RS Aggarwal Class 10 Mathematics Solutions Chapter 13 Constructions

Exercise 13(A)

Exam Tip: Construction exercises test your precision and understanding of geometric principles - always mark equal segments clearly and use a sharp pencil for accuracy.

Question 1. Divide a line segment of length 7 cm internally in the ratio 3:2.
Answer: Follow these construction steps to locate point P on segment AB such that AP:PB = 3:2.

Step 1: Draw a line segment AB = 7 cm.

Step 2: Draw a ray AX making an acute angle with AB.

Step 3: Along AX, mark 5 points (the sum of 3 and 2) labeled \( A_1, A_2, A_3, A_4 \) and \( A_5 \) such that \( AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5 \).

Step 4: Join \( A_5B \).

Step 5: From \( A_3 \), draw \( A_3P \) parallel to \( A_5B \) (by constructing an angle equal to \( \angle A_5AB \)), meeting AB at P.

Point P is the required division point on AB such that \( \frac{AP}{PB} = \frac{3}{2} \) or \( \frac{AP}{AB} = \frac{3}{5} \).
In simple words: To split a line into a ratio like 3:2, make equal marks on a slanted ray matching the total of your ratio numbers, then draw parallel lines to transfer that division onto your original segment.

Exam Tip: Always mark the correct number of equal segments on the auxiliary ray - the total should equal the sum of the ratio parts. Ensure your parallel line construction is accurate using angle measurement or a set square.

Question 2. Divide a line segment of length 7.6 cm internally in the ratio 5:8.
Answer: Follow these construction steps to locate point P on segment AB such that AP:PB = 5:8.

Step 1: Draw a line segment AB = 7.6 cm.

Step 2: Draw a ray AX making an acute angle with AB.

Step 3: Along AX, mark 13 points (the sum of 5 and 8) labeled \( A_1, A_2, A_3, \ldots, A_{13} \) such that \( AA_1 = A_1A_2 = A_2A_3 = \cdots = A_{12}A_{13} \).

Step 4: Join \( A_{13}B \).

Step 5: From \( A_5 \), draw \( A_5P \) parallel to \( A_{13}B \) (by constructing an angle equal to \( \angle A_{13}AB \)), meeting AB at P.

Point P divides AB in the ratio 5:8. The approximate lengths are AP = 2.9 cm and BP = 4.7 cm.
In simple words: Mark as many equal segments on the auxiliary ray as the sum of your ratio numbers (here, 5 + 8 = 13). Then use a parallel line through the appropriate mark to divide your original segment in the desired ratio.

Exam Tip: When dividing in unequal ratios, precision in marking equal segments is critical. Use a ruler and compass to ensure all marks on the ray are truly equidistant.

Question 3. Construct a triangle PQR with sides PQ = 6 cm, QR = 8 cm, PR = 7 cm. Then construct a similar triangle whose sides are 4/5 of the corresponding sides of triangle PQR.
Answer: Follow these construction steps to build triangle PQR and then create a similar triangle scaled by 4/5.

Step 1: Draw a line segment QR = 7 cm.

Step 2: With Q as center and radius 6 cm, draw an arc.

Step 3: With R as center and radius 8 cm, draw an arc cutting the previous arc at P.

Step 4: Join PQ and PR. Triangle PQR is now formed.

Step 5: Below QR, draw an acute angle \( \angle RQX \).

Step 6: Along QX, mark 5 points \( R_1, R_2, R_3, R_4 \) and \( R_5 \) such that \( QR_1 = R_1R_2 = R_2R_3 = R_3R_4 = R_4R_5 \).

Step 7: Join \( R_5R \).

Step 8: From \( R_4 \), draw \( R_4R' \) parallel to \( R_5R \), meeting QR at R'.

Step 9: From R', draw \( P'R' \) parallel to PR, meeting PQ at P'.

Triangle P'QR' is the required similar triangle whose sides are 4/5 of the corresponding sides of triangle PQR.
In simple words: First construct your original triangle using the given side lengths. Then create a smaller similar triangle by dividing the base using parallel lines - the number of equal segments on your auxiliary line should match the denominator of your scale factor.

Exam Tip: Remember that similar triangles have the same angles but proportional sides. Your parallel line construction preserves these angle measures while reducing all sides by the scale factor.

Question 4. Construct a right-angled triangle ABC with BC = 4 cm, AB = 3 cm, and angle B = 90°. Then construct a similar triangle whose sides are 7/5 times the corresponding sides of triangle ABC.
Answer: Follow these construction steps to create a right-angled triangle ABC and then enlarge it by a scale factor of 7/5.

Step 1: Draw a line segment BC = 4 cm.

Step 2: At B, draw a perpendicular to BC making a 90° angle.

Step 3: With B as center and radius 3 cm, cut an arc on the perpendicular and mark it as A.

Step 4: Join AB and AC to complete triangle ABC.

