The difference between the two sides adjacent to the right angle is 14 cm in a right triangle and the area is 120 cm² find the perimeter.
Answers
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
Step-by-step explanation:
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Answer:nswer:
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
Step-by-step explanation: