Constructions — Karnataka (SSLC) Class 10 Mathematics Solutions (Free)

Step 1: Draw a circle with centre O and radius 3 cm using compass. Step 2: Mark a point P on the circumference of the circle. Step 3: Draw the radius OP using a straightedge. Step 4: Place compass at P and draw two arcs on either side of OP at equal distance from O, creating points A and B on OP extended. Step 5: The tangent at P is the line perpendicular to OP passing through P. Use compass to construct this perpendicular: place compass at P with radius greater than the distance to O, draw two arcs above and below OP. From the arc intersection points, draw arcs of equal radius to find the perpendicular. Step 6: The line through P perpendicular to OP is the required tangent. Final Answer: Tangent line at P perpendicular to radius OP.

Step 1: Draw line segment AB = 6 cm using straightedge and compass. Step 2: Draw an acute angle at A, making any angle with AB. Call this ray AX. Step 3: On ray AX, mark 5 points (since ratio is 3:2, total parts = 3 + 2 = 5) at equal distance using compass. Name these points A₁, A₂, A₃, A₄, A₅ such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅. Step 4: Join A₅B using straightedge. Step 5: Through point A₃ (which is 3 parts from A), draw a line parallel to A₅B. Use compass and straightedge to construct this parallel line. Step 6: This parallel line meets AB at point P. Here P divides AB in ratio 3:2. Final Answer: Point P on AB such that AP:PB = 3:2, where AP = 3.6 cm and PB = 2.4 cm.

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 7 cm radius and draw an arc above BC intersecting the first arc at point A. Step 4: Join AB and AC to complete triangle ABC with sides AB = 5 cm, BC = 6 cm, AC = 7 cm. Step 5: Construct the circumcircle of triangle ABC by finding the circumcentre (intersection of perpendicular bisectors of two sides). Draw perpendicular bisectors of AB and BC, they meet at O. Step 6: Draw circle with centre O and radius OA (or OB or OC). Step 7: To construct tangent at vertex A: Draw radius OA. The tangent at A is perpendicular to OA. Construct perpendicular at A using compass (draw equal arcs on both sides of OA from A, then from those arc intersections, draw equal radius arcs to find perpendicular). Step 8: The line through A perpendicular to OA is the required tangent to circumcircle at A. Final Answer: Triangle ABC with circumcircle and tangent at vertex A, perpendicular to radius OA.

Step 1: Draw a circle with centre O and radius 2.5 cm. Step 2: Mark point P at distance 5 cm from O (OP = 5 cm). Step 3: Join O and P using straightedge. Step 4: Find the midpoint M of OP by constructing perpendicular bisector of OP using compass. Step 5: Draw a circle with centre M and radius MO (= MF = 2.5 cm). Step 6: This circle intersects the original circle at two points, say T₁ and T₂. These are the points of contact of tangents from P. Step 7: Join PT₁ and PT₂. These are the two required tangents. Verification: Angle OT₁P = Angle OT₂P = 90° (tangent perpendicular to radius). Final Answer: Two tangents PT₁ and PT₂ drawn from external point P to circle with centre O, where both tangents have equal length = √(OP² - r²) = √(25 - 6.25) = √18.75 ≈ 4.33 cm.

Step 1: We know that 5 = 1² + 2². So we can use Pythagoras theorem to construct √5. Step 2: Draw a perpendicular to a line at a point. Using compass, construct a right angle. Step 3: From the vertex of right angle, measure 1 cm on one arm and 2 cm on the other arm using compass. Step 4: Mark points A and B at these distances from the right angle vertex O. Step 5: Join AB using straightedge. By Pythagoras theorem, AB = √(1² + 2²) = √5 cm. Alternatively: Draw a right triangle with legs 1 cm and 2 cm. The hypotenuse will be √5 cm. Final Answer: Line segment of length √5 cm constructed as hypotenuse of right triangle with perpendicular sides 1 cm and 2 cm.

Step 1: Draw line segment PQ = 8 cm using straightedge and compass. Step 2: Draw a ray at an acute angle from P. Call this ray PR. Step 3: Using compass, set any radius and mark 8 equal divisions on ray PR starting from P (since ratio is 5:3, total = 8 parts). Mark points P₁, P₂, P₃, P₄, P₅, P₆, P₇, P₈ such that PP₁ = P₁P₂ = P₂P₃ = ... = P₇P₈. Step 4: Join P₈Q using straightedge. Step 5: From point P₅ (which is 5 parts from P), draw a line parallel to P₈Q using compass and straightedge. This parallel line meets PQ at point M. Step 6: Point M divides PQ in ratio 5:3 internally. Final Answer: Point M on PQ such that PM:MQ = 5:3, where PM = 5 cm and MQ = 3 cm.

Step 1: Draw base BC = 5 cm using straightedge and compass. Step 2: At point B, construct an angle of 60° using compass. (To construct 60°: draw an arc of any radius from B, then from the point where arc meets BC, draw another arc of same radius to intersect the first arc at a point, which gives 60° angle.) Step 3: From the 60° angle ray at B, measure 4 cm using compass and mark point A. Step 4: Join A and C using straightedge to complete triangle ABC. Verification: Measure BC = 5 cm, angle ABC = 60°, and AB = 4 cm. Final Answer: Triangle ABC with AB = 4 cm, BC = 5 cm, and angle ABC = 60°.

Step 1: Draw triangle ABC with sides AB = 4.5 cm, BC = 5.5 cm, CA = 6 cm. Step 2: Construct perpendicular bisector of side AB: mark midpoint M₁ of AB, then draw perpendicular to AB at M₁ using compass. Step 3: Construct perpendicular bisector of side BC: mark midpoint M₂ of BC, then draw perpendicular to BC at M₂ using compass. Step 4: The two perpendicular bisectors intersect at point O, which is the circumcentre. Step 5: Using compass, measure distance OA (or OB or OC) = circumradius R. Step 6: Draw circle with centre O and radius R. This circle passes through all three vertices A, B, C and is the circumcircle. Final Answer: Circumcircle of triangle ABC with circumradius R ≈ 3.5 cm and centre O at intersection of perpendicular bisectors.