Students should go through these JAC Class 9 Maths Notes Chapter 12 Heron’s Formula will seemingly help to get a clear insight into all the important concepts.

### JAC Board Class 9 Maths Notes Chapter 12 Heron’s Formula

**Mensuration:**

A branch of mathematics which concerns itself with measurement of lengths, areas and volumes of plane and solid figure is called Mensuration.

→ Perimeter: The perimeter of a plane figure is the length of its boundary. In case of a triangle or a polygon, the perimeter is the sum of the lengths of its sides.

→ Units of Perimeter: The unit of perimeter is the same as the unit of length i.e. centimetre (cm), metre (m), kilometre (km) etc.

1 centimetre (cm) = 10 milimetres (mm)

1 decimetre (dm) = 10 centimetres

1 metre (m) = 10 decimetres

= 100 centimetres

= 1000 milimetres

1 decametre (dam) = 10 metres

= 1000 centimetres

1 hectometre (hm) = 10 decametres

= 100 metres

1 kilometre (km) = 1000 metres

= 100 decametres

= 10 hectometres

**Area:**

The area of a plane figure is the measure of the surface enclosed by its boundary.

The area of a triangle or a polygon is the measure of the surface enclosed by its sides.

→ Units of Area:

The various units of measuring area are, square centimetre (cm^{2}), square metre (m^{2}), hectare etc.

1 square centrimetre (cm^{2})

= 1 cm × 1 cm

= 10 mm × 10 mm

= 100 sq.mm.

1 square decimetre (dm^{2})

= 1 dm × dm

= 10 cm × 10 cm

= 100 sq. cm.

1 square metre (m^{2})

= 1 m × 1 m

= 10 dm × 10 dm

= 100 sq. dm

1 square decametre (dam^{2})

= 1 dam × 1 dam

= 10 m × 10 m

= 100 sq.m.

1 square hectometre (hm^{2})

= 10 dam × 10 dam

= 100 sq. dam

(or 1 hectare) = 10000 sq. m.

1 square kilometre (km^{2})

= 1 km × 1 km

= 10 hm × 10 hm

= 100 sq. hm.

→ Heron’s formula:

In ΔABC if sides of triangle BC, CA, and AB are a, b, c respectively then

perimeter = 2s = a + b + c

Area = \(\sqrt{s(s-a)(s-b)(s-c)}\)

→ Perimeter and Area of a Triangle:

Right-angled triangle:

For a right-angled triangle, let b be the base, h be the perpendicular and d be the hypotenuse. Then

(A) Perimeter = b + h + d

(B) Area = \(\frac{1}{2}\)(Base × Height) = \(\frac{1}{2}\)bh

(C) Hypotenuse, d = \(\sqrt{b^2+h^2}\) [Pythagoras theorem]

Isosceles right-angled triangle:

For an isosceles right-angled triangle, let a be the equal sides, then

(A) Hypotenuse = \(\sqrt{a^2+a^2}=\sqrt{2} a\)

(B) Perimeter = 2a + \(\sqrt{2}\)a

(C) Area = \(\frac{1}{2}\)(Base × Height) = \(\frac{1}{2}\)(a × a) = \(\frac{1}{2}\)a^{2}.

Equilateral triangle:

For an equilateral triangle, let each side be a, and the height of the triangle be h, then

(A) ∠A = ∠B = ∠C = 60°

(B) ∠BAD = ∠CAD = 30°

(C) AB = BC = AC = a(say)

(D) BD = DC = \(\frac{\mathrm{a}}{2}\)

(E) \(\left(\frac{\mathrm{a}}{2}\right)^2\) + h^{2} = a^{2} ⇒ h^{2} = \(\frac{3 \mathrm{a}^2}{4}\)

∴ Height (h) = \(\frac{\sqrt{3}}{2} \mathrm{a}\)

(F) Area = \(\frac{1}{2}\)(Base × Height)

= \(\frac{1}{2} \times \mathrm{a} \times \frac{\sqrt{3}}{2} \mathrm{a}=\frac{\sqrt{3}}{4} \mathrm{a}^2\)

(G) Perimeter = a + a + a = 3a.