GSEB Class 9 Maths Solutions Chapter 11 Constructions Exercise 11.2

Question 1. Construct a triangle ABC in which BC = 7 cm, \( \angle B = 75^\circ \) and AB + AC = 13 cm.
Answer:Given: In \( \triangle ABC \), \( BC = 7 \) cm, \( \angle B = 75^\circ \) and \( AB + AC = 13 \) cm.
Required: To construct the triangle ABC.
Steps of construction:
(1) Draw the base \( BC = 7 \) cm.
(2) At point B, make an angle \( XBC = 75^\circ \).
(3) Cut a line segment \( BD = AB + AC = 13 \) cm from the ray BX.
(4) Join DC.
(5) Make \( \angle DCY = \angle BDC \).
(6) Let CY intersect BX at A. Then, ABC is the required triangle.
In simple words: First, draw a line segment BC that is 7 cm long. Next, at point B, create an angle of \( 75^\circ \) and draw a ray BX. On this ray, mark a point D such that BD is 13 cm. Connect D to C. Now, construct a line CY so that the angle it forms with DC (\( \angle DCY \)) matches \( \angle BDC \). The point where CY crosses the ray BX is your point A, completing the triangle ABC.

Exam Tip: Remember to clearly show all given values and steps. For construction problems, precision in drawing angles and lengths is vital for an accurate final shape.

Question 2. Construct a triangle ABC in which BC = 8 cm \( \angle B = 45^\circ \) and AB - AC = 3.5cm.
Answer:Given: In \( \triangle ABC \), \( BC = 8 \) cm, \( \angle B = 45^\circ \) and \( AB - AC = 3.5 \) cm.
Required: To construct the triangle ABC.
Steps of construction:
(1) Draw the base \( BC = 8 \) cm.
(2) At point B, make an angle \( XBC = 45^\circ \).
(3) Cut the line segment \( BD = AB - AC = 3.5 \) cm from the ray BX.
(4) Join DC.
(5) Draw the perpendicular bisector of DC, say PQ.
(6) Let PQ intersect BX at A. Then, ABC is the required triangle.
In simple words: First, draw a line segment BC, 8 cm long. At point B, make an angle of \( 45^\circ \) and draw a ray BX going upwards. On this ray BX, mark a point D such that the length BD is 3.5 cm. Connect D to C. Next, draw a line that cuts DC exactly in half at a \( 90^\circ \) angle (this is the perpendicular bisector). Where this bisector crosses the ray BX, that point is A. Finally, connect A to C to complete the triangle ABC.

Exam Tip: Pay close attention to whether the difference of sides is positive or negative. This determines if the segment D lies on the ray BX or on its extension in the opposite direction.

Question 3. Construct a triangle PQR in which QR = 6 cm, \( \angle Q = 60^\circ \) and PR - PQ = 2 cm.
Answer:Given: In \( \triangle PQR \), \( QR = 6 \) cm, \( \angle Q = 60^\circ \) and \( PR - PQ = 2 \) cm.
Required: To construct the \( \triangle PQR \).
Steps of construction:
(1) Draw the base \( QR = 6 \) cm.
(2) At the point Q, make an angle \( XQR = 60^\circ \).
(3) Cut the line segment \( QS = PR - PQ = 2 \) cm from the line QX extended on the opposite side of the line segment QR.
(4) Join SR.
(5) Draw the perpendicular bisector LM of SR.
(6) Let LM intersect QX at P.
(7) Join PR. Then, PQR is the required triangle.
In simple words: Start by drawing a base line QR that is 6 cm long. At point Q, draw a ray QX upwards at an angle of \( 60^\circ \). Now, from Q, draw a line downwards on the opposite side of QR and mark a point S such that QS is 2 cm long. Connect S to R. Draw a line (LM) that cuts the segment SR exactly in half at a \( 90^\circ \) angle. The spot where LM crosses the ray QX is point P. Finally, connect P to R to finish your triangle PQR.

Exam Tip: When the difference of sides is \(PR - PQ\), the segment QS should be marked on the line extended in the opposite direction of the ray forming the angle.

Question 4. Construct a triangle XYZ in which \( \angle Y = 30^\circ \), \( \angle Z = 90^\circ \) and XY + YZ + ZX = 11cm.
Answer:Given: In triangle XYZ, \( \angle Y = 30^\circ \), \( \angle Z = 90^\circ \) and \( XY + YZ + ZX = 11 \) cm.
Required: To construct the triangle XYZ.
Steps of construction:
(1) Draw a line segment \( BC = XY + YZ + ZX = 11 \) cm.
(2) Make \( \angle LBC = \angle Y = 30^\circ \) and \( \angle MCB = \angle Z = 90^\circ \).
(3) Bisect \( \angle LBC \) and \( \angle MCB \). Let these bisectors meet at a point X.
(4) Draw perpendicular bisectors DE of XB and FG of XC.
(5) Let DE intersect BC at Y and FG intersect BC at Z.
(6) Join XY and XZ. Then, XYZ is the required triangle.
In simple words: First, draw a straight line segment BC that is 11 cm long. At point B, create an angle of \( 30^\circ \) and draw a ray BL. At point C, make a \( 90^\circ \) angle and draw a ray CM. Now, split the angle at B (\( \angle LBC \)) and the angle at C (\( \angle MCB \)) exactly in half. The point where these two angle bisectors cross each other is point X. Next, draw a line that cuts the segment XB exactly in half at \( 90^\circ \) (this is DE), and another line that cuts XC exactly in half at \( 90^\circ \) (this is FG). Where DE crosses BC, that is point Y. Where FG crosses BC, that is point Z. Finally, connect X to Y and X to Z to complete the triangle XYZ.

Exam Tip: Constructing a triangle given its perimeter and two base angles involves creating a larger base, then using angle bisectors and perpendicular bisectors to find the vertices.

Question 5. Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.
Answer:Given: In right \( \triangle ABC \), base \( BC = 12 \) cm, \( \angle B = 90^\circ \) and \( AB + AC = 18 \) cm.
Required: To construct the right triangle ABC.
Steps of construction:
(1) Draw the base \( BC = 12 \) cm.
(2) At point B, make an angle \( XBC = 90^\circ \).
(3) Cut a line segment \( BD = AB + AC = 18 \) cm from the ray BX.
(4) Join DC.
(5) Draw the perpendicular bisector PQ of CD to intersect BD at a point A.
(6) Join AC. Then, ABC is the required right triangle.
In simple words: First, draw a line segment BC, which is 12 cm long. At point B, draw a ray BX straight upwards, forming a \( 90^\circ \) angle. On this ray, mark a point D such that the length BD is 18 cm. Connect D to C. Next, draw a line (PQ) that cuts the segment CD exactly in half at a \( 90^\circ \) angle. The spot where this bisector crosses the line segment BD is your point A. Finally, connect A to C to complete the right triangle ABC.

Exam Tip: For right triangles, always start by constructing the right angle accurately. The sum of hypotenuse and another side guides the length of the extended segment on the perpendicular ray.