Exercise 7.5

NCERT solutions for Class 9 Maths Triangles 

NCERT Solutions for Class 9 Mathematics Triangles

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

Ans. Let ABC be a triangle.

Draw perpendicular bisectors PQ and RS of sides AB and BC respectively of triangle ABC. Let PQ bisects AB at M and RS bisects BC at point N.

Let PQ and RS intersect at point O.

Join OA, OB and OC.

Now in AOM and BOM,

AM = MB [By construction]

AMO = BMO = [By construction]

OM = OM [Common]

AOM BOM [By SAS congruency]

OA = OB [By C.P.C.T.] …..(i)

Similarly, BON CON

OB = OC [By C.P.C.T.] …..(ii)

From eq. (i) and (ii),

OA = OB = OC

Hence O, the point of intersection of perpendicular bisectors of any two sides of ABC equidistant from its vertices.

NCERT Solutions for Class 9 Maths Exercise 7.5

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

Ans. Let ABC be a triangle.

Draw bisectors of B and C.

Let these angle bisectors intersect each other at point I.

Draw IK BC

Also draw IJ AB and IL AC.

Join AI.

In BIK and BIJ,

IKB = IJB = [By construction]

IBK = IBJ

[ BI is the bisector of B (By construction)]

BI = BI [Common]

BIK BIJ [ASA criteria of congruency]

IK = IJ [By C.P.C.T.] ……….(i)

Similarly,CIK CIL

IK = IL [By C.P.C.T.] ……….(ii)

From eq (i) and (ii),

IK = IJ = IL

Hence, I is the point of intersection of angle bisectors of any two angles of ABC equidistant from its sides.

NCERT Solutions for Class 9 Maths Exercise 7.5

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 maximum number of persons can approach it?

Ans. The parlour should be equidistant from A, B and C.

For this let we draw perpendicular bisector say of line joining points B and C also draw perpendicular bisector say of line joining points A and C.

Let and intersect each other at point O.

Now point O is equidistant from points A, B and C.

Join OA, OB and OC.

Proof: In BOP and COP,

OP = OP [Common]

OPB = OPC =

BP = PC [P is the mid-point of BC]

BOP COP [By SAS congruency]

OB = OC [By C.P.C.T.] …..(i)

Similarly, AOQ COQ

OA = OC [By C.P.C.T.] …..(ii)

From eq. (i) and (ii),

OA = OB = OC

Therefore, ice cream parlour should be set up at point O, the point of intersection of perpendicular bisectors of any two sides out of three formed by joining these points.

NCERT Solutions for Class 9 Maths Exercise 7.5

4. Complete the hexagonal rangoli and the star rangolies (See figure) but 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?

Ans. In hexagonal rangoli, Number of equilateral triangles each of side 5 cm are 6.

Area of equilateral triangle = = = sq. cm

Area of hexagonal rangoli = 6 x Area of an equilateral triangle

= = sq. cm ……….(i)

Now area of equilateral triangle of side 1 cm = = = = sq. cm …..(ii)

Number of equilateral triangles each of side 1 cm in hexagonal rangoli

= = = 150 …..(iii)

Now in Star rangoli,

Number of equilateral triangles each of side 5 cm = 12

Therefore, total area of star rangoli = 12 Area of an equilateral triangle of side 5 cm

=

=

= sq. cm ……….(iv)

Number of equilateral triangles each of side 1 cm in star rangoli

=

=

= 300 ……….(v)

From eq. (iii) and (v), we observe that star rangoli has more equilateral triangles each of side 1 cm.