Class 07 ACTIVITY 4 - TRIANGLES

 ACTIVITY 4 - TRIANGLES

Objective: 

To medians get the pass medians through of a a common triangle by point paper. folding. Also, to verify that in a triangle medians pass through a common point.

Materials Required: 

Tracing paper, colour pencils, geometry box, etc.

Procedure:

1. On a. tracing paper, trace the following triangles. Cut out each triangle from the tracing paper. Note that triangle (a) is an equilateral triangle, triangle (b) is isosceles as well as right angled triangle and  triangle(c) is as a scalene triangle. We can also say that triangle (a) is an acute angled triangle, (b) is right angled triangle, (c) is an obtuse angled triangle.

2. Fold each triangular cut out such that the vertex Q coincides with the vertex R.

3. Unfold each tracing paper. In each case mark the point of intersection of QR and the crease as X. Draw PX as dotted line.

4. Now, fold each triangular cut out such that the vertex P and R coincide.

5. Unfold each tracing paper. In each case mark the point of intersection of PR and the crease P as Y. Draw QY as a dotted line.

6. Finally fold each triangular cutout such that the vertex P coincides with the vertex Q.

7. Unfold each tracing paper. In each case mark the point of intersection of PQ and the crease as Z. Draw RZ as a dotted line. 

Observations :

In figure 3, X is the mid point of QR. So PX is a median of  each triangle.

2. Similarly in figures 5 and 7,  Y and Z are mid points PR and PQ respectively. So QY and RZ are medians of ∆PQR in each case.

3. Also  in figure 7, we see that in each case all the three medians pass through a common point O.

Conclusion

A triangle has three medians.

2. All the three medians of a triangle pass through a common point. This point is called the centroid of the angle.

Do Yourself: 

Draw an acute angled, a right angled and an obtuse angled triangle. By paper folding, verify that in each case the three medians are  concurrent.

Class 07 ACTIVITY 3 - TRIANGLES

 ACTIVITY 3 - TRIANGLES

Objective: 

To verify that an exterior angle of a triangle is equal to the sum of the two interior opposite angles by paper cutting and pasting.

Materials Required: 

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

Procedure:

On a white sheet of paper, draw a triangle ABC. Produce its side BC to D as shown in the figure. Using Colour pencils, mark its angles as shown.

2. Using a pair of scissors, cut out the two marked angular regions as shown below.

3. Paste the angular cutouts A and B over exterior angle C such that vertices A, B and C coincide as shown.

Observations:

In figure, ∠ACD is an exterior angle of ∆ABC and  ∠ A and ∠ B are its two interior opposite angles

2. In figure, we see that the angular cutouts neither overlap nor leave any gap between them. In other words, the angular cut outs A and B completely cover exterior angle C.

or ∠ A + ∠ B = exterior angle C. or ∠ A + ∠ B = ∠ ACD

Conclusion: 

From the above activity, we can say that an exterior angle of a triangle is equal to the sum of the two interior opposite angles.

Do Yourself: 

Copy each of the following triangles. In each case verify that an exterior angle of a triangle is equal to the sum of two interior opposite angles.

Class 07 ACTIVITY 2 - TRIANGLES

 ACTIVITY 2 - TRIANGLES

Objective: 

To verify the angle sum property of a triangle by paper folding.

Materials Required: 

White sheets of paper, tracing paper, geometry box, a pair of scissors, etc.

Procedure:

On a white sheet of paper, draw a fairly large triangle ABC. Using scissors, cut it out. Mark A, B and C on both sides of the cutout.

2. Fold the side BC of ∆ABC such that the folding line passes through A.

3. Unfold it and draw a line along the crease. This line cuts BC at D.

4. Fold the vertices (three corners) of ∆ABC such that A, B and C meet at D.

Observations: 

In figure 4, we see that the angles A, B and C form a straight angle 

i.e., ∠A + ∠ B + ∠ C = 180°

Conclusion: 

From the above activity, it is verified that the sum of the angles of a triangle is 180 °

Do Yourself:  

Draw an acute angled triangle, a right angled triangle. By paper folding, verify the angle sum property in each case.

Class 07 ACTIVITY - Triangles

 ACTIVITY - Triangles

Objective: 

To verify that the sum of all interior angles of a triangle is 180 °, by papercutting and pasting.

