Geometric Constructions — Maharashtra (SSC) Class 10 Mathematics Solutions (Free)

Given: Circle with center O and external point P. To construct: Tangent from P to the circle. Steps: Step 1: Join O and P. Step 2: Find midpoint M of OP. Using compass, measure distance OP, place compass at O and P alternately above and below OP line, mark two points, draw line through them to locate M. Step 3: With M as center and MO (or MP) as radius, draw a circle. Step 4: This circle intersects the original circle at two points; let them be A and B. Step 5: Join PA and PB. Both PA and PB are tangents from P to the original circle. Why it works: Angle OAP = Angle OBP = 90 degrees (angle in semicircle), confirming OA and OB are perpendicular to PA and PB respectively, making them tangents.

Given: Triangle ABC, scale factor k = 3/2. To construct: Triangle similar to ABC with sides 3/2 times larger. Steps: Step 1: Draw triangle ABC. Step 2: Extend AB to a point D such that AD = (3/2) * AB. Alternatively, divide AB into 2 parts, take 3 such parts on extension of AB. Step 3: From B, construct angle equal to angle ABC at D. Step 4: Extend line from D at angle DBA = angle ABC until it meets with line from A parallel to BC at point E. Step 5: Similarly, extend AC to F such that AF = (3/2) * AC. Step 6: Triangle AEF is the required triangle similar to ABC with scale factor 3/2. Alternative method: Using BPT (Basic Proportionality Theorem) - Draw a line DE parallel to BC such that AD/DB = AF/FC = 3/2, then DEF is similar to ABC.

Given: Circle with center O and radius 3 cm, line l. To construct: Tangent to circle parallel to line l. Steps: Step 1: Draw circle with center O and radius 3 cm. Step 2: Draw the given line l. Step 3: From O, draw a perpendicular to line l. Let the foot of perpendicular be H. Step 4: On the perpendicular OH, mark points A and B on opposite sides of O such that OA = OB = 3 cm (the radius). Step 5: Through A and B, draw lines parallel to l. These are the required tangents to the circle parallel to l. Why it works: The distance from O to any tangent equals the radius. The perpendicular distance from O to the tangent lines is 3 cm, confirming they are tangents.

Given: Side length 5 cm, scale factor 2/3. To construct: Similar triangle with sides 2/3 times original. Steps: Step 1: Construct equilateral triangle ABC with side 5 cm using compass and ruler. Step 2: To construct similar triangle with scale factor 2/3, sides should be (2/3) * 5 = 10/3 cm (approximately 3.33 cm). Step 3: On side AB, mark point P such that AP = (2/3) * AB. Divide AB into 3 equal parts using parallel lines method, take 2 parts. Step 4: Similarly, on side AC, mark point Q such that AQ = (2/3) * AC. Step 5: Join PQ. By basic proportionality theorem, PQ || BC. Step 6: Triangle APQ is similar to ABC with scale factor 2/3. Step 7: Construct the remaining part of the triangle from P and Q in the same manner to complete triangle PQR similar to ABC.

Given: Line segment AB = 8 cm, ratio 3:5. To divide: Point P on AB such that AP:PB = 3:5. Steps: Step 1: Draw line segment AB of length 8 cm. Step 2: From A, draw a ray AC making acute angle with AB. Step 3: On ray AC, starting from A, mark 3 + 5 = 8 equal segments using compass. Name the points as A1, A2, A3, ..., A8. Step 4: Join A8 to B. Step 5: Through point A3, draw a line parallel to A8B. This parallel line intersects AB at point P. Step 6: By basic proportionality theorem, AP/PB = 3/5, so P divides AB in ratio 3:5. Step 7: Measure AP and PB to verify. AP should be (3/8) * 8 = 3 cm, and PB should be 5 cm.

Given: Circle with center O, chord AB, tangent at A. To prove: Tangent-chord angle = (1/2) * arc AB. Construction: Step 1: Draw circle with center O. Step 2: Draw chord AB. Step 3: Draw tangent to circle at point A. Step 4: Let the tangent make angle with chord AB = angle PAB (where P is any point on tangent). Proof: Step 1: Join OA and OB (radii). Step 2: The tangent at A is perpendicular to radius OA (property of tangent). Step 3: In triangle OAB, let angle OAB = a and angle OBA = b. Since OA = OB (radii), triangle is isosceles, so a = b. Step 4: The arc AB subtends angle AOB = 2a at the center (inscribed angle theorem). Step 5: Angle PAB (tangent-chord angle) = 90 degrees - a (since tangent OA and radius are perpendicular). Wait, correction: Angle between tangent and chord AB = 90 - angle OAB = 90 - a, but arc angle = 2a, so tangent-chord angle = (1/2) * (2a) = a is not matching. Let me recalculate properly. Correct: Tangent-chord angle = angle between tangent PA and chord AB. By alternate segment theorem, this equals the angle in the alternate segment, which is (1/2) * arc AB.

Given: Square of side 6 cm, scale factor 1/2. To construct: Similar square with side 3 cm. Steps: Step 1: Construct square ABCD with side 6 cm using compass and ruler (right angles at all corners). Step 2: For similar square with scale factor 1/2, new side = (1/2) * 6 = 3 cm. Step 3: On side AB, mark point P such that AP = (1/2) * AB = 3 cm. Step 4: Similarly, on side AD, mark point Q such that AQ = (1/2) * AD = 3 cm. Step 5: From P, draw a line parallel to AD; from Q, draw a line parallel to AB. These lines intersect at R. Step 6: APQR is the required square with side 3 cm, similar to ABCD with scale factor 1/2. Step 7: Verify all sides = 3 cm and all angles = 90 degrees.

Given: Line segment PQ = 10 cm, ratio 2:3:5. To divide: Points A and B on PQ such that PA:AB:BQ = 2:3:5. Steps: Step 1: Draw line segment PQ of length 10 cm. Step 2: From P, draw a ray PR making acute angle with PQ. Step 3: On ray PR, starting from P, mark 2 + 3 + 5 = 10 equal segments using compass. Name the points as P1, P2, P3, ..., P10. Step 4: Join P10 to Q. Step 5: Through point P2, draw a line parallel to P10Q. This parallel line intersects PQ at point A. Step 6: Through point P5 (which is P2 + 3), draw another line parallel to P10Q. This parallel line intersects PQ at point B. Step 7: By basic proportionality theorem, PA:AB:BQ = 2:3:5. Step 8: Measure PA = 2 cm, AB = 3 cm, BQ = 5 cm to verify.