Question

Find the area of the triangle formed by the points (0,0),(4,0), (4,3) by using
Heron's formula
what is s-a​

Answer 1

Answer:

21.91 units2

Step-by-step explanation:

Length of the sides -

Suppose, side A is made of (0,0) and (4,0) points. So,

A = ((4-0)^2+(0-0)^2)^0.5

or, A = 4 units

side B is made of (4,0) and (4,3) points. So,

A = ((4-4)^2+(0-3)^2)^0.5

or, B = 3 units

and side C is made of (0,0) and (4,3) points. So,

A = ((4-0)^2+(3-0)^2)^0.5

or, C = 5 units

Now, S = (A+B+C)/2 = (3+4+5)/2 = 8 units

So, according to Heron's formula, area of the triangle is,

(S*(S-A)*(S-B)*(S-C))^0.5 = (8*(8-4)*(8-3)*(8-5))^0.5 = 21.91 units2

Answer 2

$\huge{\underline{\underline{\boxed{\sf{\purple{Answer࿐}}}}}}$

a= distance between point (0,0) and (4,0)

$\pmb{a=\sqrt{(4-0)^2 -(0-0)^2 } =4} \\ \\ \pmb{b=\sqrt{(4-4)^2 -(3-0)^2 }} =3 \\ \\ \pmb{c=\sqrt{(4-0)^2 -(3-0)^2 } =5} \\ \\ \pmb {s = \frac{a+b+c}{2}} \\ \\ \sf{ = \frac{4+3+5}{2}} \\ \\ \sf { = 6s} \\ \\ \sf {= 2a+b+c} \\ \\ \sf{= 24+3+5 =6 }\\ \\ \sf{A =\sqrt{s(s-a)(s-b)(s-c)}} \\ \\ \sf{= \sqrt{6 \times (6-4) \times (6-3) \times (6-5)}} \\ \\ \rm{ \red{=6}}$