Materials Required: 

White sheets of paper, colour pencils, a pair of scissors, geometry box, etc.

Procedure:

On a white sheet of paper, draw a triangle ABC. Using colour pencils mark its angles as shown.

2. Using a pair of scissors, cut out the three angular regions.

3. Draw a line segment XY and mark a point on it.

4. Paste the three angular cut outs so that the vertex of each falls at O as shown in the figure.

Observations:

In figure 4, we see that the angular cutouts neither overlap nor leave any gap between them. But, XOY is a straight angle So, ∠A + ∠ B + ∠C = 180 ° Or sum of the angles of a triangle is 180 °.

Conclusion: 

From the above activity, it is verified that the sum of all interior angles of a triangle is 180 °

Do Yourself: 

Draw an acute angled triangle, a right triangle and an obtuse angled triangle. By paper cutting and pasting, verify the above property for each triangle.

Class 07 ACTIVITY3 - LINES AND ANGLES

 ACTIVITY3 - LINES AND ANGLES

Objective: 

To verify that if two parallel lines are cut by a transversal, then each pair of interior angles on the same side of the transversal are supplementary by paper cutting and pasting

Materials Required: 

White sheets of paper, colour pencils, geometry box, a pair of scissors, gluestick, etc. 

Procedure:

1. cutting On a white sheet of paper, draw two parallel lines AB and CD and a transversal EF cutting them at P and Q resp., Mark a point O on PQ. 

2. Cut the angular regions ∠OQC, ∠ OQD ∠ OPA and ∠ OPB.

3. Draw a line segment and take a point X on it. Paste the angular cut out OQD such that Q coincides with X and QD falls along the Straight line.

Now, paste the angular cut out OPB such that P coincides with X and PO falls along QO.

4. Draw another line segment and take a point Y on it. Paste the angular cut out OQC such that Q coincides with Y  and QC falls along the straight line. Now, paste the angular cut out OPA such that P coincides with Y and PO falls along QO.

Observations:

In figure 1, AB || CD and EF is a transversal.

So, (∠ APO, ∠ CQO) and (∠ BPO, ∠ DRO) are two pairs of interior angles on the same side of the transversal.

2. In figure 3, BD is a straight line. So, ∠ BXD = 180 °or

 ∠ BPO + ∠ DRO = 180 °

3. In figure 4, AC is a straight line.

So, ∠ AYC = 180 °or ∠ APO + ∠ CQO = 180 °

Conclusion: 

From the above activity, we can say that if two parallel lines are cut by a transversal, then each pair of interior angles on the same side of the transversal are supplementary.

Class 07 ACTIVITY2 - LINES AND ANGLES

 ACTIVITY2 - LINES AND ANGLES

Objective:

 To verify that if two parallel lines are cut by a transversal, then each pair of alternate interior angles are equal by paper cutting and pasting

Materials Required : 

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

Procedure:

On a white sheet of paper, draw a pair of parallel lines AB and CD. Also, draw a transversal EF cutting them at P and Q resp., Mark a point O somewhere in the middle of PQ. Mark the angles as shown.

2 . Cut the angles ∠ OQD and ∠ OQC.

3. Paste the angular cutouts ∠ CQO and ∠ DQO over ∠BPO and ∠ APO respectively such that in each case the vertex Q coincides with vertex P and one arm of each angle falls along one arm of the corresponding angles.

Observations: 

In figure 1, AB || CD and EF is a transversal. So, (∠APO, ∠ DRO) and (∠ BPO, ∠ CQO) are two pairs of alternate interior angles.

2. In figure 3, we see that if vertex Q of ∠ CQO coincides with vertex P of ∠ BPO and arm QC falls along PB, then QO  falls along PO, i.e., ∠ CQO completely overlaps ∠ BPO.

So, ∠ CQO = ∠ BPO 

Similarly, ∠ DQO completely overlaps ∠ APO.

So, ∠ DQO = ∠ APO transversal 

Conclusion : 

From the above activity, interior, we can say that if two parallel lines are cut by a transversal, then each pair of alternate interior angles are equal.

