SESSION:2013-2014 ROLL NO:08

ACTIVITY: To verify basic proportionality theorem using parallel lines board and triangle cut-outs.

NAME: ARPIT GUPTA

CLASS/SEC: X-B

MADE BY: ARPIT GUPTA

SUBMITTED TO: MR. S.C. VERMA

SIGNATURE:

S.NO PARAMETERSMARKS

OBTAINED

1.Identification and

statement of the project

2.Origin/design

Of the Project

3.Procedure/

ProcessDeveloped

4.Write up/

Presentation

5.Presentation

TOTAL

PARENT’ SIGNATURE:

TEACHER’S SIGNATURE:

To verify the basic proportionality theorem using parallel line board and triangle cut-outs.

BASIC PROPORTIONALITY THEOREM:

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

MATERIAL REQUIRED:

Coloured paper, pair of scissors, parallel line board, ruler, sketch pens.

PRE- REQUISITE KNOWLEDGE:

Drawing parallel lines on a rectangular sheet of the paper.

PROCEDURE:

1. Cut two different triangles from a coloured paper. Name them as triangle ABC, PQR.

2. Take the parallel line board(a board on which parallel lines are drawn).

3. Place triangle ABC on the board such that any one side of the triangle is placed on one of the lines of the board as shown on next pages

4. Mark the pts.p1,p2,p3,p4 on triangle ABC. Join p1p2 and p3p4. p1p2 || BC and p3p4||BC.

5. Note the following by measuring the lengths of the respective segments using a ruler.

6. Repeat this activity for the triangle PQR.

On the next three pages:

Fig.1= parallel line board.

Fig.2= triangle ABC

Fig.3= triangle PQR.

SOLUTION:

AP1= 5cm AP2= 5cm

P1B= 7cm P2C= 7cm

AP3= 9cm AP4= 9cm

P3B= 2cm P4C= 2cm

In triangle ABC

P1P2 || BC P3P4 ||BC

AP1/P1B = 5/7AP2/P2C = 5/7AP3/P3B = 9/2AP4/P4C = 9/2 AP1/P1B = AP2/P2C AP3/P3B = AP4/P4C

Hence verified.....

RATIOS VALUE AP1/P1B 5/7 AP2/P2C 5/7 AP3/P3B 9/2 AP4/P4C 9/2

LEARNING OUTCOME:

By doing this activity we learnt that how ratios of sides of the triangle are taken and how their values are obtained.

I also understood the mechanism of basic proportionality theorem of the triangle that is:

“If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided

in the same ratio”.

LIST OF RESOURCES WITH REFERENCE:

- I took the reference of class-x mathematics NCERT book.

- Chapter – 6(Triangles)

Somewhere I found difficulty to do this activity that how the ratios of different two triangles would be applied.

But my cousin helped me out to solve this problem. Firstly he asked me to do some examples of this theorem from R.D Sharma book, but I stuck at one of the example he helped me to do it.

He also helped me to understand the concept of the ratios and the proportions of the triangle. Then I felt comfortable and did this activity with ease and perfection. My experience was good and by doing this activity my confidence level has also risen up.

RATIOS VALUE QS1/S1P 9/2

QS2/S2R 9/2

QS3/S3P 2/9

QS4/S4R 2/9

QS1= 9cm S1P= 2cm

QS2= 9cm S3P= 9cm

QS3= 2cm S2R= 2cm

QS4= 2cm S4R= 9cm

In triangle PQR,

S1S2 || PR

S3S4 || PR

QS1/S1P= 9/2 QS3/S3P= 2/9

QS2/S2R= 9/2 QS4/S4R= 2/9

Hence verified.....

We observed that:

In triangle ABC,

AP1/P1B= AP2/P2C

AP3/P3B= AP4/P4C

In triangle PQR,

QS1/S1P= QS2/S2R

QS3/S3P= QS4/S4R

We observed that in both two triangles:

ABC and PQR

The basic proportionality theorem is verified or proved.