LIST OF ACTIVITIES CLASS 9 TH

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1 LIST OF ACTIVITIES CLASS 9 TH S.N. ACTIVITIES 1) 2) To create a wheel of THEODOROUS that demonstrates spiral in real number world. To verify algebraic identity (a + b + c) 2 = a 2 + b 2 + c ab + 2 bc + 2ca To verify exterior angle property of triangle 3) To verify triangle inequality property. 4) 5) To verify Mid Point Theorem. 6) 7) Central angle is double the inscribed angle of a circle. To verify CSA and TSA of cylinder by paper cutting and pasting. 8) To verify that the area of a triangle is half the area of parallelogram if they have same base and lie between same parallel lines.

2 ACTIVITY 1 AIM - To create a wheel of Theorodus that demonstrates spiral in real world. PROCEDURE 1) using a template for a particular unit length and a right angle, create an isosceles right triangle. 2) using your template again, add another unit length and right angle to the hypotenuse of original right triangle. 3) make a right triangle out of the new unit lengths and previous hypotenuse. 4) keep adding a new unit length to the previous hypotenuse at right angles to build new right triangles. 5) when we get to the stage where new right triangles will overlap previous right triangles, draw hypotenuse towards the centre of the spiral. 6) label the figure with all dimensions and mark the hypotenuse having length with a rational number. ACTIVITY 2 Aim To verify algebraic identity ( a+b+c ) 2 = a 2 + b 2 +c 2 + 2ab +2bc +2ca geometrically. Material Required Origami sheets, a pair of scissors, geometry box, fevicol. Procedure 1) Cut three squares with sides a=6 cm, b=4cm and c=2cm. 2) Cut six rectangles with following measurements: 2 rectangles with length (a)= 6 cm, breadth (b) =4 cm. 2 rectangles with length (a)= 4 cm, breadth (b) =2 cm. 2 rectangles with length (a)= 2 cm, breadth (b) =6 cm. each side 3) Arrange and paste all squares and rectangles to form a bigger square with equal to a+b+c = = 12cm. Observation Area of bigger square = ar( 3 squares ) + ar ( 6 rectangles) Side side = =

3 144 cm 2 = 144 cm 2 Conclusion - ( a+b+c ) 2 = a 2 + b 2 +c 2 + 2ab +2bc +2ca ACTIVITY - 3 Aim: To explore the relationship between the exterior and interior angles of a triangle by paper cutting and pasting. Material required : Coloured paper, pair of scissors, pencil, glue, ruler. Procedure: Step 1; On a coloured paper, draw any triangle and name its vertices as A,B and C. Step 2: Label the interior angles. Extend any one of the sides say, BC to form a ray BD. Step 3: Make the replica of interior angles CAB and ABC. Step 4: Place the two cut outs adjacent to each other on the exterior angle ACD. Observation: On placing the interior opposite angles, adjacent to each other On the exterior angleacd. We observe that it fully covers angleacd. Result: Sum of two interior opposite angles is equal to exterior angle. ACTIVITY 4 Aim To verify that the sum of any two sides of a triangle is always greater than the third side. Material Required - pencil, ruler, broom stick. Procedure Get sticks of different lengths. Take three at a time. Set 1 4cm, 5cm,10cm Set 2 5cm,5cm, 10cm Set 3 5cm, 6cm,10cm

4 For each triplet of numbers in a given set above, try to form triangle. Observation - Set 1 Conclusion Set 2 Set 3 (observation and conclusion will be written in class ) ACTIVITY - 5 Objective To verify the mid point theorem for a triangle, using paper cutting and pasting. Pre-requisite knowledge Two lines are parallel if for a transversal cutting them, the corresponding angles are equal. Material Required Coloured paper, a pair of scissors, gum. Procedure From a sheet of paper, cut a triangle ABC. Find the mid points P and Q of AB and AC respectively by paper folding. Join P and Q by folding and making a crease PQ. [Fig 7 (a)] Cut APQ. Superimpose AQ over QC so that QP falls along CB as shown in Fig 7 (b). Observations 1. Angle APQ is now renamed as (A)(P)(Q). A falls on Q since Q is the mid point of AC. 2. Triangle AQP is superimposed on triangle QCB and the two angles are seen to be equal. They are the corresponding angles made on PQ and BC by AC. 3. Therefore, PQ is parallel to BC. 4. Also (P) is seen to be the mid point of BC by paper folding method already described. ACTIVITY - 6 Aim - By paper cutting and pasting verify that angle subtended by an arc at the centre of a circle is twice the angle subtended by the same arc at any other point on the remaining part of the circle. Material required coloured paper, a pair of scissors, geometry box, fevistick, carbon paper. Procedure step 1 Draw a circle of any radius with centre O on a coloured paper and cut it

5 step 2 Take a rectangular sheet of paper and paste the circle cut out on it step 3 Mark two points A and B on the circle to get arc AB. step 4 Form a crease joining OA and draw OA. step 5 Form a crease joining OB and draw OB. step 6 Take a point C on the remaining part of the circle. step 7 Form a crease joining AC and draw AC step 8 Form a crease joining BC and draw BC. step 9 Make two replicas of angle ACB using carbon paper. step 10 Place the two replicas of angle ACB adjacent to each other on angle AOB Observations 1. Arc AB subtends angle AOB at the centre and angle ACB at the point C on the remaining part of the circle. 2. Two replicas of angle ACB completely cover angle AOB. Result angle AOB = 2 (angle ACB) ACTIVITY 7 Aim To obtain formula for LSA and TSA of a right circular cylinder in terms of radius(r) of its base and its height (h). Material Required coloured paper, a pair of scissors, ruler, fevicol. Procedure - 1) Draw a rectangle with length (l) and breadth (b) on a coloured paper. 2) Roll the rectangular sheet along its length and paste the edges to form a hollow cylinder. Observation length of rectangular sheet ( l )= circumference of base of cylinder ( C ) Breadth of rectangular sheet (b )= height of cylinder (h) CSA of cylinder = Area of rectangular sheet = length breadth

6 = C h = 2 rh TSA = CSA + area of two circular ends = 2 rh + 2 r 2 Conclusion - CSA of a right circular cylinder = 2 rh TSA of a right circular cylinder = 2 rh + 2 r 2 ACTIVITY 8 Aim - To verify that the area of a triangle is half the area of parallelogram if they have same base and lie between same parallel lines. Material Required - coloured paper, geometry box, fevicol. Procedure - 1) Construct a parallelogram ABCD with given measurement : AB = 10cm, AD = 6cm, angle DAB = 30. 2) Drop a perpendicular DM on AB. 3) Calculate area of parallelogram using formula base height. 4) Take any point P on DC and join AP, PB to form a triangle APB. 5) Calculate are of triangle using formula ½ b h. Observation Ar ( ǁgm ) = b h = Ar ( ) = ½ b h = Ar ( ǁgm ) = 2 Ar ( ) Conclusion - Area of a triangle is half the area of parallelogram if they have same base and lie between same parallel lines.

Project Maths Geometry Notes

The areas that you need to study are: Project Maths Geometry Notes (i) Geometry Terms: (ii) Theorems: (iii) Constructions: (iv) Enlargements: Axiom, theorem, proof, corollary, converse, implies The exam