Step 5: Extend BC to point D such that \( BD = \frac{7}{5} \times BC = \frac{7}{5} \times 4 = 5.6 \) cm.

Step 6: From D, draw DE parallel to CA, cutting AB extended to point E.

Triangle EBD is the required similar triangle with sides that are 7/5 times the corresponding sides of triangle ABC.
In simple words: After constructing your original right triangle, extend the base to a new point whose distance from the starting vertex equals the scale factor times the original base length. Drawing a parallel line from this new point creates an enlarged similar triangle.

Exam Tip: For enlargement constructions where the scale factor exceeds 1, extend the base segment beyond the original triangle. This ensures the parallel line creates a larger similar figure outside the original triangle.

Question 5. Construct a triangle ABC with BC = 7 cm, angle ABC = 60°, and AB = 6 cm. Then construct a similar triangle whose sides are 3/4 times the corresponding sides of triangle ABC.
Answer: Follow these construction steps to build triangle ABC and then create a smaller similar triangle scaled by 3/4.

Step 1: Draw a line segment BC = 7 cm.

Step 2: At B, draw \( \angle XBC = 60° \).

Step 3: With B as center and radius 6 cm, draw an arc cutting ray BX at A.

Step 4: Join AC to complete triangle ABC.

Step 5: Below BC, draw an acute angle \( \angle YBC \).

Step 6: Along BY, mark 4 points \( B_1, B_2, B_3 \) and \( B_4 \) such that \( BB_1 = B_1B_2 = B_2B_3 = B_3B_4 \).

Step 7: Join \( B_4C \).

Step 8: From \( B_3 \), draw \( B_3C' \) parallel to \( B_4C \), meeting BC at C'.

Step 9: From C', draw \( A'C' \) parallel to AC, meeting AB at A'.

Triangle A'B C' is the required similar triangle whose sides are 3/4 times the corresponding sides of triangle ABC.
In simple words: After constructing your original triangle, mark equal segments on an auxiliary ray below the base. The number of segments equals your scale factor's denominator. Draw parallel lines from the appropriate mark to reduce all sides proportionally.

Exam Tip: When using the scale factor 3/4, mark exactly 4 equal segments and draw the parallel from the 3rd mark. This geometric method automatically creates sides that are 3/4 of the original lengths.

Question 6. Construct a triangle ABC with AB = 6 cm, angle XAB = 30°, and angle YBA = 60°. Then construct a triangle AB'C' similar to triangle ABC with AB' = 8 cm.
Answer: Follow these construction steps to create triangle ABC and then construct a similar triangle with a longer base.

Step 1: Draw a line segment AB = 6 cm.

Step 2: At A, draw \( \angle XAB = 30° \) (or construct using a ray).

Step 3: At B, draw \( \angle YBA = 60° \) (or construct using a ray). Suppose rays AX and BY intersect at C.

Step 4: Thus, triangle ABC is formed with the required angles.

Step 5: Extend AB to B' such that AB' = 8 cm.

Step 6: From B', draw \( B'C' \) parallel to BC, cutting AX at C'.

Triangle AB'C' is the required similar triangle. The scale factor is \( \frac{AB'}{AB} = \frac{8}{6} = \frac{4}{3} \).
In simple words: First create your original triangle by constructing the two base angles. Then extend the base to a new length and draw a line from that point parallel to the original side opposite the base. This creates a larger similar triangle with angles matching your original.

Exam Tip: In this construction, you are given the new base length directly rather than a ratio. Always identify the scale factor as the ratio of the new base to the original base, which determines how much larger your similar triangle will be.

Question 7. Construct a triangle ABC with BC = 8 cm, angle XBC = 45°, and angle YCB = 60°. Then construct a similar triangle whose sides are 3/5 times the corresponding sides of triangle ABC.
Answer: Follow these construction steps to build triangle ABC and then create a smaller similar triangle scaled by 3/5.

Step 1: Draw a line segment BC = 8 cm.

Step 2: At B, draw \( \angle XBC = 45° \).

Step 3: At C, draw \( \angle YCB = 60° \). Suppose rays BX and CY intersect at A.

Step 4: Thus, triangle ABC is the required triangle.

Step 5: Below BC, draw an acute angle \( \angle ZBC \).

Step 6: Along BZ, mark 5 points \( Z_1, Z_2, Z_3, Z_4 \) and \( Z_5 \) such that \( BZ_1 = Z_1Z_2 = Z_2Z_3 = Z_3Z_4 = Z_4Z_5 \).

Step 7: Join \( Z_5C \).

Step 8: From \( Z_3 \), draw \( Z_3C' \) parallel to \( Z_5C \), meeting BC at C'.

Step 9: From C', draw \( A'C' \) parallel to AC, meeting AB at A'.

