RS Aggarwal Class 9 Mathematics Solutions Chapter 12 Geometrical Constructions

Question 1. Construct the perpendicular bisector of a line segment AB = 5 cm.
Answer: Follow these steps to build the perpendicular bisector:
(i) Draw a line segment AB = 5 cm
(ii) Using A as the centre and a radius greater than half of AB, draw two arcs - one above AB and one below AB.
(iii) Using B as the centre and the same radius, draw two more arcs that intersect the previously drawn arcs at points C and D.
(iv) Join the points C and D, cutting AB at point P.

CD is now the perpendicular bisector of AB at point P. This line passes through the midpoint of AB and meets it at a right angle.
In simple words: Draw two intersecting arcs from each endpoint using the same radius. Connect where the arcs meet. This line is perpendicular to AB and divides it equally.

Exam Tip: Always ensure the radius is more than half the segment length - if it's too small, the arcs won't intersect, making it impossible to find the perpendicular bisector.

Question 2. Construct an angle of 45 degrees.
Answer: Follow these construction steps:
(i) Draw a line segment OA.
(ii) At point A, construct \( \angle AOE = 90° \) using a ruler and compass.
(iii) Using B as the centre and a radius greater than half of BD, draw an arc.
(iv) Using D as the centre and the same radius, draw another arc intersecting the previous arc at F.
(v) Join OF. This gives \( \angle AOF = 45° \).
(vi) Now using B as the centre and a radius greater than half of BC, draw an arc.
(vii) Using C as the centre and the same radius, draw another arc intersecting the previous arc at X.
(viii) Join OX. OX is the angle bisector of \( \angle AOF \).

By bisecting the right angle, you obtain 45 degrees.
In simple words: First make a 90-degree angle. Then split that 90-degree angle in half using the angle bisector method. You get 45 degrees.

Exam Tip: Ensure the arc radius is consistently more than half the distance between the two points whose arcs you are drawing - this guarantees the arcs will meet.

Question 3. Construct an angle of 60 degrees and then bisect it to create a 30-degree angle.
Answer: Follow these steps to accomplish this construction:
(i) Draw a line segment OA.
(ii) Using O as the centre and any suitable radius, draw an arc that intersects OA at point B.
(iii) Using B as the centre and the same radius, cut the previously drawn arc at point C.
(iv) Using C as the centre and the same radius, cut the arc again at point D.
(v) Using C as the centre and a radius greater than half of CD, draw an arc.
(vi) Using D as the centre and the same radius, draw another arc that cuts the previous arc at E.
(vii) Join E. Now \( \angle AOE = 90° \).
(viii) Using B as the centre and a radius greater than half of CB, draw an arc.
(ix) Using C as the centre and the same radius, draw an arc that cuts the previous one at F.
(x) Join OF.
(xi) F is the bisector of the right angle \( \angle AOE \), giving a 45-degree angle. Further bisection yields 30 degrees.
In simple words: Mark equal spaces along an arc from a baseline to create 60 degrees. Then bisect that angle by finding where equal arcs from two points on the angle's sides meet.

Exam Tip: When marking equal arc distances, keep the compass opening exactly the same for each step - any variation will throw off the angle measurement.

Question 4. Build an equilateral triangle with side length 5 cm.
Answer: Use the following construction method:
(i) Draw a line segment BC = 5 cm.
(ii) Using B as the centre and radius equal to BC, draw an arc.
(iii) Using C as the centre and the same radius, draw another arc that cuts the first arc at point A.
(iv) Join AB and AC.

Triangle ABC is now the required equilateral triangle with all sides measuring 5 cm and all angles equal to 60 degrees.
In simple words: Draw a base line of 5 cm. From each endpoint, swing an arc with radius 5 cm. Where the arcs meet is your third corner. Connect to finish the triangle.

Exam Tip: The compass radius must equal the base segment length exactly - use the same opening to measure the base and to draw both arcs for consistency.

Question 5. Build an equilateral triangle using angle construction on a baseline with specific angle requirements.
Answer: Construct the triangle through these steps:
(i) Draw a line XY.
(ii) Mark any point P on it.
(iii) From P, draw PQ perpendicular to XY.
(iv) From P, set off PA = 5.4 cm along the baseline (or as specified).
(v) Construct \( \angle PAB = 30° \) and \( \angle PAC = 30° \), meeting XY at points B and C respectively.

Triangle ABC is the required equilateral triangle. The angle construction ensures all three angles equal 60 degrees and all sides are equal.
In simple words: Draw angles of 30 degrees on each side of a line from a point. These angles create a triangle where all three sides are the same length.

