot of perpendicular from external point P to line AB.

Step 1: Draw a ray OA using straightedge. Step 2: Construct a 60° angle at O using compass. (Set compass to radius r, draw arc from O intersecting OA at P. From P with same radius r, draw arc intersecting the first arc at Q. Angle QOA = 60°.) Step 3: Bisect the 60° angle using angle bisector. From O, set compass to same radius and draw arc intersecting both rays at points C and D. Step 4: From C and D with radius more than half CD, draw intersecting arcs at point R inside the angle. Step 5: Ray OR bisects 60° angle. Angle ROA = 30°. Final Answer: Angle of 30° constructed as half of 60° angle using angle bisector.

Step 1: Draw triangle ABC with sides AB = 5 cm, BC = 6 cm, CA = 7 cm. Step 2: Construct angle bisector of angle A using compass: from A set compass to radius r, draw arc intersecting AB and AC at points P and Q. From P and Q with same radius, draw intersecting arc inside angle at point R. Ray AR is angle bisector of angle A. Step 3: Similarly, construct angle bisector of angle B. Step 4: The two angle bisectors intersect at point I, the incentre of triangle. Step 5: From I, draw perpendicular to any side (say AB) to find point of tangency. Measure distance from I to AB (inradius r). Step 6: Draw circle with centre I and radius r. This circle is tangent to all three sides and is the incircle. Final Answer: Incircle of triangle ABC with centre I at intersection of angle bisectors and inradius r ≈ 1.55 cm.