Class 07 Project : Value of 𝝅

 Project : Value of 𝝅

Objective: 

To find the value of it by activity method.

Materials Required: 

Coloured paper, geometry box, different coloured threads, a pair of scissors, sketch pens, fevicol, etc.

Principle:

 1. 𝝅 is defined as the ratio of circumference and diameter of a circle,

 i.e., 𝝅  = ^{"Circumference"} /_"Diameter"  

2. 𝝅 is an irrational number.

3. For practical purposes we take the value of 𝝅  as 𝟐𝟐/𝟕 or  3.14 ... (approx.)

Procedure:

1. Take a coloured paper and draw five circles with different radii. Name them as A, B, C, D and E as shown in Fig.

2. Cut the circles A, B, C, D and E from the coloured paper.

3. Take a coloured paper and fix a thread on it with fevicol and mark its end points as P and Q as shown in Fig. 

4. Take a circle and put a mark on the circumference 

of it as shown in Fig. 

5. Now place the circle on the thread in such a way the mark touches the point P as shown in Fig. 

6. Now roll the circle along the line PQ, till the mark touches the thread again, mark this point X, as shown in Fig. 

7. Measure the distance PX with the help of a ruler and denote it by C.

8. Calculate the diameter of the circle and denote it by d.

9. Take the ratio "C" /"d"  to calculate the value of 𝝅.

10. Repeat the steps from 5 to 9 for other circles and record your observations.

Observations : 

2. Average value of 𝝅 is _________

3. The length of the thread gives the ____________of the circle.

Class 07 Activity – 3 Congruence of triangles

 Activity – 3 Congruence of triangles

Objective: 

To establish the congruency of triangles by making paper models and superimposing.

Materials Required : 

Paper, pencil ruler and scissors. 

Procedure:

Draw two triangles PQR and LMN where PQ = LM, PR = LN, and ∠P = ∠ L. Cut out these triangles 

1. Overlap PQ of triangle I with LM of triangle II. It will coincide exactly as they are equal.

2. Overlap ∠ P on to ∠ L. They too will coincide as their measures are equal.

3. Now, since PR and LN are equal, they will coincide exactly and the open end of PR will coincide with ∠ N.

4. This means that QR and MN too will fall along each other. Thus triangle I coincides with triangle II.

Observations: Record your observations in the following table.

Conclusion: 

From the above activity, it is verified that if two sides and the included angle of one triangle respectively are equal to two sides and the included angle of another triangle, the triangles are congruent.

Class 07 Activity – Area of Circle

 Activity – Area of Circle

Objective: 

To verify the formula for area of a circle. Or to verify the following formula.

Area of a circle = 𝜋 x (radius)²

Materials Required: 

Thick sheets of paper, colour pencils, a pair of scissors, glue stick, geometry box, etc.

Procedure:

On a thick sheet of paper, draw a circle of any convenient 

radius. Colour one-half of the circle. Using a pair of scissors, 

cut it out.

2. Divide the circular cutout into eight equal parts. This can be done by drawing two pairs of perpendicular diameters. Using a pair of scissors, cut out the sectors along the radii.

3. Arrange the 8 sectors of the circle as shown below, which is roughly a parallelogram.

4. Repeat step 1. Now, divide the circular cut out into 16 equal parts. Using a pair of scissors, cut out the sectors along the radii.

5. Arrange the 16 sectors of the circle as shown below.

6. Repeat the procedure. As we go on increasing the number of sectors, the length of arcs go on decreasing. The more the sectors we have the nearer we reach a rectangle.

Observations:

 1.In figure , AB = CD = radius of the circle.

 Also the whole circle is divided into 8 sectors and oneach side there are  4 sectors. The length of the parallelogram is the length of 4 sectors, which is half of the circumference of the circle. 

2. Similarly, in figure, AB = CD =  radius of the circle  and the length of the parallelogram is equal to half of the circumference of the circle.

3. In figure, AB = CD = Breadth of the rectangle ABCD = radius of the circle and BC =  AD = length of the rectangle, which is half of the circumference of the circle. 

Area of the circle = area of the rectangle 

= length x breadth

= 1/2 x 2 𝜋 r x r , where r is the radius of the circle.

= 𝜋 r²

Conclusion: 

From the above activity, it is verified that 

area of a circle= 𝜋 x (radius)² 

Do Yourself: 

Draw a circle of radius 5 cm. Verify the formula for area of a circle by paper cutting and Pasting method.

 

Class 07 Activity – Area of Parallelogram

Activity – Area of Parallelogram 

Objective : 

To verify the formula for Area of a parallelogram or to verify the following formula. Area of a parallelogram = base x height

Materials Required : 

Squared paper, colour pencils,  a pair Of scissors, glue stick, geometry box, etc.,

Procedure:

On a  squared paper, draw a parallelogram ABCD. Draw. DE (height corresponding to the base AB). Shade the two parts using different colours.

2. Using a pair of scissors, cut out the triangle ADE.

3. Paste the triangular cutout ADE to the right side of BCDE such that AD and BC coincide.

Observations: 

In figure, we see that the resulting figure becomes a rectangle. 

So, area of the parallelogram = area of the rectangle = length x breadth

= CD X DE

= AB X DE  [AB = CD, opposite sides of a parallelogram are equal]

= Base x height 

Conclusion: 

From the above activity we find that Area of a parallelogram = base x height

Do Yourself: 

Draw three different parallelograms on squared papers. In each case, verify the formula for area of a parallelogram by paper cutting and pasting.