Exam Tip: Use a protractor or compass bisector method to ensure the 30-degree angles are precise - any deviation will distort the equilateral triangle property.

Question 6. Construct a triangle with sides 5 cm, 3.8 cm, and 2.6 cm (this is an isosceles triangle).
Answer: Follow this construction procedure:
(i) Draw a line segment BC = 5 cm (the base).
(ii) Using B as the centre and radius 3.8 cm, draw an arc.
(iii) Using C as the centre and radius 2.6 cm, draw another arc that intersects the first arc at point A.
(iv) Join AB and AC.

Triangle ABC is the required triangle. Since two sides are equal (both approximately 3.8 and 2.6 in the figure shown), it displays the isosceles property, though the exact side lengths are determined by the arc intersections.
In simple words: Draw the longest side as your base. From one end, swing an arc with one given length. From the other end, swing an arc with the second length. Where they meet is your third corner.

Exam Tip: Always verify that the two arc radii sum to more than the base length - if their sum is less than or equal to the base, the arcs will not intersect and no triangle can form.

Question 7. Construct a triangle with a base of 4.7 cm and base angles of 60 degrees and 30 degrees.
Answer: Use these construction steps:
(i) Draw a line segment BC = 4.7 cm.
(ii) At B, draw \( \angle XBC = 60° \).
(iii) At C, draw \( \angle YCB = 30° \).
(iv) Let the rays BX and CY intersect at point A.

Triangle ABC is the required triangle. The base angles sum to 90 degrees, so the angle at A equals 90 degrees, making this a right-angled triangle. The third angle (at A) is automatically 90 degrees since angles in a triangle sum to 180 degrees.
In simple words: Draw your base line. At one end, mark a 60-degree angle. At the other end, mark a 30-degree angle. Where the two angled lines meet is your third corner.

Exam Tip: Use a protractor or angle bisector method to construct the specified angles - ensure they are on the same side of the base to form a valid triangle.

Question 8. Construct an isosceles triangle with base QR = 5 cm and equal sides measuring 4.5 cm each.
Answer: Follow these construction steps:
(i) Draw a line segment QR = 5 cm (the base).
(ii) Using Q as the centre and radius 4.5 cm, draw an arc.
(iii) Using R as the centre and the same radius 4.5 cm, draw another arc that cuts the previous arc at point P.
(iv) Join PQ and PR.

Triangle PQR is the required isosceles triangle where PQ = PR = 4.5 cm and QR = 5 cm. The two equal sides meeting at P create the characteristic isosceles triangle shape.
In simple words: Draw a base. From each endpoint of the base, swing equal-length arcs toward each other. Where they cross is your top corner. Connect these three points.

Exam Tip: Confirm that the arc radius (4.5 cm) is longer than half the base (2.5 cm) - this ensures the arcs will intersect above the base to form the triangle apex.

Question 9. Construct an isosceles triangle with base BC = 4.8 cm, where the two equal sides meet at a right angle.
Answer: Apply these construction steps:
(i) Draw a line segment BC = 4.8 cm.
(ii) At B, construct \( \angle CBX = 80° \) below the line segment BC.
(iii) At B, construct \( \angle ABX = 90° \).
(iv) Draw the perpendicular bisector of BC, intersecting BY at point O.
(v) Using O as the centre and radius OB, draw a circle that intersects the constructed angle line at point A.
(vi) Join AB and AC.

Triangle ABC is the required isosceles triangle where AB = AC and \( \angle BAC = 90° \). The circle ensures both equal sides are equidistant from the base midpoint.
In simple words: Draw your base. Create a right angle at one endpoint. Use a circle centred on the base's midpoint to locate the third corner so both slant sides are equal length.

Exam Tip: The perpendicular bisector and circle arc together guarantee that the two non-base sides remain equal - this is the key to maintaining the isosceles property while achieving the right angle.

Question 10. Construct a triangle using the angle bisector method and circle construction.
Answer: For this advanced construction, follow these detailed steps. The method involves drawing a base segment, constructing specific angles, finding bisectors, and using circle arcs to locate the third vertex. The exact configuration depends on the given measurements and angle specifications. Key principles include:

- Use the perpendicular bisector to find equidistant points from two known vertices
- Apply angle bisector techniques to create precise angular relationships
- Employ circle arcs centred at strategic points to intersect construction lines
- Verify that the final triangle satisfies all given constraints (side lengths and angles)

The construction ensures geometric accuracy through the combination of perpendicular and angle bisector methods along with circular arc intersection points.
In simple words: Use arcs and angle bisectors together to guide where your third corner should go. The combination creates a triangle matching your requirements.

Exam Tip: Always test your final triangle by measuring the constructed sides and angles against the original requirements - this verification step catches any accumulated compass or straightedge errors.

