Question

the difference between the sides containing a right angle in a right angled triangle is 14 cm. the area of the triangle is 120 cm².
calculate the perimeter of a triangle

Answer 1

Let sides are x and (x - 14) cm,

We know,

$area \: of \: right \: angled \: triangle \: = \frac{1}{2} \times product \: of \: sides \: containing \: right \: angle$


Then,


$120 = \frac{1}{2} \times x \times (x - 14) \\ \\ 120 \times 2 = x(x - 14) \\ \\ 240 = {x}^{2} - 14x \\ \\ 0 = {x}^{2} - 14x - 240 \\ \\ 0 = {x}^{2} - (24 - 10)x - 240 \\ \\ 0 = {x}^{2} - 24x + 10x - 240 \\ \\ 0 = x(x - 24) + 10( x - 24) \\ \\ 0 = (x - 24)(x + 10) \\ \\ \\ then \\ x = 24 \: or \: x = - 10 \: \: \: \: \: \: \: \: \: (sides \: cannot \: ne \: negative ) \\ so \: \\ x = 24$



Now, sides are :

x = 24
x - 14 = 24 - 14 = 10 cm



××××××××××××××××××××××××


By Pythagoras theorem,


${24}^{2} + {10}^{2} = hypotenuse ^{2} \\ \\ 576 + 100 = hypotenuse^{2} \\ \\ \sqrt{676} = hypotenuse \\ \\ 26 \: cm = hypotenuse$




Perimeter = sum of all sides

Here



Perimeter = 24 + 10 + 26

Perimeter = 60 cm





I hope this will help you


(-:

Answer 2

Answer:

let one side at right angle be x

the other must be x-14

area of a triangle =(1/2) × (height) (base)

120 = (1/2)(x)(x-14)

120 = (x^2 -14x)/2

240 = x^2 -14x

x^2 -14x -240 = 0

x^2 -24x +10x -240 =0

x (x-24) + 10 (x-24) =0

(x-24)(x+10)=0

now, find x

x-24 =0

x = 24

again

x+10 =0

x = -10

as the side couldn't be -v so the value of x as a side must be 24cm

so the measure of one side we got is x =24cm and the other side = x-14 = 24 -14 =10cm

apply pythagoras theorem for third side as it is a right angled triangle

[let hypotaneous be a]

(24)^2 + (10^2) = a^2

576 +100 = a^2

676 = a^2

a = 26 cm

so the hypataneous = 26cm

perimeter of a triangle = sum of all sides

= (24 + 10 + 26 ) cm

= 60cm

have a great day