Activity Of Maths CLass X

Objective
To prove Distance formula = by experimentally
Pre-knowledge
We know Pythagoras Theorem
Area of triangle
Some Knowledge about coordinate
Rules for signs of Co-ordinates
Axes of Co-ordinates
Geometrical Representation of quadratic polynomials
Material Required
Coloured Glazed paper
Pair of scissors
Geometry box
Graph paper
Drawing sheet
Colour stick
Pencil colour
Fevistick/ Gum
Procedure
Let two points P(x1,y1) and Q(x2,y2) on graph sheet.
And draw a set of perpendicular axes on the graph paper. As PM and QN on the axis.
From P draw PL perpendicular to QN as shown in figure below
Then
OM = x1, MP = y1
ON = x2 , NQ = y2

Calculation
So PL = MN QL= QN
Now,
PL = MN = ON – OM = x2 – x1 QL = QN – LN = QN – PM = y2 – y1
Applying Pythagoras in Δ PQL we get
PQ2 = PL2 + QL2
PQ2 = (x2 – x1)2 + (y2 – y1)2
PQ =
:- Note that since distance is always non-negative, We take only positive square root. So the distance between the points P(x1,y1) and Q (x2,y2)is
PQ =
Which is called the distance formula
Result
The distance between two point can be find out by distance formula =
Remark
In particular, the distance of a point P( x, y) from the origin O(0,0) is given by
OP =
We can also write ,PQ =

Objective
To prove section formula by experimentally.

Pre-Knowledge
We know Pythagoras Theorem
Area of triangle
Some Knowledge about coordinate
Rules for signs of Co-ordinates
Axes of Co-ordinates
Geometrical Representation of quadratic polynomials
Now we also know Distance formula
Basic Proportionality Theorem.
Ratio of sides
Similar triangle Properties
Material Required
Coloured Glazed paper
Pair of scissors
Geometry box
Graph paper
Drawing sheet
Colour stick
Pencil colour
Fevistick/ Gum
Procedure
1. A point C divides the Straight line joining two points A(x1,y1) and B(x2,y2) internally in the ratio m1:m2 to find the co-ordinates of C. Let the