Samacheer Kalvi Class 9 Maths Solutions Chapter 4 Geometry Exercise 4.5

Question 1. Construct the ∆LMN such that LM = 7.5 cm, MN = 5cm and LN = 8cm. Locate its centroid.
Answer: Follow these steps to construct the triangle and find its centroid. First, draw the triangle LMN using the given side lengths: LM = 7.5 cm, MN = 5 cm, and LN = 8 cm. Next, find the midpoints of any two sides by drawing their perpendicular bisectors. For example, draw perpendicular bisectors for LM and MN. These bisectors will help locate the midpoints P on LM and Q on MN. Then, draw the medians by connecting a vertex to the midpoint of the opposite side. Draw median LQ and median PN. These two medians will intersect at a point, which we label G. This point G is the centroid of the triangle ∆LMN, representing its geometric center.

In simple words: Draw the triangle with the given sides. Then, draw lines from each corner to the middle of the opposite side. These lines are called medians, and where they cross is the centroid.

🎯 Exam Tip: Always use a compass and ruler for accurate construction. Mark all vertices and the centroid clearly with labels and show the given measurements on the diagram for full marks.

Question 2. Draw and locate the centroid of the triangle ABC where right angle at A, AB = 4cm and AC = 3 cm.
Answer: To construct the right triangle ABC and find its centroid, start by drawing a right angle at vertex A. Mark side AB as 4 cm and side AC as 3 cm. Connect points B and C to complete the triangle. Next, find the midpoints of any two sides. For instance, draw perpendicular bisectors for side AB and side AC to find their midpoints, P and Q respectively. Then, draw the medians: connect C to P (median CP) and B to Q (median BQ). The point where these two medians intersect is the centroid, which we label G. This geometric center is crucial for understanding a triangle's balance point.

In simple words: Draw a right-angled triangle. Find the middle points of two sides. Draw lines from the corners to the middle of the opposite sides. Where these lines meet is the centroid.

🎯 Exam Tip: Remember that for a right-angled triangle, the right angle must be accurately drawn first, often using a protractor or set square, before marking the side lengths.

Question 3. Draw the ∆ABC, where AB = 6cm, ∠B = 110° and AC = 9cm and construct the centroid.
Answer: To draw triangle ABC and find its centroid, begin by drawing a line segment AB of 6 cm. At point B, use a protractor to draw an angle of 110°. Now, from point A, draw an arc with a radius of 9 cm, which is the length of AC. This arc will intersect the ray from B at point C, completing the triangle. Next, construct the perpendicular bisectors of two sides, for example, AB and BC. These bisectors will give you the midpoints, P on AB and Q on BC. Then, draw the medians by connecting a vertex to the midpoint of its opposite side. Draw median PC and median AQ. The point where these medians cross is the centroid, G. The centroid is always inside the triangle and divides each median in a 2:1 ratio.

In simple words: Draw the triangle using the given side and angle. Find the midpoints of two sides. Connect each midpoint to the opposite corner. The point where these connecting lines meet is the centroid.

🎯 Exam Tip: When given an angle, use a protractor precisely. Also, ensure the arc for the third side correctly intersects the ray to define the third vertex.

Question 4. Construct the ∆PQR such that PQ = 5cm, PR = 6cm and ∠QPR = 60° and locate its centroid.
Answer: To construct triangle PQR and its centroid, first draw line segment PQ measuring 5 cm. At point P, use a protractor to draw an angle of 60°. Along this ray, measure and mark point R such that PR is 6 cm. Connect Q and R to complete the triangle. Next, find the midpoints of two sides, for example, PQ and QR. Do this by drawing their perpendicular bisectors to find midpoints M on PQ and N on QR. Then, draw the medians by connecting a vertex to the midpoint of the opposite side. Draw median PN and median MR. The point where these medians intersect is the centroid, G. The centroid is a unique point inside every triangle, reflecting its balance.

In simple words: Draw the triangle using the given sides and angle. Find the middle points of two sides. Draw lines from the corners to the middle of the opposite sides. Where these lines meet is the centroid.

🎯 Exam Tip: An included angle (like ∠QPR) means the angle is formed between the two given sides (PQ and PR). Always measure this angle at the common vertex (P).