Class 07 ACTIVITY1 - LINES AND ANGLES

 ACTIVITY1 - LINES AND ANGLES

Objective: 

To verify that if two parallel lines are cut by a transversal, then each pair of corresponding angles are equal, by paper cutting and pasting.

Materials Required: 

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

Procedure:

1. On a white sheet of paper, draw a pair of parallel lines AB and CD. Also, draw a transversal EF, cutting AB and CD at P and Q respectively. Mark the angles as shown in the figure. Mark a point O somewhere in the middle of PQ.

2. Cut the figure along the dotted lines to get four angular cut outs as shown below.

3. Paste the angular cutout ∠ DQO over ∠ BPE such that Q coincides with P and QD falls along PB.

4. Similarly, paste the angular cut outs ∠DQF,   ∠ CQF and ∠ CQO over ∠ BPO, ∠ APO and ∠ APE resp., such that in each case Q coincides with P and one arm of each angle falls along one arm of the corresponding angle.

Observations:

In figure1, AB ∥ CD and EF is Transversal. So, (∠ EPB, ∠ DQO), (∠ BPO, ∠ DQF), (∠ APE, ∠CQP) and (∠ APQ, ∠ CQF ) are four pairs of corresponding angles.

2. In figure, we see that if vertex Q of ∠ DQO coincides with the vertex P of ∠ BPE and QD falls along PB, then QO falls along PE, i.e., DQO completely overlaps BPE. So, ∠ DQO = ∠ BPE.

3. Similarly, in fig, we see that ∠ DQF, ∠ CQF and ∠ CQO completely overlap ∠ BPO, ∠ APO and ∠ APE resp.,

So, ∠ DQF = ∠ BPO, ∠ CQF = ∠ APO and ∠ CQO = ∠ APE

Conclusion : 

From the above activity, we can say that if two parallel lines are cut by a transversal, then each pair of corresponding angles are equal.

Do yourself: 

Verify the above property by drawing a pair of parallel lines which are 5 cm apart.

Class 07 Fun Activity – Algebraic Expression CROSS NUMBER PUZZLE

 Fun Activity – Algebraic Expression
CROSS NUMBER PUZZLE

Find the root of the given equations and complete the following cross number puzzle

Fun Activity – Algebraic Expression -  solution

CROSS NUMBER PUZZLE

Find the root of the given equations and complete the following cross number puzzle

CROSS NUMBER PUZZLE

Find the root of the given equations and complete the following cross number puzzle

Class 07 Fun Activity – Algebraic Expression

 Fun Activity – Algebraic Expression

Given the values a = 5, b = 3, c = 6, and d = 10.

Move from Start to finish and evaluate the algebraic expressions on the way using the given values. Take the path going from one odd number to another.

Fun Activity – Algebraic Expression - Solution

Given the values a = 5, b = 3, c = 6, and d = 10.

Move from Start to finish and evaluate the algebraic expressions on the way using the given values. Take the path going from one odd number to another.

3ab = 3 x 5 x 3 = 45

2ab= 2 x 5 x 3 = 30

9a = 9 x 5 = 45 5ac / 2 = (5 x 5 x6)/2 =150/2=75

3ac/ 2= (3 x 5 x 6)/2 = 90/2 = 45 5b2  = 5 x 3 x 3= 45 

6ab = 6 x 5 x 3 =90 abd/2 =( 5 x 3 x 10)/2=150/2=75

bcd/2 = (3x6x10)/2 =180/2=90 3acd/4= (3x5x6x10)/4=900/4=225

3cd/4=(3x6x10)/4=180/4=45 2a2 = 2x5x5=50

7ab = 7 x 5 x 3 = 105 abd = 5x3x10=150

c/2 = 6/2=3 abc/2 = (5 x 3x6)/2 = 90 /2=45

3ad/2 = 3x 5 x 10/2 = 150/2=75 bcd/4 = (3x6x10)/4= 180/4=45

7ac = 7 x 5x 6=210 6ad/4 = (6x 5x 10)/4=300/4=75

c2 + d2 = 36+100=136 a2 + b2 = 25+9 =34

2ad = 2 x 5 x 10 =100 acd/4 = 5 x 6 x 10 /4 = 300 /4 = 75

bd/6 = 3x10 /6 = 30 /6 =5 d2 /4 = 100 / 4 =25

a2 = 5 x 5 = 25 b2 = 3 x 3 =9

5b = 5 x 3 =15 6ab = 6x 5 x 3 = 90

Fun Activity – Algebraic Expression - solution

Class 07 Activity - Rational Numbers

 Activity - Rational Numbers

GROUP ACTIVITY

This activity involves finding rational numbers between two rational numbers.