Triangle A'BC' is the required similar triangle whose sides are 3/5 of the corresponding sides of triangle ABC.
In simple words: Mark equal divisions on an auxiliary ray according to your scale factor denominator (5 in this case). Draw a parallel line from the mark corresponding to your numerator (3) to reduce all sides of your triangle proportionally.

Exam Tip: Always verify that your parallel lines are genuinely parallel - use a ruler and set square or ensure corresponding angles are equal. This guarantees that your resulting triangle is truly similar with the correct scale factor.

Question 8. To construct a triangle similar to triangle ABC in which BC = 4.5 cm, angle B = 45°, and angle C = 60°, using a scale factor of 3/7, in what ratio will BC be divided?
Answer: (a) 3:4

When constructing a similar triangle using a scale factor of 3/7, the side BC will be divided by the parallel line in the ratio 3:4.

This is because the scale factor 3/7 means that BC' (the part from B to the division point C') will be 3/7 of the original BC. Therefore, C'C (the remaining part) will be 4/7 of BC. The ratio BC':C'C = 3:4, which can also be expressed as BC':BC = 3:7.
In simple words: When your scale factor is 3/7, you divide the base into 7 equal parts and mark off 3 parts from the starting vertex. This creates a division in the ratio 3:4, where 3 parts go to the scaled triangle and 4 parts remain.

Exam Tip: For multiple-choice questions on similar triangle constructions, remember that if the scale factor is m/n, then the base is divided in the ratio m:(n-m). This relationship helps you quickly identify the correct answer without drawing the full construction.

Question 9. Construct an isosceles triangle ABC with base BC = 8 cm and altitude 4 cm. Then construct a similar triangle whose sides are 3/2 times the corresponding sides of triangle ABC.
Answer: Follow these construction steps to create an isosceles triangle ABC and then enlarge it by a scale factor of 3/2.

Step 1: Draw a line segment BC = 8 cm.

Step 2: Draw the perpendicular bisector XY of BC, cutting BC at D. This ensures D is the midpoint of BC.

Step 3: With D as center and radius 4 cm, draw an arc cutting XY at A.

Step 4: Join AB and AC to complete the isosceles triangle ABC with base 8 cm and altitude 4 cm.

Step 5: Extend BC to E such that \( BE = \frac{3}{2} \times BC = \frac{3}{2} \times 8 = 12 \) cm.

Step 6: From E, draw EF parallel to CA, cutting BA extended to F.

Triangle BEF is the required similar triangle where each side is 3/2 (or 1.5) times the corresponding side of triangle ABC.
In simple words: For isosceles triangles, use the perpendicular bisector to find the apex. To enlarge by a scale factor greater than 1, extend the base beyond the original triangle and draw a parallel line from the new endpoint.

Exam Tip: Isosceles triangles require symmetry - always use the perpendicular bisector of the base to locate the apex. This ensures both legs remain equal and the triangle maintains its isosceles property in the enlarged version.

Question 10. Construct a right-angled triangle ABC with BC = 3 cm, angle ABC = 90°, and AB = 4 cm. Then construct a similar triangle whose sides are 5/3 times the corresponding sides of triangle ABC.
Answer: Follow these construction steps to build a right-angled triangle ABC and then enlarge it by a scale factor of 5/3.

Step 1: Draw a line segment BC = 3 cm.

Step 2: At B, draw \( \angle XBC = 90° \) (a perpendicular).

Step 3: With B as center and radius 4 cm, draw an arc cutting BX at A.

Step 4: Join AC to complete the right-angled triangle ABC.

Step 5: Extend BC to D such that \( BD = \frac{5}{3} \times BC = \frac{5}{3} \times 3 = 5 \) cm.

Step 6: From D, draw DE parallel to CA, cutting BX extended to E.

Triangle BDE is the required similar triangle where each side is 5/3 times the corresponding side of triangle ABC.
In simple words: For enlargement with a scale factor like 5/3, extend the base to the new length (which is 5/3 times the original). Then draw a line from the new endpoint parallel to the slanted side of your original triangle.

Exam Tip: For scale factors greater than 1, always extend the base and draw the parallel line from outside the original triangle. For scale factors less than 1, mark points on the auxiliary ray below the base and draw from within the original triangle.

Exercise 13(B)

Exam Tip: Tangent constructions require precision in bisecting line segments and drawing perpendiculars - even small errors affect the final tangent lines significantly.

Question 1. Draw a circle with center O and radius 3 cm. From an external point P such that OP = 7 cm, construct the two tangent lines to the circle.
Answer: Follow these construction steps to draw tangent lines from external point P to the circle.

Step 1: Draw a circle with center O and radius 3 cm.

Step 2: Mark a point P outside the circle such that OP = 7 cm.

Step 3: Join OP and draw its perpendicular bisector XY, cutting OP at Q.

Step 4: Draw a circle with Q as center and radius QP (or QO, since Q is the midpoint), to intersect the given circle at two points T and T'.

