Question

if the difference between sides of a right angle triangle is 3 cm and area is 54 cm find its perimeter by heron's formula​

Answer 1

Step-by-step explanation:

let that triangle =ABC

GIVEN THAT:

DIFFERENCE BETWEEN 2 SIDES OF TRIANGLE =3CM

And area of triangle =54cm²

solution:

let that 2 sides =AB and BC ( as here area is given and to find area we use height and base)

let value of BC=x

and value of AB=x+3

AREA OF TRIANGLE=

$\frac{1}{2} \times base \times height$

=ATQ

$\frac{1}{2} \times x \times( x + 3) = 54 {cm}^{2} \\ \frac{1}{2} \times {x}^{2} \times 3x = 54 {cm}^{2} \\ {x}^{2} \times 3x = 54 \times 2 {cm}^{2} \\ {x}^{2} \times 3x = 108 {cm}^{2} \\ {x}^{2} \times 3x - 108 = 0$

Find two factors of  -108  whose sum equals the coefficient of the middle term,which is 3

$-108   +   1   =   -107     \\  -54   +   2   =   -52    \\   -36   +   3   =    -33    \\   -27   +   4   =   -23   \\    -18   +   6   =   -12   \\    -12   +   9   =   -3     \\  -9   +   12   =   3   \\ = {x}^{2} - 9x + 12x - 108 = 0$

$= x(x - 9) + 12(x - 9) = 0 \\ = (x + 12)(x - 9) = 0$

here there are 2 no. (x+12) and (x-9) whose product is 0

it means any one of these no. is 0

either( x+12) is 0 or( x-9) is 0 both can't be 0 because there will be some value of all side of triangle

let's solve it separately to find which term is 0

x+12=0

x=-12

x-9=0

x=9

since value of side of triangle cannot have negative value the real value of x=9

BC=9

AB=9+3=12

hypotenious =

${ac}^{2} = {9}^{2} + {12}^{2} \\ ac = \sqrt{81 + 144} \\ a c = 15$

PERIMETER OF TRIANGLE =9+12+15=36cm

Answer 2

Answer:

sry for late

and thanku for points