Teacher will write about 20 pairs of rational numbers on the blackboard (at least half of them should be negative).

The student will pick out 2 pairs from these to work on.

(For example,) ^{(− 1)}/₂ and ^{( 1 )}/₄  is one pair and  ^{( −7 )}/₁₁  and ^{( −6 )}/₁₁  is another pair.

The student  will find  two rational numbers that lie between the two rational numbers of the first pair, like ^{(− 1)}/₄ and ⁰/₄ lie between ^{(− 1)}/₂ and ^{( 1 )}/₄. 

Now, he has to find two rational numbers that lie between ^{(− 1)}/₄ and ⁰/₄ . Suppose he writes ^{(− 2)}/₁₂  and ^{(− 1)}/₁₂  lie between and ^{(− 1)}/₄ and ⁰/₄ .

 Now he has to find two rational numbers that lie between ^{(− 2)}/₁₂  and ^{(− 1)}/₁₂. He can then find that ^{(− 5)}/₃₆  and ^{(− 4)}/₃₆ lie between ^{(− 2)}/₁₂  and ^{(− 1)}/₁₂. 

 

Similarly, he has to find two rational numbers between the two rational numbers of the second chosen pair.

Then again he has to find two  rational numbers between those two and again two more numbers between those two numbers.

Since the number of pairs (of rational numbers) given to the students to choose from are limited, many other students too would pick the same pair of numbers. 

Thus, a comparison could be made of their results. 

A whole array of numbers can thus be found between two rational numbers which are given. 

All the results for a particular pair could be written on a chart and then displayed in the room.

Class 07 PROJECT - TANGRAM

 PROJECT - TANGRAM

Objective : 

To understand the concept of ratio by activity, method.

Materials Required: 

Different coloured   triangle cutouts, geometry box, chart paper, etc.

Procedure: 

Arrange the different triangle cutouts as shown in Fig. 1.

2. Count the: different coloured triangles and record your observations.Observations :

Red triangles: Yellow triangles = ____________

2. Red triangles: Blue triangles = ______________

3. Red triangles: Green triangles = ______________

4. Red triangles: (Yellow + Blue) triangles = ______________

5. Red triangles: (Yellow + Green) triangles = ___________

6. Red triangles: (Yellow + Blue + Green) triangles = ________

Class 07 To use a 10 x 10 grid as a model for solving various types of per cent problems

 ACTIVITY – Percent Problems

Objective: 

To use a 10 x 10 grid as a model for solving various types of per cent problems.

Materials Required: 

Sketch pens, squared paper, colour pencils, geometry box, etc.

Procedure:

1. To represent the actual per cent.

1. Draw a 10 x 10 grid on a squared paper as shown in Fig. 

In the figure, one whole square represents 100 per cent and one small square represents 1 per cent. 

2. In the bottom of the big square, shade 30 small squares horizontally as shown in fig to depict 30 %

2. In the bottom of the big square, shade 30 small squares horizontally as shown in fig to depict 30 %

II. To represent the less than 1 %.

To show less than 1 per cent mark as shown in Fig. 3.

III. To represent more than 100%.

1. Make two grids of 10 x 10 each on squared papers.

2. To represent 150%, we shade one grid in full and the other 5 rows ( or 50 small squares) as shown in Fig.

100% + 50 % = 150%

Class 07 To solve a puzzle of linear equations in one variable by activity method

 ACTIVITY – Linear Equations in one Variable

Objective : 

To solve a puzzle of linear equations in one variable by activity method.

Materials Required :  

Calendar of any year. 

Procedure:

Choose four dates mentally to form a box as shown 

in the table below:

2. Find the sum of the chosen dates. Here, the sum of four chosen dates is 72. Divide 72 by 4 to get 18.

3. Now subtract 4 from 18 to get the first date, i.e., 14. To obtain other dates add 1, 7 and 8 respectively to 14.

Therefore, desired dates are 14, 15, 21 and 22.

