Question

the difference between the sides at right angles in a right angled triangle is 14 cm .the area of a triangle is 120 cm^2 calculate the perimeter of the triangle

Answer 1

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

be brainly ,

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

★Assumption :-

Sides of right angle be p and (p - 4)

★We know that :-

$\tt{\rightarrow Area\;of\;Triangle=\dfrac{1}{2}\times p\times(p-14)}$

★Given - Area = 120 cm²

$\tt{\rightarrow\dfrac{1}{2}\times p(p-14)=120}$

= p² - 14p - 240 = 0

★Splitting the middle term :-

p² - 24p + 10p - 240 = 0

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

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

(p - 24) = 0

p = 24

(p + 10) = 0

p = -10 (Not applicable)

★Therefore :-

One side = 24 cm

★Other side :-

= 24 - 14

= 10 cm

$\tt{\rightarrow Hypotenuse=\sqrt{(24)^2+(10)^2}}$

$\tt{\rightarrow Hypotenuse=\sqrt{576+100}}$

$\tt{\rightarrow Hypotenuse=\sqrt{676}}$

= 26 cm

★Perimeter of Triangle :-

= 24 + 10 + 26

= 60 cm