Welcome, Class 9 Scholars!

Welcome to this comprehensive study guide for Class 9 Mathematics Chapter 7 (Triangles) Exercise 7.5. This exercise is exceptionally unique and interesting as it introduces you to the fascinating world of Locus—the path traced by a point moving under specific geometric conditions. Understanding these concepts will not only help you score high in your MP Board exams but will also lay a solid foundation for higher geometry and competitive examinations like NTSE and IIT-JEE foundation courses. Let's dive in!

🔑 Important Formulas, Rules & Key Points

• Locus: The set of all points, and only those points, which satisfy a given geometric condition.
• Perpendicular Bisector Locus: The locus of a point equidistant from two given points is the perpendicular bisector of the line segment joining the two points.
• Angle Bisector Locus: The locus of a point equidistant from two intersecting lines is the bisectors of the angles between the two lines.
• Circumcentre: The point in the interior of a triangle which is equidistant from all its vertices. It is the point of intersection of the perpendicular bisectors of the sides of the triangle.
• Incentre: The point in the interior of a triangle which is equidistant from all its sides. It is the point of intersection of the internal angle bisectors of the angles of the triangle.

Q1. ABC is a triangle. Locate a point in the interior of \(\triangle ABC\) which is equidistant from all the vertices of \(\triangle ABC\).

Answer & Construction Steps:

To find a point equidistant from all the three vertices \(A\), \(B\), and \(C\) of \(\triangle ABC\), we use the concept of the perpendicular bisector.

1. Draw any triangle \(ABC\).
2. Draw the perpendicular bisector of the side \(AB\). (The points on this bisector are equidistant from \(A\) and \(B\)).
3. Draw the perpendicular bisector of the side \(BC\). (The points on this bisector are equidistant from \(B\) and \(C\)).
4. Let these two perpendicular bisectors intersect at point \(O\).

Conclusion: Since point \(O\) lies on the perpendicular bisector of \(AB\), we have \(OA = OB\). Similarly, since \(O\) lies on the perpendicular bisector of \(BC\), we have \(OB = OC\). Therefore, \(OA = OB = OC\).
Thus, Point \(O\) (called the Circumcentre) is the required point in the interior of \(\triangle ABC\) which is equidistant from all its vertices.

Q2. In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

Answer & Construction Steps:

To find a point equidistant from all three sides of a triangle, we utilize the property of angle bisectors.

1. Draw a triangle \(ABC\).
2. Draw the angle bisector of \(\angle B\). (Any point on this line is equidistant from the sides \(AB\) and \(BC\)).
3. Draw the angle bisector of \(\angle C\). (Any point on this line is equidistant from the sides \(BC\) and \(AC\)).
4. Let these two angle bisectors intersect at point \(I\).

Conclusion: Point \(I\) is equidistant from sides \(AB\) and \(BC\), and also from sides \(BC\) and \(AC\). Hence, \(I\) is equidistant from all three sides.
Therefore, Point \(I\) (called the Incentre) is the required point in the interior of the triangle.

Q3. In a huge park, people are concentrated at three points:
A: where there are slides and swings for children,
B: near which a man-made lake is situated,
C: which is near to a large parking and exit.
Where should an icecream parlour be set up so that maximum number of persons can access it?

Answer & Explanation:

To make the ice-cream parlour easily and equally accessible to the maximum number of people concentrated at points \(A\), \(B\), and \(C\), the parlour must be set up at an equal distance from all three locations.

If we join points \(A\), \(B\), and \(C\), they form a triangle \(ABC\). The point equidistant from three vertices of a triangle is its Circumcentre.

Step-by-step Solution:

1. Join points \(A\), \(B\), and \(C\) to form \(\triangle ABC\).
2. Draw perpendicular bisectors of any two sides, say \(AB\) and \(BC\).
3. Let these perpendicular bisectors meet at a point, say \(P\).

Conclusion: The icecream parlour should be set up at point \(P\) (the circumcentre of \(\triangle ABC\)), because it is equidistant from the three concentrated points \(A\), \(B\), and \(C\).

Q4. Complete the hexagonal and star-shaped Rangolis [see book figures] by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?

Answer & Mathematical Calculation:

Let's analyze the areas of both Rangoli designs mathematically to find the number of 1 cm equilateral triangles that can fit into each.

Case I: Hexagonal Rangoli

• The hexagonal shape is a regular hexagon of side \(5\text{ cm}\).
• It consists of 6 equilateral triangles of side \(5\text{ cm}\).
• Area of one equilateral triangle of side \(5\text{ cm}\) = \(\frac{\sqrt{3}}{4} \times (5)^2 = \frac{25\sqrt{3}}{4}\text{ cm}^2\).
• Total area of the hexagon = \(6 \times \frac{25\sqrt{3}}{4} = \frac{150\sqrt{3}}{4} = 37.5\sqrt{3}\text{ cm}^2\).
• Area of one small equilateral triangle of side \(1\text{ cm}\) = \(\frac{\sqrt{3}}{4} \times (1)^2 = \frac{\sqrt{3}}{4}\text{ cm}^2\).
• Number of small triangles in Hexagonal Rangoli = \(\frac{\text{Total Area of Hexagon}}{\text{Area of 1 cm Triangle}} = \frac{37.5\sqrt{3}}{\frac{\sqrt{3}}{4}} = 37.5 \times 4 = 150\).

Case II: Star-shaped Rangoli

• The star shape is formed by taking a regular hexagon of side \(5\text{ cm}\) and constructing equilateral triangles of side \(5\text{ cm}\) outwardly on each of its 6 sides.
• Therefore, the total area of the star shape = Area of the central hexagon + Area of 6 external equilateral triangles of side \(5\text{ cm}\).
• Total Area of Star = \(6 \times \frac{25\sqrt{3}}{4} + 6 \times \frac{25\sqrt{3}}{4} = 12 \times \frac{25\sqrt{3}}{4} = 75\sqrt{3}\text{ cm}^2\).
• Number of small triangles in Star Rangoli = \(\frac{\text{Total Area of Star}}{\text{Area of 1 cm Triangle}} = \frac{75\sqrt{3}}{\frac{\sqrt{3}}{4}} = 75 \times 4 = 300\).

Final Verdict: The star-shaped Rangoli contains 300 triangles, which is double the number of triangles in the hexagonal Rangoli (150). Thus, the star-shaped Rangoli has more triangles.

🚀 Aptitude Corner (For NTSE & Olympiad Aspirants)

Did you know? The position of the Circumcentre changes depending on the type of triangle:

• Acute-angled Triangle: The circumcentre lies inside the triangle.
• Right-angled Triangle: The circumcentre lies exactly at the midpoint of the hypotenuse.
• Obtuse-angled Triangle: The circumcentre lies outside the triangle.

Quick Challenge: If a right-angled triangle has sides of length 6 cm, 8 cm, and 10 cm, what is the distance of its vertices from its circumcentre?
Hint: The circumcentre is the midpoint of the hypotenuse (10 cm). Thus, the distance (circumradius) is \(10 / 2 = 5\text{ cm}\).

Congratulations! More solutions are waiting for you to discover.