RBSE Solutions Class 9 Maths Chapter 8 Construction of Triangles Exercise 8.6

Chapter 8 Construction of Triangles Ex 8.6

Question 1. Construct \( \triangle ABC \) in which BC = 7 cm, \( \angle C = 50^\circ \) and AC + AB = 8 cm.
Answer: We need to construct a triangle \( \triangle ABC \) where the base BC is 7 cm, the angle at C is \( 50^\circ \), and the sum of the other two sides (AC + AB) is 8 cm. This is a common type of construction when the sum of two sides is given. The key idea is to extend one of the sides to create a longer segment equal to the sum, then use a perpendicular bisector.
Steps of construction:
1. Draw a line segment BC of length 7 cm. This will be the base of the triangle.
2. At point C, construct a ray CY making an angle of \( 50^\circ \) with CB. This angle helps define one of the vertices.
3. From ray CY, cut off a line segment CD of length 8 cm. This length CD represents the sum of the other two sides (AC + AB).
4. Join points B and D with a straight line.
5. Draw the perpendicular bisector of the line segment BD. Let this bisector intersect CD at point A. This step is crucial as it ensures that A is equidistant from B and D, so AB = AD.
6. Join points A and B. Then \( \triangle ABC \) is the required triangle. In this construction, since AB = AD, we have AC + AB = AC + AD = CD = 8 cm, which matches the given condition.
In simple words: First, draw the base line BC. Then, make an angle at C and extend that line out to a point D, so CD is as long as the sum of the other two sides. Connect B to D. Now, draw a line that cuts BD exactly in half and at a right angle. Where this line crosses CD, that's your point A. Join A to B, and you have your triangle.

🎯 Exam Tip: Always label your points clearly and make sure your construction lines (like the perpendicular bisector) are faint but visible, as they show your working steps.

Question 2. Construct a triangle PQR in which PQ = 6 cm, \( \angle Q = 60^\circ \) and QR + PR = 8 cm.
Answer: We need to construct a triangle \( \triangle PQR \) where the base PQ is 6 cm, the angle at Q is \( 60^\circ \), and the sum of the other two sides (QR + PR) is 8 cm. This construction is similar to the previous one, involving the sum of two sides.
Steps of construction:
1. Draw a line segment PQ of length 6 cm. This will be the base of the triangle.
2. At point Q, construct a ray QY making an angle of \( 60^\circ \) with QP. This angle helps define one of the vertices.
3. From ray QY, cut off a line segment QM of length 8 cm. This length QM represents the sum of the other two sides (QR + PR).
4. Join points P and M with a straight line.
5. Draw the perpendicular bisector of the line segment PM. Let this bisector intersect QM at point R. This step ensures R is equidistant from P and M, so RP = RM.
6. Join points P and R. Then \( \triangle PQR \) is the required triangle. Here, because RP = RM, we have QR + PR = QR + RM = QM = 8 cm, which fulfills the condition.
In simple words: Draw the base PQ. At Q, draw an angle of 60 degrees. Mark a point M on this new line so that QM equals the total length of the two unknown sides. Join P to M. Now, draw a line that cuts PM exactly in half and at a right angle. Where this line crosses QM, that is point R. Connect P and R to complete the triangle.

🎯 Exam Tip: When constructing triangles based on the sum of two sides, the auxiliary line segment (CD or QM) should always be equal to the given sum, and the perpendicular bisector ensures the equal segments needed for the sum condition.

Question 3. Construct a triangle PQR in which QR = 5 cm, \( \angle R = 40^\circ \) and PR – PQ = 1 cm.
Answer: We need to construct a triangle \( \triangle PQR \) where the base QR is 5 cm, the angle at R is \( 40^\circ \), and the difference between two sides (PR – PQ) is 1 cm. This type of construction involves using the difference of two sides.
Steps of construction:
1. Draw a line segment QR of length 5 cm. This will be the base.
2. At point R, construct a ray RT making an angle of \( 40^\circ \) with RQ. Since PR - PQ is positive, point P will be on the ray RT or its extension.
3. From ray RT, cut off a line segment RS of length 1 cm. This length RS represents the difference between PR and PQ.
4. Join points Q and S with a straight line.
5. Draw the perpendicular bisector of the line segment QS. Let this bisector intersect ray RT (or its extension) at point P. This step is key, as it ensures that PQ = PS. Since P is on RT, PR = PS + SR. Substituting PS = PQ, we get PR = PQ + SR, or PR - PQ = SR, which is the given difference.
6. Join points P and Q. Then \( \triangle PQR \) is the required triangle.
In simple words: Draw the base QR. At R, draw an angle of 40 degrees. On this new line, mark a point S so that RS is the difference (1 cm). Connect Q to S. Now, draw a line that cuts QS exactly in half and at a right angle. Where this line crosses the ray from R, that is point P. Join P to Q to finish the triangle.

