GSEB Class 9 Maths Solutions Chapter 7 Triangles Exercise 7.5

Question 1. ABC is a triangle. Locate a point in the interior of ∆ABC which is equidistant from all the vertices of ∆ABC.
Answer: To find a point inside \( \triangle ABC \) that is equally far from all its vertices, we need to draw perpendicular bisectors for any two sides of the triangle. The spot where these perpendicular bisectors meet will be equidistant from all the vertices. This point, often named O, becomes the center of a circle that passes through all the triangle's vertices.
In simple words: To find a point that is the same distance from all corners of a triangle, draw lines that cut two sides in half at a right angle. Where these lines cross is the spot.

Exam Tip: Remember that the point equidistant from the vertices of a triangle is its circumcenter, and it is found by the intersection of perpendicular bisectors of the sides.

Question 2. In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
Answer: To find a point inside a triangle that is equally far from all its sides, you must draw the angle bisectors of any two angles of the triangle. The spot where these angle bisectors meet will be equidistant from all three sides. So, this point, commonly known as O, is the incenter of the triangle.
In simple words: To find a point inside a triangle that is the same distance from all its edges, draw lines that cut two of the triangle's angles in half. Where these lines cross is the point you're looking for.

Exam Tip: Remember that the point equidistant from the sides of a triangle is its incenter, and it is found by the intersection of angle bisectors of the angles.

Question 3. In a huge park, people are concentrated at three points (See figure) A: where there are different 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 ice-cream parlour be set up so that the maximum number of persons can approach it? (Hint: the parlour should be equidistant from A, B and C)
Answer: If an ice-cream shop is positioned at the circumcenter of the triangle created by points A, B, and C, then the greatest number of people will be able to reach it. This spot will be equally far from each vertex: A, B, and C.
In simple words: The ice-cream shop should be placed at the center of the circle that goes through all three points (A, B, C). This way, it's the same distance from everyone at A, B, and C.

Exam Tip: To maximize access from three distinct points, the ideal location is the circumcenter, which is equidistant from all three points.

Question 4. Complete the hexagonal and star shaped Rangolies (See figure (i) and (ii)) 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:
(i) For the hexagonal-shaped Rangoli:
Area of each equilateral triangle forming the hexagon, with side \( s = 5 \) cm, is \( \frac { \sqrt{3} }{ 4 } s^2 \).
So, Area of \( \triangle OAB = \frac { \sqrt{3} }{ 4 } \times 5^2 = \frac { 25\sqrt{3} }{ 4 } \).
A regular hexagon consists of 6 such equilateral triangles. Therefore, the total area of the hexagonal Rangoli is \( 6 \times \frac { 25\sqrt{3} }{ 4 } = \frac { 150\sqrt{3} }{ 4 } = \frac { 75\sqrt{3} }{ 2 } \).
The area of a smaller equilateral triangle with a side of 1 cm is \( \frac { \sqrt{3} }{ 4 } \times 1^2 = \frac { \sqrt{3} }{ 4 } \text{ cm}^2 \).
To find the number of 1 cm equilateral triangles in the hexagonal Rangoli, we divide the total area of the Rangoli by the area of one small triangle:
Number of triangles \( = \frac { \frac { 75\sqrt{3} }{ 2 } }{ \frac { \sqrt{3} }{ 4 } } = \frac { 75\sqrt{3} }{ 2 } \times \frac { 4 }{ \sqrt{3} } = 75 \times 2 = 150 \).
So, the hexagonal Rangoli contains 150 equilateral triangles of side 1 cm.

(ii) For the star-shaped Rangoli:
The star-shaped Rangoli consists of 12 equilateral triangles, each with a side of 5 cm. This means the shape is made up of 12 larger triangles.
The total area of the star-shaped Rangoli is \( 12 \times (\frac { \sqrt{3} }{ 4 } \times 5^2) = 12 \times \frac { 25\sqrt{3} }{ 4 } = 3 \times 25\sqrt{3} = 75\sqrt{3} \).
To find the number of 1 cm equilateral triangles in the star-shaped Rangoli, we divide its total area by the area of one small triangle (which is \( \frac { \sqrt{3} }{ 4 } \text{ cm}^2 \)):
Number of triangles \( = \frac { 75\sqrt{3} }{ \frac { \sqrt{3} }{ 4 } } = 75\sqrt{3} \times \frac { 4 }{ \sqrt{3} } = 75 \times 4 = 300 \).
So, the star-shaped Rangoli contains 300 equilateral triangles of side 1 cm.

Comparing the two, the star-shaped Rangoli has 300 triangles, while the hexagonal Rangoli has 150 triangles. Therefore, the star-shaped Rangoli has more triangles.
In simple words: First, calculate the area of each big Rangoli shape. Then, find the area of one tiny 1 cm triangle. Divide the big area by the small area to see how many small triangles fit. The star shape holds more little triangles than the hexagon.

Exam Tip: When comparing areas of shapes made of smaller units, calculate the area of the larger composite shape and the area of the smallest unit, then divide the larger area by the smaller unit area to find the count. Ensure all units are consistent.