RBSE Solutions Class 9 Maths Chapter 8 Construction of Triangles Important Questions

Question 1. Construct a ∆ABC in which AB = AC, BC = 3.8 cm and AD the altitude from A to BC = 4.3 cm.
Answer:
Given:
ABC is an isosceles triangle with AB = AC.
Base BC = 3.8 cm.
Altitude AD from A to BC = 4.3 cm.
AD is perpendicular to BC.

Steps of construction:
1. First, draw a straight line segment BC that is 3.8 cm long.
2. Next, draw the perpendicular bisector of BC. This line will pass through the midpoint of BC. Let's call this bisector AX.
3. The altitude AD is 4.3 cm. So, from point D (the midpoint of BC where AX crosses BC), measure 4.3 cm upwards along AX and mark point A.
4. Finally, connect point A to point B and point A to point C. This completes the triangle ABC.
This construction effectively uses the properties of an isosceles triangle where the altitude to the base also acts as the median and perpendicular bisector.
In simple words: Draw the base line. Find its middle and draw a straight line up from there. Measure the height on this upward line and mark the top point of the triangle. Connect the top point to the ends of the base.

🎯 Exam Tip: When constructing isosceles triangles with an altitude, remember that the altitude from the vertex angle to the base also bisects the base and the vertex angle. This simplifies the construction process significantly.

Question 2. Construct a ΔΑΒC in which BC = 3.8 cm, (AB + AC) = 5 cm and ∠ABC = 60°.
Answer:
Given:
In triangle ABC, BC = 3.8 cm.
The sum of sides AB + AC = 5 cm.
Angle ABC = 60°.

Steps of construction:
1. First, draw a straight line segment BC that is 3.8 cm long.
2. At point B, use a protractor or compass to construct an angle of 60 degrees. Extend this line from B to a point X.
3. Now, measure 5 cm (which is AB + AC) along the line BX from point B. Mark this point as D.
4. Draw a straight line connecting point D to point C.
5. To find point A, we need to make sure AB + AC is 5 cm. To do this, construct an angle at C, say ∠BCY, such that ∠BCY = ∠BDC. Extend CY to intersect BD at point A. Alternatively, construct the perpendicular bisector of CD, which will intersect BD at A.
Therefore, ∆ABC is the triangle we needed to construct. This method is useful when the sum of two sides and one angle are given.
In simple words: Draw the base and make the given angle. Extend this angle line and mark a point that is the sum of the other two sides away from the angle's vertex. Connect this new point to the third vertex of the base. Then, find the third vertex of the actual triangle by using angle properties or a perpendicular bisector.

🎯 Exam Tip: When constructing a triangle with the sum of two sides given, remember to extend one side and mark the total length. The key is finding the third vertex by creating an isosceles triangle with the remaining segments.

Question 3. Construct a ∆ABC in which BC = 3.6 cm, (AC - AB) = 1.6 cm and ∠ACB = 30°.
Answer:
Given:
In triangle ABC, BC = 3.6 cm.
The difference of sides AC - AB = 1.6 cm.
Angle ACB = 30°.

Steps of construction:
1. Begin by drawing a straight line segment BC that is 3.6 cm long.
2. At point C, construct an angle of 30 degrees using a protractor or compass. Extend this line downwards (or upwards, depending on the orientation of the difference construction) from C, and label it CX.
3. Now, measure 1.6 cm (which is AC - AB) along the line CX from point C. Mark this point as D.
4. Draw a straight line connecting point D to point B.
5. To locate point A, you need to create a line such that any point on it is equidistant from D and B. This is the perpendicular bisector of BD. The intersection of this perpendicular bisector with CX (extended if necessary) will give you point A.
This specific construction method is used when the difference of two sides and one angle are provided, helping to convert the difference into a usable length in the construction.
In simple words: Draw the base line. Make the given angle at one end and draw a line. From that same end, measure the 'difference' length along the angle line and mark a point. Connect this new point to the other end of the base. Then, draw a special line (perpendicular bisector) for this new connection to find the third point of the triangle.

🎯 Exam Tip: When the difference of two sides (AC - AB) is given, be careful with the direction of construction for the difference length. If AC > AB, you extend the angle line, but if AB > AC, you extend it on the other side of the vertex. Ensure you correctly apply the perpendicular bisector to find the final vertex.

Question 4. Construct a triangle ABC in which AB = 3.3 cm, AC = 2.8 cm and altitude AD = 2.3 cm.
Answer:
Given:
In triangle ABC, AB = 3.3 cm, AC = 2.8 cm, and altitude AD = 2.3 cm.

Steps of construction:
1. First, draw a straight line PQ. This line will act as the base on which the triangle rests. Mark a point D anywhere on this line.
2. At point D, draw a line DX perpendicular to PQ. This line represents the path for the altitude.
3. Measure 2.3 cm (the altitude AD) along DX from D, and mark point A.
4. Now, from point A, use a compass to draw an arc with a radius of 3.3 cm (for AB). This arc will intersect line PQ at point B.
5. From point A, again use a compass to draw another arc with a radius of 2.8 cm (for AC). This arc will intersect line PQ at point C.
6. Finally, connect point A to point B and point A to point C. This completes the triangle ABC. This method allows you to construct a triangle when the lengths of two sides and the altitude to the third side are known.
In simple words: Draw a straight line and pick a spot on it. Draw another straight line straight up from that spot. Measure the height (altitude) on the upward line and mark point A. From point A, draw two arcs that cut the first line. One arc uses the length of side AB, and the other uses the length of side AC. Connect A to where the arcs cut the line to finish the triangle.

🎯 Exam Tip: When constructing with an altitude and two side lengths, accurately drawing the perpendicular for the altitude first is crucial. Ensure your compass settings for the two arcs are precise to get the correct positions for points B and C.