Definitions [1]
Area of a circle: The area of a circle is the region occupied by the circle in a two-dimensional plane.
Formulae [5]
Area of parallelogram = base x height
- Area of a rhombus = `1/2` x product of lengths of diagonals.
Area of the trapezium = `1/2` × sum of the lengths of parallel sides × height.
Area of triangle = `(1/2) × "base" × "height" = 1/2 × b × h`.
Area of the circle = πr2
Theorems and Laws [4]
Prove that the points A(a, 0), B(0, b) and C(1, 1) are collinear, if `(1/a + 1/b) = 1`.
Consider the points A (a,0), B( 0,b) and C (1,1) .
` Here (x_1=a,y_1=0).(x_2 = 0,y_2=b) and (x_3=1,y_3=1).`
It is given that the points are collinear. So,
`x_1 (y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2) =0`
`⇒ a(b-1)+0(1-0)+1(0-b)=0`
`⇒ ab-a-b=0`
Dividing the equation by ab:
`⇒ 1-1/b-1/a=0`
`⇒ 1-(1/a+1/b)=0`
`⇒(1/a+1/b)=1`
Therefore, the given points are collinear if `(1/a+1/b)=1`
Prove that the points A(7, 10), B(–2, 5) and C(3, –4) are the vertices of an isosceles right triangle.
The given points are A (7, 10), B(-2, 5) and C(3, -4).
`AB= sqrt((-2-7)^2 +(5-10)^2) = sqrt((-9)^2 +(-5)^2) = sqrt((81+25)) = sqrt(106)`
`BC = sqrt((3-(-2))^2 +(-4-5)^2) = sqrt((5)^2 +(-9)^2 )= sqrt((25+81) )= sqrt(106)`
`AC = sqrt((3-7)^2 +(-4-10)^2) = sqrt(( -4)^2 +(-14)^2) = sqrt(16+196) = sqrt(212)`
Since, AB and BC are equal, they form the vertices of an isosceles triangle
Also,`(AB)^2 + (BC)^2 = ( sqrt(106))^2 +( sqrt(106)^2) = 212`
and `(AC)^2 = (sqrt(212))^2 = 212.
`Thus , (AB)^2 + (BC)^2 = (AC)^2`
This show that ΔABC is right- angled at B. Therefore, the pointsA (7, 10), B(-2, 5) and C(3, -4). are the vertices of an isosceles rightangled triangle.
Prove that the points A(2, 4), B(2, 6) and `C(2 + sqrt(3), 5)` are the vertices of an equilateral triangle.
The given points are A(2, 4), B(2, 6) and C(2 +`sqrt(3)`,5) Now
`AB =sqrt(((2-2)^2 +(4-6)^2 )) = sqrt((0)^2 +(-2)^2)`
`= sqrt((0+4) =2`
`BC = sqrt((2-2- sqrt(3))^2 + (6-5)^2 ) = sqrt((- sqrt(3))^2 +(1)^2)`
`= sqrt(3+1) = 2`
`AC = sqrt((2-2-sqrt(3))^2 + (4-5)^2 ) = sqrt((- sqrt(3))^2 +(-1)^2)`
`= sqrt(3+1) =2`
Hence, the points A(2, 4), B(2, 6) and C(2 +`sqrt(3)`,5) are the vertices of an equilateral triangle
A(7, –3), B(5, 3) and C(3, –1) are the vertices of a ΔABC and AD is its median. Prove that the median AD divides ΔABC into two triangles of equal areas.
The vertices of the triangle are A(7, -3), B(5,3) and C(3,-1)
`"Coordinates of" D = ((5+3)/2,(3-1)/2) = (4,1)`
For the area of the triangle ADC, let
`A (x_1,y_1)=A(7,-3), D(x_2,y_2) =D(4,1) and C (x_3,y_3) = C(3,-1)`. Then
`"Area of" Δ ADC = 1/2 [ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]`
`=1/2 [7(1+1)+4(-1+3)+3(-3-1)]`
`=1/2[14+8-12}=5` sq. unit
Now, for the area of triangle ABD, let
`A(x_1,y_1) = A(7,-3), B(x_2,y_2) = B(5,3) and D (x_3,y_3) = D (4,1). `Then
`"Area of" Δ ADC = 1/2 [ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]`
`=1/2 [7(3-1)+5(1+3)+4(-3-3)]`
`=1/2[14+20-24] = 5` sq. unit
Thus, Area (ΔADC) = Area (ΔABD) = 5. sq units
Hence, AD divides ΔABC into two triangles of equal areas.