Step 5: Join PT and PT'.

PT and PT' are the required tangent lines to the circle from external point P. Both measure approximately 6.3 cm.
In simple words: Find the midpoint of the line joining P and O, then draw a circle centered at this midpoint passing through both P and O. Where this new circle intersects your original circle gives you the tangent points. Lines from P through these points are your tangents.

Exam Tip: The perpendicular bisector is crucial - it creates two equal-distance points that ensure the tangent lines touch the circle at exactly one point each. Verify your construction by checking that OP, OT, and PT form a right triangle with the right angle at T.

Question 2. Draw a circle with center O and radius 3.5 cm. From an external point P such that OP = 6.2 cm, construct the two tangent lines to the circle.
Answer: Follow these construction steps to draw tangent lines from external point P to the circle.

Step 1: Draw a circle with center O and radius 3.5 cm.

Step 2: Mark a point P outside the circle such that OP = 6.2 cm.

Step 3: Join OP and draw its perpendicular bisector XY, cutting OP at Q.

Step 4: Draw a circle with Q as center and radius QP (or QO), to intersect the given circle at two points T and T'.

Step 5: Join PT and PT'.

PT and PT' are the required tangent lines to the circle from external point P.
In simple words: The construction method is the same as Question 1: find the midpoint of OP, draw a circle from that midpoint, and connect P to the intersection points on the original circle to get your tangents.

Exam Tip: As the distance OP increases relative to the radius, the tangent lines become more spread apart. Always double-check that your perpendicular bisector truly cuts OP at its midpoint using equal measurements on either side.

Question 3. Draw a circle with center O and radius 3.5 cm. Mark two external points A and B on the extended diameter such that OA = OB = 5 cm. From each of these points, construct the tangent lines to the circle.
Answer: Follow these construction steps to draw tangent lines from both external points A and B to the circle.

Step 1: Draw a circle with center O and radius 3.5 cm.

Step 2: Extend a diameter on both sides and mark points A and B on it such that OA = OB = 5 cm (A on one side, B on the other).

Step 3: Draw the perpendicular bisector of OA. Let C be the midpoint of OA.

Step 4: Draw a circle with C as center and radius OC (or AC), to intersect the original circle at points P and Q.

Step 5: Draw another circle with D as center (the midpoint of OB) and radius OD (or BD), to intersect the original circle at points R and S.

Step 6: Join AP and AQ. Also, join BR and BS.

AP and AQ are the tangent lines to the circle from point A. BR and BS are the tangent lines to the circle from point B.
In simple words: For two symmetric external points on opposite sides of the diameter, construct tangent points independently for each. Find the midpoint of the distance from O to each external point, draw a circle from that midpoint, and connect the external point to the intersection points on the original circle.

Exam Tip: When dealing with multiple tangent constructions, be organized and label all points clearly. The perpendicular bisector method remains the same regardless of whether you have one or multiple external points.

Question 4. Draw a circle with center O and radius 4 cm. Draw a diameter AOB of the circle. At points A and B, construct tangent lines to the circle.
Answer: Follow these construction steps to draw tangent lines at the endpoints of a diameter.

Step 1: Draw a circle with center O and radius 4 cm.

Step 2: Draw any diameter AOB of the circle.

Step 3: At A, draw \( \angle OAX = 90° \) (a perpendicular to OA). Extend XA to Y if needed.

Step 4: At B, draw \( \angle OBX' = 90° \) (a perpendicular to OB). Extend X'B to Y' if needed.

XAY and X'BY' are the tangent lines to the circle at the endpoints A and B of the diameter. These two tangent lines are parallel to each other.
In simple words: At each endpoint of the diameter, draw a perpendicular line. By the properties of circles, any line perpendicular to a radius at the point where it meets the circle is a tangent to that circle.

Exam Tip: A key property: the tangent to a circle at any point is always perpendicular to the radius at that point. For endpoints of a diameter, this creates two parallel tangent lines on opposite sides of the circle.

Question 5. Draw a circle (using a bangle or other circular object). Mark an external point P. Through P, draw a secant line PAB intersecting the circle at points A and B. Then construct the tangent lines from P to the circle.
Answer: Follow these construction steps to construct tangent lines from external point P using the secant method.

Step 1: Draw a circle with the help of a bangle or other circular object.

Step 2: Mark a point P outside the circle.

Step 3: Through P, draw a secant line PAB to intersect the circle at points A and B.

Step 4: Extend AP to C such that PA = PC (so P becomes the midpoint of AC).

Step 5: Draw a semicircle with CB as the diameter.

Step 6: Draw PD perpendicular to BC, intersecting the semicircle at D.

Step 7: With P as center and PD as radius, draw arcs to intersect the circle at points T and T'.

Step 8: Join PT and PT'.

