Question

Let 'P' be the perimeter and 'a' be the length of unequal side of an isosceles triangle and it's area is 'A', then prove that 16A² = a²p(p-2a)

Answer 1

Answer:

a is unequal side

P=2x+a

A=1/2*(a/2)*x

A=ax/4

16(a2x2)/16

so a2x2

from above equation of P substituting value of x

16A²=a²p(p-2a)

proved

Answer 2

Step-by-step explanation:

Let the unequal side be c = a

s = (a+b+c)/2 = (2b + a)/2

2b = 2s-a

b = (2s-a)/2 = (p-a)/2

A = √s(s-a)(s-b)(s-c)

= √(s)(s-b)(s-b)(s - a)

= (s-b) √[s(s-a)]

= [s - (p-a)/2] √[s(s-a)]

= [p/2 - (p-a)/2] √[s(s-a)]

= (a/2) √[p/2(p/2-a)]

= (a/2) √[p/2(p/2 - a)]

2A= a √[p/2(p/2 - a)]

4A^2 = a²[ (p^/2)/4 - ap/2)]

16A^2 = a² p (p-2a)