Question 5. Draw ∆PQR with sides PQ = 7 cm, QR = 8 cm and PR = 5 cm and construct its Orthocentre.
Answer: To construct triangle PQR and its orthocentre, first draw the triangle using the given side lengths: PQ = 7 cm, QR = 8 cm, and PR = 5 cm. Then, you need to draw at least two altitudes. An altitude is a line segment from a vertex that is perpendicular to the opposite side. For example, draw an altitude from vertex P to side QR, and another altitude from vertex Q to side PR. These altitudes might fall inside or outside the triangle depending on its angles. The point where these two altitudes intersect is the orthocentre, which we label H. The orthocentre is an important point in triangle geometry.

In simple words: Draw the triangle. From each corner, draw a straight line that goes exactly perpendicular (at 90 degrees) to the opposite side. Where these two lines cross is the orthocentre.

🎯 Exam Tip: Use a set square or protractor to ensure the altitudes are perfectly perpendicular to the opposite sides. Clearly mark the right angle symbol where the altitude meets the side.

Question 6. Draw an equilateral triangle of sides 6.5 cm and locate its Orthocentre.
Answer: To draw an equilateral triangle and its orthocentre, first draw a line segment of 6.5 cm. Using a compass, draw arcs from both ends of this segment with a radius of 6.5 cm. The intersection of these arcs will be the third vertex, forming an equilateral triangle. For an equilateral triangle, the orthocentre, centroid, circumcenter, and incenter all coincide at the same point. To find the orthocentre, draw altitudes from any two vertices, for example, from A to BC and from C to AB. The point where these altitudes intersect is the orthocentre, H. In an equilateral triangle, this point is exactly in the center.

In simple words: Draw an equilateral triangle where all sides are 6.5 cm. Draw lines from two corners that go straight down (at 90 degrees) to the opposite side. The point where these lines meet is the orthocentre.

🎯 Exam Tip: For an equilateral triangle, all altitudes are also medians and angle bisectors. Thus, the orthocentre, centroid, and circumcenter are the same point.

Question 7. Draw ∆ABC, where AB = 6 cm, ∠B = 110° and BC = 5 cm and construct its Orthocentre.
Answer: To construct triangle ABC and its orthocentre, first draw line segment AB = 6 cm. At point B, draw an angle of 110° using a protractor. Along this ray, mark point C such that BC = 5 cm. Connect points A and C to complete the triangle. To find the orthocentre, draw at least two altitudes. An altitude is a perpendicular line from a vertex to the opposite side. For an obtuse triangle (like this one, with ∠B = 110°), the orthocentre will lie outside the triangle. Extend sides AC and BC if needed. Draw an altitude from vertex B to side AC (extending AC if necessary) and another altitude from vertex C to side AB (extending AB if necessary). The point where these altitudes (or their extensions) intersect is the orthocentre, H.

In simple words: Draw the triangle. From each corner, draw a line that is straight down to the opposite side, making a 90-degree angle. If the triangle has an angle bigger than 90 degrees, you might need to extend the sides to draw these lines. The point where these lines meet is the orthocentre.

🎯 Exam Tip: For obtuse triangles, the orthocentre always lies outside the triangle. Remember to extend the sides as dashed lines to find the intersection point of the altitudes.

Question 8. Draw and locate the Orthocentre of a right triangle PQR where PQ = 4.5 cm, QR = 6 cm and PR = 7.5 cm.
Answer: To construct the right triangle PQR and its orthocentre, first draw the triangle using the given side lengths. Since \( 4.5^2 + 6^2 = 20.25 + 36 = 56.25 \) and \( 7.5^2 = 56.25 \), this is a right-angled triangle with the right angle at Q (opposite the longest side PR). Draw PQ = 4.5 cm. At Q, draw a right angle and mark QR = 6 cm. Connect P and R to complete the triangle. For a right-angled triangle, two of its altitudes are actually the sides that form the right angle (PQ and QR). The altitude from P is QR, and the altitude from R is PQ. Both intersect at the vertex where the right angle is, which is Q. Therefore, the orthocentre, H, of a right triangle is always the vertex that has the right angle.

In simple words: First, draw the right-angled triangle. For a right triangle, the two sides that form the 90-degree angle are also its altitudes. So, the point where these two sides meet (the corner with the right angle) is the orthocentre.

🎯 Exam Tip: Always check if the given side lengths satisfy the Pythagorean theorem \( a^2 + b^2 = c^2 \) to confirm if it's a right triangle. If it is, the orthocentre is simply the vertex at the right angle.