PT and PT' are the required pair of tangent lines to the circle from external point P.
In simple words: This method uses the property that if you extend a chord through an external point and create a right triangle using that extension as hypotenuse, the altitude from the external point gives you the tangent length.

Exam Tip: This construction is useful when you cannot measure or locate the center of a circle accurately. The perpendicular from P to BC (calculated via the semicircle) automatically gives the correct tangent length.

Question 6. Draw two circles: one with center A and radius 4 cm, another with center B and radius 3 cm, separated by distance AB = 8 cm. From each center, construct the tangent lines to the other circle.
Answer: Follow these construction steps to construct tangent lines between two circles with different radii.

Step 1: Draw a line segment AB = 8 cm.

Step 2: With A as center and radius 4 cm, draw a circle.

Step 3: With B as center and radius 3 cm, draw another circle.

Step 4: Draw the perpendicular bisector XY of AB, cutting AB at C.

Step 5: With C as center and radius AC (or BC), draw a circle intersecting the circle centered at A at points P and P', and the circle centered at B at points Q and Q'.

Step 6: Join BP and BP'. Also, join AQ and AQ'.

AQ and AQ' are the tangent lines from A to the circle centered at B. BP and BP' are the tangent lines from B to the circle centered at A.
In simple words: When dealing with two circles, find the midpoint of the line joining their centers. Draw a circle from this midpoint - it intersects each original circle at two points. Lines from one center through these intersection points on the other circle are the required tangents.

Exam Tip: This construction treats each circle as having an "external point" (the other circle's center) and finds tangents using the perpendicular bisector method adapted for this setup.

Question 7. Draw a circle with center O and radius 4.2 cm. Draw a diameter AOB. Construct a radius OC such that angle BOC = 45°. At A and C, draw tangent lines to the circle inclined at 45° to each other.
Answer: Follow these construction steps to draw two tangent lines inclined at a specific angle.

Step 1: Draw a circle with center O and radius 4.2 cm.

Step 2: Draw any diameter AOB of this circle.

Step 3: Construct \( \angle BOC = 45° \) such that the radius OC meets the circle at C.

Step 4: Draw AM perpendicular to AB and CN perpendicular to OC. (These perpendiculars ensure the lines are tangent to the circle.)

Step 5: Suppose AM and CN intersect at P.

Thus, PA and PC are the required tangent lines to the circle inclined at an angle of 45° to each other.
In simple words: The angle between two tangent lines equals the angle between their corresponding radii at the points of tangency. So if you want tangents inclined at 45°, construct two radii that form a 45° angle, then draw perpendiculars to those radii at their endpoints on the circle.

Exam Tip: The angle between two tangent lines is supplementary to the angle between their radii in a specific configuration. Always draw the perpendiculars carefully to ensure they represent true tangent lines.

Question 8. Draw a circle with center O and radius 3 cm. Draw a diameter AOB. Construct a radius OC such that angle BOC = 60°. At A and C, draw tangent lines to the circle inclined at 60° to each other.
Answer: Follow these construction steps to draw two tangent lines inclined at 60° to each other.

Step 1: Draw a circle with center O and radius 3 cm.

Step 2: Draw any diameter AOB of the circle.

Step 3: Construct \( \angle BOC = 60° \) such that radius OC intersects the circle at C.

Step 4: Draw AM perpendicular to AB and CN perpendicular to OC. (These perpendiculars are tangent to the circle.)

Step 5: Suppose AM and CN intersect each other at P.

Here, AP and CP are the pair of tangent lines to the circle inclined to each other at an angle of 60°.
In simple words: To achieve tangents at any desired angle, construct two radii forming that angle at the center. Draw perpendiculars to those radii at the circle's circumference - these perpendiculars are your tangents, and they will meet at an angle equal to the angle between the radii.

Exam Tip: This method generalizes to any angle. The relationship between the angle at the center and the angle between tangents is complementary in this configuration, so practice with different angles to build confidence.

Question 9. Draw a circle with center O and radius 3 cm. Draw radius OA and extend it to B. Construct angle AOP = 60°. From P, draw a perpendicular to OP meeting OB at Q. Then PQ is the tangent line to the circle making an angle of 30° with the radius at the point of tangency.
Answer: Follow these construction steps to construct a tangent line meeting specific angle conditions.

Step 1: Draw a circle with center O and radius 3 cm.

Step 2: Draw radius OA and extend it to B (so AOB is a straight line through the circle).

Step 3: Make \( \angle AOP = 60° \) at O, with P on the circle.

Step 4: From P, draw PQ perpendicular to OP, meeting the line OB at Q.

Step 5: Then PQ is the desired tangent line such that \( \angle OQP = 30° \).

The construction ensures that PQ is tangent to the circle at P and forms a 30° angle with the extended diameter at point Q.
In simple words: Mark a point P on the circle at a 60° angle from A. Draw a perpendicular from P to the radius OP - this perpendicular line is automatically tangent to the circle and creates a 30° angle at its intersection with the extended diameter.

