RBSE Solutions Class 9 Maths Chapter 8 Construction of Triangles Exercise 8.5

Question 1. Construct a \( \Delta XYZ \) where \( \angle XYZ = 60^\circ \), XY = 5 cm and XZ = 4.5 cm. How many such triangles are possible?
Answer: Here are the steps to construct the triangle:
1. First, draw a straight line, which we will call YA.
2. From point Y, draw a ray (let's call it BY) making an angle of \( 60^\circ \) with YA. This forms \( \angle BYA = 60^\circ \).
3. Using Y as the center, draw an arc with a radius of 5 cm. This arc will cut the ray BY at point X, so XY = 5 cm.
4. Now, using X as the center, draw another arc with a radius of 4.5 cm. This arc will cut the base line YA at point Z, so XZ = 4.5 cm.
5. Finally, connect points X and Z to complete the triangle.
The triangle XYZ is now constructed. In this specific construction case, only one unique triangle is possible.
In simple words: First, draw a base line and an angle of 60 degrees. Mark a point X on the angle line 5 cm from the corner. Then, from point X, draw a curved line 4.5 cm long to hit the base line at Z. Connect X and Z to finish the triangle. For these measurements, only one such triangle can be made.

🎯 Exam Tip: When constructing a triangle with Side-Angle-Side (SAS) or specific Side-Side-Angle (SSA) conditions, always check if the given measurements lead to a unique triangle or an ambiguous case.

Question 2. Construct a \( \triangle PQR \) where \( \angle PQR = 45^\circ \), PQ = 6 cm and PR = 5 cm.
Answer: Here are the steps to construct the triangle:
1. Begin by drawing a line segment PQ, which will be the base of the triangle. Make sure its length is 6 cm.
2. At point Q, construct an angle of \( 45^\circ \). This means you will draw a ray (let's call it QX) such that \( \angle PQX = 45^\circ \).
3. Now, using P as the center, open your compass to a radius of 5 cm. Draw an arc that cuts the ray QX. Mark the point where the arc intersects QX as R.
4. Connect points P and R to form the side PR.
You have now successfully constructed \( \triangle PQR \).
In simple words: Draw a 6 cm line called PQ. At point Q, draw a line going up at a 45-degree angle. From point P, draw a curved line 5 cm long that crosses the 45-degree line. Where they cross is point R. Join P and R.
🎯 Exam Tip: Always use a sharp pencil and precise measurements with a ruler and protractor to ensure the constructed triangle is accurate.

Question 3. Construct \( \triangle ABC \) where a = 5.4 cm, b = 6.8 cm and \( \angle A = 45^\circ \). Can two triangles be drawn in this case?
Answer: Yes, in this specific situation, two different triangles can be drawn. This is known as the ambiguous case in triangle construction.
Here are the steps for construction:
1. Draw a straight line and label it AX.
2. At point A, draw a ray (let's call it AY) forming an angle of \( 45^\circ \) with AX. So, \( \angle XAY = 45^\circ \).
3. With A as the center, draw an arc of radius 6.8 cm. This arc will intersect the ray AY at point C. So, AC = 6.8 cm.
4. Now, using C as the center, draw another arc with a radius of 5.4 cm (since 'a' refers to the side opposite angle A, which is BC). This arc will intersect the base line AX at two distinct points. Label these points as B and B'.
5. Finally, connect C to both B and B'.
This results in two required triangles: \( \triangle ABC \) and \( \triangle AB'C \). Both triangles fit the given conditions, making this an ambiguous case for construction.
In simple words: Start by drawing a straight line and an angle of 45 degrees at point A. Mark point C on the angled line, 6.8 cm from A. Then, from point C, draw a curved line 5.4 cm long. This curved line will hit the first straight line in two different spots, B and B'. You can then draw two different triangles (ABC and AB'C) that both match the problem.

🎯 Exam Tip: The ambiguous case (SSA - Side-Side-Angle) occurs when the given side opposite the angle is shorter than the adjacent side, but longer than the altitude from the vertex to the opposite side, allowing for two possible triangles.