Note: 1. This activity is done with the help of a friend.

2. Your friend would tell only the sum of the four dates.

3. After finding the dates, ask the friend whether the answer is correct or not.

Mathematical Explanation:

Let the first number (date) be x.

Then the next numbers would be x + 1, x + 7 and x + 8.

 x + x + 1 + x + 7 + x + 8 = 4x + 16

Let us say that the sum of four dates = 72

4x + 16 = 72

4 (x + 4) = 72

x + 4 = 18

x = 14

That gives you the first date. The other dates are 15, 21 and 22.

Class 07 Finding the values of 3^0, 3^1, 3^2, ........

 ACTIVITY –Exponents and Powers

Objective:

 Finding the values of 3^0, 3^1, 3^2, ........

Materials Required: 

A coloured chart paper, a scale, a pencil, an eraser, a pair of scissors.

Procedure:

1. Any rectangular piece of paper represents the base as shown in Fig. 

Number of times the rectangular piece is folded represents the exponent of the base. Cut a few equal size rectangular pieces from the chart paper. 

Let each rectangular piece represent the base 3.

2. Take one of the rectangular pieces. This piece is not folded i.e., it has been folded zero times. It represents 3^0, which is equal to 1. Hence, 3^0 = 1

3. Take another rectangular piece of chart paper.

Fold it into three equal parts along the width. 

We have folded it one time. It represents 3^1. Cut along the folds. The rectangular piece is divided into three equal parts. 

Hence, 3^1 = 3

4. Take one more rectangular piece of chart paper. Fold it once as folded in step 2 above.

Fold it once more along the width dividing it further into three equal parts. We have folded the original piece two times (Fig. 4), so it represents 3^2. Unfold it and cut along the folds. We get 9 equal pieces. Hence, 3^2 = 9

5. Take another rectangular piece of chart paper. 

Fold it as folded in step 3 above. 

Fold it now once along the length, dividing it further into three equal parts. We have folded the original piece three times , so it represents 3^3.

Unfold it and cut along the folds. We get 27 equal pieces. Hence, 3^3 = 27

Class 07 To understand the exponential growth of the triangles by activity method

 MATHS ACTIVITIESCLASS 7Based on CHAPTERs

9.Rational Numbers

13.Exponents and Powers

12.Algebraic Expression

4.Simple Equations

8. Comparing Quantities

ACTIVITY –Exponents and Powers

Objective: 

To understand the exponential growth of the triangles by activity method.

Materials Required: 

Coloured glazed papers, a pair of scissors, pencil, white chart sheet.

Procedure:

1. Cut out a square from a coloured glazed paper.

2. Draw the diagonal. Fold it to get two triangles and paste it on a white chart sheet as shown in Fig.

3. Now, make 2 and 3 folds and count the number of triangles formed in each case.

4. Also make 4 and 5 folds. Then note down the observations in a table as shown below:

We see that the number of triangles increase exponentially corresponding to the number of folds.

Class 07 FUN ACTIVITY - FRACTION

FUN ACTIVITY - FRACTION 

You are only allowed to move from one stone to another named by a smaller fraction.

1. Copy the diagram and colour your route.

2. Can you find a different route? (Use a different colour.)

FUN ACTIVITY – FRACTION - Solution

You are only allowed to move from one stone to another named by a smaller fraction.

1. Copy the diagram and colour your route.

2. Can you find a different route? (Use a different colour.)

Class 07 VEDIC MATHEMATICS

 VEDIC MATHEMATICS

We can find the products like 568 x 998, where the numbers to be multiplied are close to 1000, easily.

568 -432 →    568 is 432 below 1000, so we put 432 next to it.

998 -2  →  998 is 2 below 1000, so we put -2 next to it.

566 864                       We call the 432 and 2 deficiencies as the numbers 568 and 998 are deficient from 1000 by 432 and 2.

The answer 566864 is in two parts: 566 and 864.

The 566 is found by subtracting one of the deficiencies from the other number.

For example, 568-2 = 566 or 998 – 432 = 566 (whichever you like). 

And 864 is simply the product of the deficiencies: 432 x 2 = 864

So, 568 x 998 = 566864.