Solve Herons Formula

We are familiar with the general formula for finding the area of a triangle when the base and height of the triangle is given. For scalene traingle we do not have any area formula. Heron's formula proof is a general formuls used to find area of triangles of all types.

Heron a famous mathematician gave simple formula for finding the area of any triangle based on its three sides. Thus, this formula for area is known as Heron's formula and is stated as below,

If p, q, r, denote the lengths of sides of a `Delta` PQR, then,

Heron's formula of a triangle

Area of `Delta` PQR = `sqrt(s(s-p)(s-q)(s-r))` , where   s = `(p+q+r)/(2) = ("perimeter of triangle")/(2)`

Here, s = semiperimeter of `Delta` PQR.

Heron's formula can used to find area of all types of triangles, quadrilateral, trapezoid, etc.
Solved Examples Using Heron's Formula

Ex 1: Find the area of a triangle whose sides are respectively 150 cm, 120 cm, and 200 cm.

Sol: Given three sides, p = 150 cm, q = 120 cm and r = 200 cm.

Step 1:  s = `(p + q + r)/(2) = (150 + 120 + 200)/(2)`

= `(470)/(2) = 235`

Step 2:  (s-p) = (235 -150) = 85

(s-q) = (235 - 120) = 115

(s -r ) = (235 - 200) = 35

Step 3: Area of triangle = `sqrt(s(s-p)(s-q)(s-r))`

= `sqrt(235xx 85xx 115xx 35)`

= `sqrt(5xx 47 xx 5xx 17 xx 5 xx 23 xx 5 xx 7)`

=  25`sqrt(47 xx 17 xx 23 xx 7)`

= 8966.56 cm2

Ex 2: Find the area of a triangle whose two sides are 18 cm and 10 cm and the perimeter is 42 cm.

Sol:  Given p = 18 cm, q = 10 cm and perimeter = 42 cm

Step 1:  perimeter = p + q + r

42 = 18 + 10 + r

42 = 28 + r

subtract 28 on both sides

42 - 28 = 28 - 28 + r

14 = r

Step 2 :  s = `("perimeter)/(2) = (42)/(2)`

s = 21

Step 3:  (s - p) = 21 - 18 = 3

(s - q) = 21 - 10 = 11

(s - r)  = 21 - 14 = 7

Step 4:  Area of triangle = `sqrt(s(s-p)(s-q)(s-r))`

= `sqrt ( 21 xx 3 xx 11 xx 7)`

= `sqrt( 3 xx 7 xx 3 xx 7 xx 11)`

= 3 x 7 `sqrt(11)`

= 21`sqrt(11)` cm2
Practice Problems on Heron's Formula

Pro 1: Find the area of a triangle whose sides are respectively 100 cm, 80 cm, and 60 cm

Ans: 2400 cm2

Pro 2: Find the area of a triangle whose two sides are 12 cm and 8 cm and the perimeter is 30 cm.

Ans: 39.68 cm2

Adding Cosine Functions

Trigonometry is one of the part of mathematics managing with angles, triangles and trigonometry functions such as sine, cosine, and tangent which are abbreviated as sin, cos, and tan respectively. There are several formulas for adding trigonometric functions such as sine, cosine, and tangents. The formulas for adding cosine functions are used to find the ratios when two angles are given. Those formulas for adding cosine functions are shown below.


cosine (A + B) = cosine A cosine B − sine A sine B.
Proof for Adding Cosine Functions:

To prove: cosine (A + B) = cosine A cosine B − sine A sine B.

Proof: We know that, cosine (A − B) = cosine A cosine B + sine A sine B.

Replacing B with (−B), these identities becomes,

sine (A − (− B)) = cosine A cosine (− B) + sine A sine (− B)

Since we know that, cosine (−A) = cosine A, and

sine (−A) = − sine A.

cosine (A + B) = cosine A cosine B − sine A sine B.

Hence proved that, sine (A – B) = sine A cosine B – cosine A sine B.
Solved Examples for Adding Cosine Functions:

Example 1: If sineA = 14, sineB = 6, cosineA = 12, and cosineB = 8, then find cosine(A + B).

Solution: Given that, sineA = 14, sineB = 6, cosineA = 12, and cosineB = 8.

We know that, cosine (A + B) = cosineA cosineB − sineA sineB,

=> cosine (A + B) = (12 × 8) − (14 × 6),

=> cosine (A + B) = 96 − 84,

=> cosine (A + B) = 12.

Answer: cosine (A + B) = 12.

Example 2: If sineA = 5, sineB = 9, cosineA = 17, and cosineB = 3,  then find cosine(A + B).

Solution: Given that, sineA = 5, sineB = 9, cosineA = 17, and cosineB = 3.

We know that, cosine (A + B) = cosineA cosineB − sineA sineB,

=> cosine (A + B) = (17 × 3) − (5 × 9),

=> cosine (A + B) = 51 − 45,

=> cosine (A + B) = 6.

Answer: cosine (A + B) = 6.

Rules for Division

In mathematics, especially in elementary arithmetic, division (÷) is the arithmetic operation that is the inverse of multiplication.

Understanding Rules of Divisibility is always challenging for me but thanks to all math help websites to help me out

Specifically, if c times b equals a, written:

C = b x a

where b is not zero, then a divided by b equals c, written:

a/b = c

For instance,

6/3 = 2

since

2 x 3 = 6

In the above expression, a is called the dividend, b the divisor and c the quotient.
Basic Rules for Division:

The basic rules for division are given as below:

Rule 1:

Division of 2 numbers with the same sign should be ‘+’ ve (positive sign) sign.

‘+’ve (Positive number) ÷’+’ve ( positive number) = ‘+’ve ( positive number)
‘-‘ve (Negative number) ÷ ‘-‘ ve (negative number) = =’+’ ve (positive number)

Rule 2:

Division of 2 numbers with different signs should be ‘-’ve negative

‘+’ve (Positive number) ÷(‘-‘ve negative number) = ‘-‘ve (negative number)
‘-‘ve (Negative number) ÷ ‘+’ve (positive number) = ‘-‘ve (negative number)

Examples for division:

Examples:

54 ÷ 9 = 6 (same signs)

(-32) ÷ (-2) = 16 (same signs)

10 ÷ (-2) = -5 (different signs)

(-12) ÷ 12 = -1 (different signs)


Rules for Division on Dividing Variable:

In this operation division represents dividing the variables based on the presence of values.

Example:

Solve the following division: 49 ÷ p, given that p = -7

Solution:

49 ÷ p

Substitute p = -7,

= 49 ÷ (-7)

= -7 (dissimilar signs).
Division on Dividing Decimal on Rules for Division:

Rules for decimal division solving problems,

Rule 1:

To create the decimal divisor as whole number by changing the decimal point to the right side.

Rule 2:

To change the same decimal point in the dividend to the right side to create as whole number

Rule 3:

After divide the new dividend or whole number by new divisor or whole number


Example:

Divide the following decimal function:  24.24 ÷ 0.20

Solution:

24.24 ÷ 0.20 = 24.24 / 0.20

= 242.4 / 2 (Take the decimal divisor as whole number)

= 2424 / 20 (Take the decimal dividend as whole number)

= 121.2 (Divide the new dividend by new divisor).