Question 11. Construct a triangle with base angles of 30 degrees and 60 degrees, with PA = 4.8 cm.
Answer: Follow this construction sequence:
(i) Draw any line XY.
(ii) Select any point P on XY and construct PQ perpendicular to XY.
(iii) Along PQ, mark off PA = 4.8 cm.
(iv) Through A, draw a line LM parallel to XY.
(v) Construct \( \angle LAB = 30° \) and \( \angle MAC = 60° \), with these angles meeting XY at points B and C respectively.

Triangle ABC is the required triangle. The angle and length specifications ensure the geometric configuration matches the construction requirements exactly. The perpendicular height of 4.8 cm combined with the specified base angles determines the triangle's dimensions completely.
In simple words: Draw a vertical line 4.8 cm high from a baseline. From the top, create two angles (30 and 60 degrees) that meet the baseline. This forms your triangle.

Exam Tip: The parallel line through A ensures the angles you construct relate correctly to the baseline - without it, the angle measurements would not produce the intended triangle.

Question 12. Construct a triangle on a semicircle using a specific base and arc method.
Answer: Apply these construction steps:
(i) Draw a line segment BC = 5.3 cm.
(ii) Find the midpoint O of BC.
(iii) Using O as the centre and radius OB, draw a semicircle on BC (this is called the semicircle on the hypotenuse).
(iv) Using B as the centre and radius 4.5 cm, draw an arc that cuts the semicircle at point A.
(v) Join AB and AC.

Triangle ABC is the required triangle. By Thales' theorem, any angle inscribed in a semicircle is a right angle, so \( \angle BAC = 90° \). The arc from B with radius 4.5 cm ensures that one of the legs (AB) has the specified length. This construction elegantly guarantees a right angle without explicitly constructing one.
In simple words: Draw a base and find its middle. Make a semicircle with that middle as centre. From one endpoint, swing an arc to meet the semicircle. Connect all three points - you get a right-angled triangle automatically.

Exam Tip: The semicircle method is powerful because it automatically creates a 90-degree angle - any point on a semicircle (except the endpoints) always forms a right angle with the diameter endpoints.

Question 13. Construct a triangle using base-angle construction and perpendicular bisector method for finding the third vertex.
Answer: Follow these construction procedures:
(i) Draw BC = 4.5 cm as the base.
(ii) Construct \( \angle CBX = 60° \) at point B.
(iii) Along BX, mark off BP = 8 cm.
(iv) Join C to P (forming segment CP).
(v) Draw the perpendicular bisector of CP; this line intersects BP at point A.
(vi) Join AC to complete the triangle.

Triangle ABC is the required triangle. The perpendicular bisector of CP ensures that A is equidistant from both C and P (meaning AC = AP). This constraint, combined with the specified base and angle, uniquely determines the triangle's shape. The construction is particularly useful when you need to create a triangle where two sides have a specific relationship to given segments.
In simple words: Draw a base and an angled line from one endpoint. Mark a point far along this angled line. Find the perpendicular bisector of the segment connecting this point to the base's other endpoint. The bisector meets the angled line at your third corner.

Exam Tip: The perpendicular bisector method is elegant because it converts an angle/distance requirement into a pure geometric condition (equidistance) that is easy to construct accurately.

Question 14. Construct a triangle with base BC = 5.2 cm and base angle at B equal to 30 degrees, with BP = 3.5 cm measured along the angled side.
Answer: Execute these construction steps:
(i) Draw BC = 5.2 cm as the base.
(ii) At B, construct \( \angle CBX = 30° \).
(iii) Along BX, mark off BP = 3.5 cm.
(iv) Join P to C (creating segment PC).
(v) Draw the right (perpendicular) bisector of PC; allow it to meet BP extended at point A.
(vi) Join AC to complete the triangle.

Triangle ABC is the required triangle where the perpendicular bisector of PC intersects the extension of BP. This ensures that A lies on the perpendicular bisector, meaning AC = AP. The construction satisfies all the specified constraints: the 30-degree angle at B, the base length of 5.2 cm, and the relationship defined by the 3.5 cm measurement and the perpendicular bisector condition. Note that A may lie beyond P on the line BP, depending on the geometric configuration.
In simple words: Draw your base and a 30-degree angled line from one end. Mark 3.5 cm along this angled line. Construct the perpendicular bisector of the segment from this mark to the other base endpoint. Extend the angled line if needed until it meets the bisector - that's your third corner.

Exam Tip: When the perpendicular bisector intersects the extended line (not the original segment), ensure your straightedge extends far enough beyond the marked point - this is a common mistake that leads to incomplete construction.