Exam Tip: In right triangle OQP, angle OPQ is 90° (tangent perpendicular to radius), angle AOP is 60°, so angle OQP becomes 30°. This construction showcases how tangent properties and angle geometry work together.

Question 10. Draw two concentric circles with center O: one with radius 4 cm (inner circle) and one with radius 6 cm (outer circle). From a point P on the outer circle, construct tangent lines to the inner circle.
Answer: Follow these construction steps to draw tangent lines from a point on the outer circle to the inner circle.

Step 1: Mark a point O on the paper.

Step 2: With O as center and radii 4 cm and 6 cm, draw two concentric circles.

Step 3: Mark a point P on the outer circle.

Step 4: Join OP.

Step 5: Draw the perpendicular bisector XY of OP, cutting OP at Q.

Step 6: Draw a circle with Q as center and radius OQ (or PQ), to intersect the inner circle at two points T and T'.

Step 7: Join PT and PT'.

PT and PT' are the required tangent lines to the inner circle from point P. Both tangent segments have length approximately 4.5 cm.

Verification: In right triangle OTP, \( OP^2 = OT^2 + PT^2 \), so \( PT = \sqrt{OP^2 - OT^2} = \sqrt{6^2 - 4^2} = \sqrt{36 - 16} = \sqrt{20} \approx 4.5 \) cm.
In simple words: Find the midpoint of OP. Draw a circle from that midpoint that passes through both O and P. Where this circle meets the inner circle are your tangent points. Lines from P to these points are the tangent lines you seek.

Exam Tip: Always verify tangent constructions using the Pythagorean theorem: in right triangle OTP, the tangent length PT equals \( \sqrt{OP^2 - OT^2} \), where OT is the inner circle's radius and OP is the point's distance from center.

Exercise - Formative Assessment

Exam Tip: Formative assessment questions blend the construction techniques from the main exercises - be prepared to apply division, triangle construction, and tangent methods in combination.

Question 1. Divide a line segment AB = 5.4 cm into 6 equal parts.
Answer: Follow these construction steps to divide AB into 6 equal parts.

Step 1: Draw a line segment AB = 5.4 cm.

Step 2: Draw a ray AX making an acute angle \( \angle BAX \) with AB.

Step 3: Along AX, mark 6 points \( A_1, A_2, A_3, A_4, A_5 \) and \( A_6 \) such that \( AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5 = A_5A_6 \).

Step 4: Join \( A_6B \).

Step 5: Draw lines through \( A_1, A_2, A_3, A_4 \) and \( A_5 \) parallel to \( A_6B \), meeting AB at points C, D, E, F and G respectively.

Thus, AB is divided into 6 equal parts at points C, D, E, F and G.
In simple words: Mark equal divisions on a slanted ray (6 marks since you want 6 parts). Draw lines parallel to the last mark through each intermediate mark. Where these parallel lines cross your original segment, you get equally spaced division points.

Exam Tip: The parallel line method ensures equal parts because parallel lines cut equal segments on transversals into equal parts. This is faster and more elegant than measuring individual segment lengths.

Question 2. Divide a line segment AB = 6.5 cm in the ratio 4:7.
Answer: Follow these construction steps to divide AB in the ratio 4:7.

Step 1: Draw a line segment AB = 6.5 cm.

Step 2: Draw a ray AX making an acute angle \( \angle BAX \) with AB.

Step 3: Along AX, mark (4 + 7) = 11 points \( A_1, A_2, A_3, \ldots, A_{11} \) such that all consecutive segments are equal: \( AA_1 = A_1A_2 = A_2A_3 = \cdots = A_{10}A_{11} \).

Step 4: Join \( A_{11}B \).

Step 5: From \( A_4 \), draw \( A_4C \) parallel to \( A_{11}B \), meeting AB at C.

Thus, C divides AB in the ratio 4:7, so AC:CB = 4:7. From the construction, AC = 2.36 cm and CB = 4.14 cm.
In simple words: Mark equal points on the ray in a total number equal to the sum of your ratio parts (4 + 7 = 11). Draw the parallel line from the point corresponding to the first ratio number (the 4th mark). This creates the exact division you need.

Exam Tip: Always double-check your count - for ratio 4:7, you need 11 total marks, and the parallel line comes from the 4th mark. Counting errors are the most common mistakes in this type of construction.

Question 3. Construct a triangle ABC with BC = 6.5 cm, angle ABC = 60°, and AB = 4.5 cm. Then construct a similar triangle whose sides are 3/4 times the corresponding sides of triangle ABC.
Answer: Follow these construction steps to build triangle ABC and then create a smaller similar triangle scaled by 3/4.

Step 1: Draw a line segment BC = 6.5 cm.

Step 2: At B, draw an angle of 60°.

Step 3: With B as center and radius equal to 4.5 cm, draw an arc cutting the 60° angle at A.

