Haryana State Board HBSE 9th Class Maths Solutions Chapter 7 Triangles Ex 7.5 Textbook Exercise Questions and Answers.

## Haryana Board 9th Class Maths Solutions Chapter 7 Triangles Exercise 7.5

Question 1.

ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.

Solution:

(i) Draw a triangle ΔABC.

(ii) Draw the perpendicular bisector PQ and RS on AB and BC respectively, intersecting each other at O.

(iii) Join OA, OB and OC. Here OA = OB = OC.

Hence, point O is the point which is equidistant from the vertices of ΔABC.

Question 2.

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

Solution:

(i) Draw a triangle ΔABC.

(ii) Draw the bisectors BO and CO of ∠B and ∠C respectively, intersecting each other at O.

(iii) Draw OL ⊥ BC, OM ⊥ CA and ON ⊥ AB, meeting BC at L. CA at M and AB at N respectively.

(iv) Here OL = OM = ON

Hence, point O is the point which is equidistant from the all sides of the triangle.

Question 3.

In a huge park, people are concentrated at three points (see figure 7.81).

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 maximum number of persons can approach it?

[Hint: The parlour should be equidistant from A, B and C.]

Solution :

Join A to B and B to C. Parlour should be equidistant from A, B and C.

Let PO and OQ be the perpendicular bisector of AB and BC respectively. They intersect at O. Since OP ⊥ bisector of AB. Therefore will equidistant from A and B.

Again, OQ ⊥ bisector of BC, therefore o will equidistant from B and C. Thus, O is the required point which is equidistant from A, B and C.

Question 4.

Complete the hexagonal and star-shaped Rangolies (see figure 7.83 (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?

Solution:

We divide the given hexagon into six equilateral triangles whose each side is 5 cm. Now take one such triangle and divide it into many equilateral of side 1 cm, as shown in the figure :

On counting number of triangles of side 1 cm each are :

Number of Δs in Ist row = 9

Number of Δs in Ilnd row = 7

Number of Δs in Illrd row = 5

Number of Δs in IVth row = 3

Number of Δs in Vth row = 1

Total number of triangles = 25

∴ Total number of triangles in the hexagon = 6 × 25 = 150.

(ii) Total number of triangles in figure 7.172

= 12 × 25

= 300.

Clearly, star-shaped Rangoli has more triangles.