🎯 Exam Tip: For constructions involving the difference of two sides, the point S is marked on the angle ray (or its extension). If the longer side is the first in the difference (e.g., PR - PQ), the point S is on the ray; if the shorter side is first (PQ - PR), the point S is on the extended ray in the opposite direction.

Question 4. Construct a triangle ABC in which the perimeter BC + CA + AB = 12 cm, \( \angle B = 50^\circ \) and \( \angle C = 70^\circ \).
Answer: We need to construct a triangle \( \triangle ABC \) where the sum of all three sides (perimeter) is 12 cm, and two base angles are given as \( \angle B = 50^\circ \) and \( \angle C = 70^\circ \). This is a construction based on the perimeter and two base angles.
Steps of construction:
1. Draw a line segment PQ of length 12 cm. This segment represents the total perimeter of the triangle.
2. At point P, construct a ray PY making an angle of \( 25^\circ \) with PQ. This angle is half of \( \angle B \) (i.e., \( \frac{1}{2} \times 50^\circ \)).
3. At point Q, construct a ray QZ making an angle of \( 35^\circ \) with QP. This angle is half of \( \angle C \) (i.e., \( \frac{1}{2} \times 70^\circ \)).
4. Let the rays PY and QZ intersect at point A. This point is a vertex of the required triangle.
5. Draw the perpendicular bisector of the line segment AP. Let this bisector intersect PQ at point B. This ensures BP = AB.
6. Draw the perpendicular bisector of the line segment AQ. Let this bisector intersect PQ at point C. This ensures CQ = AC.
7. Join AB and AC. Then \( \triangle ABC \) is the required triangle. In this construction, because BP = AB and CQ = AC, we have BC + CA + AB = BC + CQ + BP = PQ = 12 cm, which satisfies the perimeter condition.
In simple words: Draw a long line equal to the triangle's total perimeter. At each end of this line, draw half of the given base angles. Where these lines meet is the top point of your triangle (A). Now, draw lines that cut the other two segments (AP and AQ) exactly in half and at right angles. Where these cutting lines meet the base line, those are points B and C. Connect A to B and A to C to form the triangle.

🎯 Exam Tip: Remember to use half of the given base angles when constructing a triangle with a given perimeter. The perpendicular bisectors ensure the equality of segments needed to match the perimeter condition.

Question 5. Construct \( \triangle ABC \) in which BC = 8 cm, and medians AD and CF are 6 cm and 7.5 cm respectively.
Answer: We need to construct a triangle \( \triangle ABC \) given its base BC and the lengths of two medians, AD and CF. The medians intersect at the centroid G, which divides each median in a 2:1 ratio. This property is crucial for the construction.
Steps of construction:
1. Draw a line segment BC of length 8 cm. Find its midpoint and label it D. (D is the midpoint of BC, so AD is a median).
2. Calculate the lengths of the segments of the medians from the centroid. The centroid G divides median AD in a 2:1 ratio (AG:GD = 2:1). So, `GD = (1/3) * AD = (1/3) * 6 cm = 2 cm`.
3. Similarly, G divides median CF in a 2:1 ratio (CG:GF = 2:1). So, `CG = (2/3) * CF = (2/3) * 7.5 cm = 5 cm`. (We are calculating CG to construct triangle GDC).
4. From point D, draw an arc with a radius of 2 cm. This arc represents possible locations for the centroid G.
5. From point C, draw an arc with a radius of 5 cm. This arc will intersect the previous arc at point G, which is the centroid of the triangle.
6. Extend the line segment DG to a point A such that `AG = 2 * GD = 2 * 2 cm = 4 cm`. (Since AD is the median and D is the midpoint of BC, extending DG to make AG twice GD gives us the vertex A).
7. Join points A and B, and points A and C. This forms the required triangle \( \triangle ABC \).
8. To show the median CF, extend line segment CG to a point F such that CF = 7.5 cm. (F will be the midpoint of AB).
In simple words: Start by drawing the base BC and finding its middle point, D. Then, figure out where the centroid (G) should be by using parts of the given median lengths (one-third of AD from D, and two-thirds of CF from C). Draw arcs to find where these meet, which gives you point G. Now, extend the line from D through G, and measure out the remaining part of the median AD to find the top point A. Connect A to B and C to complete your triangle.

🎯 Exam Tip: The key to constructing a triangle using medians is to remember that the centroid divides each median in a 2:1 ratio. This allows you to construct a smaller triangle involving the centroid and a side, and then extend to find the missing vertices.