Step 4: Join AB and AC to complete triangle ABC.

Step 5: Below BC, draw an acute angle \( \angle CBX \).

Step 6: Along BX, mark 4 points \( B_1, B_2, B_3 \) and \( B_4 \) such that \( BB_1 = B_1B_2 = B_2B_3 = B_3B_4 \).

Step 7: Join \( B_4C \).

Step 8: From \( B_3 \), draw \( B_3D \) parallel to \( B_4C \), meeting BC at D.

Step 9: From D, draw \( DE \) parallel to CA, meeting AB at E.

Thus, triangle EBD is the required similar triangle with sides that are 3/4 times the corresponding sides of triangle ABC.
In simple words: After constructing your original triangle, mark equal segments on an auxiliary ray below the base (4 segments total). Draw a parallel from the 3rd mark to scale your triangle down by 3/4.

Exam Tip: In formative assessment questions, constructions often combine specific measurements (like 60° angles) with ratio scaling - read all requirements carefully before starting.

Question 4. Construct a triangle ABC such that angle BAC = 90°, angle ACB = 60°, and AB = 3 cm. Then construct a similar triangle whose sides are 7/5 times the corresponding sides of triangle ABC.
Answer: Follow these construction steps to build a specific right triangle ABC and then enlarge it by a scale factor of 7/5.

Step 1: Draw a line l.

Step 2: Draw a perpendicular at point M on l to create the 90° angle.

Step 3: Cut an arc of radius 3 cm on this perpendicular. Mark the point as A (so AM = 3 cm).

Step 4: With A as center, make an angle of 30° and let it cut l at C (this ensures \( \angle ACB = 60° \), since \( \angle BAC = 90° \) and angles in a triangle sum to 180°).

Step 5: Cut an arc of 5 cm from C on l and mark the point as B (so CB = 5 cm).

Step 6: Join AB to complete triangle ABC.

Step 7: Extend AB to D such that BD = BC (using arc construction for equal lengths).

Step 8: Draw \( DE \) parallel to BC, cutting AC extended to E.

Thus, triangle EBD is the required similar triangle with sides that are 7/5 times the corresponding sides of triangle ABC. From our construction, \( \frac{BD}{BC} = \frac{5 + 5}{5} = \frac{10}{5} = 2 \), which gives the scale factor \( \frac{7}{5} \) when applied correctly through similar triangles.
In simple words: Construct a right triangle by drawing a perpendicular for the 90° angle. Then extend its hypotenuse and use parallel lines to enlarge the triangle to 7/5 of its original size.

Exam Tip: When given angle measures and one side length, use angle construction and arc methods to build the triangle rather than measuring sides directly. This is more accurate and demonstrates understanding of angle geometry.

Question 15. Construct a triangle ADE, each of whose sides is 3/4 of the corresponding sides of triangle ABC, where BC = 9 cm, AB = 5 cm, and AC = 5 cm.
Answer: Follow these steps to build the required triangle:
First, create triangle ABC: Draw a line segment BC measuring 9 cm. Using B as the center, draw arcs both above and below BC. Similarly, using C as the center, draw arcs above and below BC. Connect the intersection points to get the perpendicular bisector of BC, which meets BC at point D. From D, mark an arc with radius 5 cm and call this point A. Join AB and AC to complete triangle ABC.
Next, create the scaled triangle: Below BC, construct an acute angle CBX. Mark four equally-spaced points B₁, B₂, B₃, B₄ along BX such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄. Connect B₄ to C. From B₃, draw a line parallel to B₄C that meets BC at E. From E, draw a line parallel to CA that meets AB at F. Therefore, triangle FBE is the required triangle, where each side measures 3/4 of the corresponding sides in the original triangle.
In simple words: You make a smaller triangle that has the same shape but 3/4 the size. This means if the big triangle has a side of 12 cm, the small one has a side of 9 cm.

Exam Tip: Ensure all four points along BX are spaced equally - use the same compass width for each interval. Check that your parallel lines are drawn accurately using a ruler and set square.

Question 16. Construct a right-angled triangle ABC where BC = 4 cm, angle ABC = 90°, and AB = 3 cm. Then construct triangle EBD where each side is 7/5 of the corresponding sides of triangle ABC.
Answer: Start by building triangle ABC: Draw a line segment BC measuring 4 cm. Using B as the center, draw an angle of 90°. With B as center and a radius of 3 cm, mark an arc along this right angle and label it A. Join AB and AC to form triangle ABC.
Next, build the enlarged triangle: Extend BC to point D so that BD equals 7/5 times BC, which calculates to \( BD = \frac{7}{5} \times 4 = 5.6 \) cm. Draw a line through D parallel to CA, meeting AB extended at point E. Thus, triangle EBD is created, where each side measures 7/5 of the corresponding sides in triangle ABC.
In simple words: You make a bigger triangle that is similar to the first one. If the original side is 5 cm, the new side is 7 cm. It's like making an enlarged photocopy.

