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Let a , b and c are the sides of traingle in which a < b< c
P is the perimeter of ∆ .and A is the area of ∆
A/C to question,
P = 5a
we know,
perimeter of ∆ = sum of all sides length
P = a + b + c
a + b + c = 5a
b + c = 4a ------(1)
again,
A = 15a
area of right angled ∆ = 1/2 × base × perpendicular
A = 1/2 × a × b
so,
1/2ab = 15a
b = 30 -----(2)
we also know,
according to Pythagoras theorem
a² + b² = c²
from eqn (1)
a² + b² = (4a - b)²
from eqn (2)
a² + (30)² = (4a - 30)²
a² + 900 = 16a² + 900 -240a
15a² - 240a = 0
a = 0 and 16
but a ≠ 0 so, a = 16
now , put a and b value in eqn (1)
b + c = 4a
30 + c = 64
c = 34
hence, a = 16 , b = 30 and c = 34
P is the perimeter of ∆ .and A is the area of ∆
A/C to question,
P = 5a
we know,
perimeter of ∆ = sum of all sides length
P = a + b + c
a + b + c = 5a
b + c = 4a ------(1)
again,
A = 15a
area of right angled ∆ = 1/2 × base × perpendicular
A = 1/2 × a × b
so,
1/2ab = 15a
b = 30 -----(2)
we also know,
according to Pythagoras theorem
a² + b² = c²
from eqn (1)
a² + b² = (4a - b)²
from eqn (2)
a² + (30)² = (4a - 30)²
a² + 900 = 16a² + 900 -240a
15a² - 240a = 0
a = 0 and 16
but a ≠ 0 so, a = 16
now , put a and b value in eqn (1)
b + c = 4a
30 + c = 64
c = 34
hence, a = 16 , b = 30 and c = 34
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