Class 9 Maths notes for Heron Formula | Mensuration

Mensuration

• It is branch of mathematics which is concerned about the measurement of length ,area and Volume of plane and Solid figure

Perimeter

• The perimeter of plane figure is defined as the length of the boundary
• It units is same as that of length i.e. m ,cm,km

1 Meter	10 Decimetre	100 centimetre
1 Decimetre	10 centimetre	100 millimetre
1 Km	10 Hectometre	100 Decameter
1 Decameter	10 meter	1000 centimetre

Area

• The area of the plane figure is the surface enclosed by its boundary
• It unit is square of length unit. i.e. m² , km²

1 square Meter	100 square Decimetre	10000 square centimetre
1 square Decimetre	100 square centimetre	10000 square millimetre
1 Hectare	100 squareDecameter	10000 square meter
1 square myraimeter	100 square kilometre	108 square meter

Perimeter and Area of Different Figure

N	Shape	Perimeter/height	Area
1	Right angle triangle Base =b, Height =h Hypotenuse=d	$P=b+h+d$ Height =h	$A=\frac {1}{2}BH$
2	Isosceles right angled triangle Equal side =a	Height=a	$A=\frac {1}{2} \sqrt {a}$
3	Any triangle of sides a,b ,c	$P=a+b+c$	$A=\sqrt {s(s-a)(s-b)(s-c)}$ $s=\frac {a+b+c}{2}$ This is called Heron's formula (sometimes called Hero's formula) is named after Hero of Alexandria
4	Square Side =a		$A=a^2$
5	Rectangle of Length and breath L and B respectively	$P=2L +2B$	$A=L \times B$
6	Parallelograms Two sides are given as a and b	$P=2a+2b$	$A= Base \times height$ When the diagonal is also given ,say d Then $A= \sqrt {s(s-a)(s-b)(s-d)}$ $s=\frac {a+b+d}{2}$
7	Rhombus Diagonal d1 and d2 are given	$p=2 \sqrt {d_1^2+d_2^2}$ Each side=$ \frac {1}{2} \sqrt {d_1^2+d_2^2} $	$A=\frac {1}{2} d_1 d_2$
8	Quadrilateral a. All the sides are given a,b,c ,d b. Both the diagonal are perpendicular to each other c. When a diagonal and perpendicular to diagonal are given	a. $P=a+b+c+d$	a. $A=\sqrt {s(s-a)(s-b)(s-c)(s-d)}$ $s=\frac {(a+b+c+d)}{2}$ b. $A=\frac {1}{2} d_1 d_2$ where d1 and d2 are the diagonal c. $A=\frac {1}{2} d(h_1+h_2)$ where d is diagonal and h1 and h2 are perpendicular to that

How to solve the Area and Perimeter problems

1. We must remember the formula for all the common figures as given above the table
2. Find out what all is given in the problem
3. Convert all the given quantities in the same unit
4. Sometimes Perimeter is given and some side is unknown,So you can calculate the sides using the Perimeter
5. If it is a complex figure ,break down into common know figures like square,rectangle,triangle
6. Sometimes we can find another side using Pythagoras theorem in the complex figure
7. If common figure, apply the formula given above and calculate the area.
8. If complex figure, calculate the area for each common figure in it and sum all the area at the end to calculate the total area of the figure

Solved Examples

1. A right angle triangle has base 20 cm and height as 10 cm, What is the area of the triangle?
Solution
Given values B=20 cm
H=10 cm
Both are in same units
A=(1/2)BH=100 cm²
2. Sides of triangles are in the ratio 12:17:25. The perimeter of the triangle is 540 cm. Find out the area of the triangle?
Solution
Let the common ration between the sides be y,then sides are 12y,17y,25 y
Now we know the perimeter of the triangle is given by
P=a+b+c
540=12y+17y+25y
or y=10 cm
Now Area of triangle is =[s(s-a)(s-b)(s-c)]^1/2
Where s=(a+b+c)/2
Here s=270 cm
a=120 cm
b=170cm
c=250cm
Substituting all these values in the area equation,we get
A=9000cm²
3. A equilateral triangle is having side 2 cm. What is the area of the triangle?
Solution:
We know that Are of equilateral triangle is given by
A=[(3)^1/2 a²]/4
Substituting the values given above
A=(3)^1/2

We can summarize various method to calculate the Area of the Triangle

If you know the altitude and Base	Area =(1/2)BH
If you all the three sides	A=[s(s-a)(s-b)(s-c)]1/2 s=(a+b+c)/2
If it is isosceles triangle with equal side a	A=(1/2)a1/2
If it is equilateral triangle with equal side a	A=[(3)1/2 a2]/4
If it is right angle triangle with Base B and Height H	Area =(1/2)BH

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