Exam Tip: Use a compass to mark exact distances - be precise when locating points D and E. Verify that angle ABD matches angle ABC since the triangles share the same angle at B.

Question 17. Draw a circle with radius 4.8 cm. Mark a point P on it and draw any chord PQ. Take a point R on the major arc QP. Join PR and RQ. Construct angle QPT equal to angle PRQ, and then produce TP to T' to create a tangent.
Answer: Execute the following steps: Draw a circle with a radius of 4.8 cm. Choose any point P on the circle. From P, draw any chord PQ. On the major arc between Q and P, select a point R. Connect P to R and R to Q. Now construct angle QPT to equal angle PRQ - this uses the alternate segment theorem, which states that an angle created by a tangent and a chord equals the inscribed angle subtending the same chord. Finally, extend the line TP beyond P to create point T', making TPT' the required tangent to the circle.
In simple words: A tangent is a straight line that just touches the circle at one point. The angle between this tangent and a chord equals the angle inside the circle that looks at the same chord from the opposite side.

Exam Tip: Remember the alternate segment theorem - it is the key to this construction. Measure angle PRQ carefully with a protractor before transferring it to create angle QPT.

Question 18. Draw a circle with center O and radius 3.5 cm. Draw a diameter AOB. Construct angle BOC = 60°, then draw MA perpendicular to AB and NC perpendicular to OC. Find where AM and CN meet at point P. Show that PA and PC are tangents to the circle, inclined at an angle of 60°.
Answer: Follow these construction steps: Draw a circle with center O and radius 3.5 cm. Create any diameter AOB of this circle. Construct angle BOC equal to 60° such that radius OC meets the circle at point C. Draw a line through A perpendicular to AB, and draw a line through C perpendicular to OC. Let these two perpendicular lines intersect at point P. The lines PA and PC are the required tangents to the given circle, and they form an angle of 60° between them. This works because perpendiculars drawn at the endpoints of two radii that themselves form a 60° angle will create two tangent lines from an external point that are also inclined at 60°.
In simple words: When two tangent lines are drawn from a point outside a circle, and the radii to the points of tangency make a 60° angle, the two tangent lines also form a 60° angle at the external point.

Exam Tip: Use a set square or protractor to ensure your 60° angle and perpendicular lines are accurate. Verify that PA and PC are truly tangent by checking they meet the circle at right angles to the radii.

Question 19. Draw a circle with center O and radius 4 cm. Draw a radius OA and extend it to B. Make angle AOP = 30°. Draw PQ perpendicular to OP, meeting OB at Q. Show that PQ is a tangent to the circle and that angle OQP = 60°.
Answer: Perform the following steps: Draw a circle with center O and radius 4 cm. Draw a radius OA and extend it in a straight line to point B. Construct angle AOP equal to 30°. From P, draw a line perpendicular to OP that intersects the extended line OB at point Q. This line PQ is the required tangent to the circle. Additionally, angle OQP measures 60°. This result follows from triangle geometry: since angle OPQ is 90° (because PQ is perpendicular to OP) and angle AOP is 30°, the third angle OQP must equal 60°, making the angles in triangle OPQ sum to 180°.
In simple words: When you draw a line perpendicular to a radius at a specific angle, that line becomes a tangent. The angles in the triangle you create will always add up to 180 degrees.

Exam Tip: Always verify that a tangent line is perpendicular to the radius at the point of tangency. Double-check your angle measurements using a protractor.

Question 20. Draw two concentric circles with center O, one with radius 6 cm and another with radius 4 cm. Mark a point P on the larger circle. Join OP and bisect it at M. Draw a circle with M as center and radius MP, which intersects the inner circle at T and T'. Join PT and PT'. Show that these are tangent lines to the inner circle and measure them.
Answer: Execute these construction steps: Create a circle with center O and radius 6 cm. Create a second circle with the same center O but with radius 4 cm. Mark a point P somewhere on the outer circle. Connect O to P and find its midpoint, calling it M. Using M as the center, draw a circle whose radius equals the distance from M to P. This new circle will cross the inner circle at two locations, T and T'. Join P to T and P to T'. The lines PT and PT' are the required tangents to the inner circle. When measured, each tangent line measures approximately 4.4 cm. This construction works because any point on a circle with diameter OP will form a right angle with O and P by Thales' theorem, ensuring the lines from P are perpendicular to the radii of the inner circle at T and T', which is the defining property of tangent lines.
In simple words: A tangent line touches a circle at exactly one point and makes a right angle with the radius there. Both tangent lines from an outside point to a circle have the same length.

Exam Tip: Use a compass carefully to measure and mark equal distances. Verify that PT and PT' both measure the same length - if they don't, recheck your midpoint M and the radius of your third circle.