NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.5

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.5 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.5.

• Triangles Class 9 Ex 7.1
• Triangles Class 9 Ex 7.2
• Triangles Class 9 Ex 7.3
• Triangles Class 9 Ex 7.4
• Triangles Class 9 Ex 7.5

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 7
Chapter Name	Triangles
Exercise	Ex 7.5
Number of Questions Solved	4
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.5

Question 1.
ABC is a triangle. Locate a point in the interior of Δ ABC which is equidistant from all the vertices of Δ ABC.
Solution:

Let OD and OE be the perpendicular bisectors of sides BC and CA of Δ ABC.
∴ 0 is equidistant from two ends B and C of line-segment BC as 0 lies on the perpendicular bisector of BC.
Similarly, 0 is equidistant from C and A. Thus, the point of intersection 0 of the perpendicular bisectors of sides BC, CA and AB is the required point.

Question 2.
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
Solution:
Let BE and CF be the bisectors of ∠ABC and ∠ACB respectively intersecting AC and AB at E and F respectively.

Since 0 lies on BE, the bisector of ∠ABC, hence O will be equidistant from AB and BC. Again, O lies on the bisector CF of ∠ACB. Hence, 0 will be equidistant from BC and AC. Thus, 0 will be equidistant from AB, BC, and CA.

Question 3.
In a huge park, people are concentrated at three points (see figure)
A.where there are different sides 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 parlor be set up so that maximum number of persons can approach it?
[Hint: The parlor should be equidistant from A, B and C.]
Solution:
The ice-cream parlor should be equidistant from A, B, and C for which the point of intersection of perpendicular bisectors of AB, BC, and CA should be situated.
So, 0 is the required point which is equidistant from A, B, and C.

Question 4.
Complete the hexagonal and star shaped Rangolies [see fig. (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

Solution:
We first divide the hexagon into six equilateral triangles of side 5 cm as follows:

We take one triangle from six equilateral triangles as shown above and ‘ make as many equilateral triangles of one side 1 cm as shown in the figure given below.

The number of equilateral triangles of side 1cm = 1+ 3 + 5 + 7 + 9 = 25
So, the total number of triangles in the hexagon = 6 x 25 = 150.
To find the number of triangles in Fig. (ii), we adopt the same procedure.
So, the number of triangles in Fig. (ii) = 12 x 25 = 300 Hence, Fig. (ii) has more triangles.

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