Question

the difference between the sides at right angled triangle is 14cm . The area of  the triangle is 120 cm² .  calculate the perimeter of the triangle

Answer 1

the ans is 80 cm while sides are 30 , 16 and 34 

Answer 2

let a right triangle ΔABC,in which angle B =90
AB = x cm, BC = y cm
a/q, 
       x-y = 14 ----------(1)
and, 
       $Area = \frac{1}{2}(BC)(AB)$
$120 = \frac{1}{2} (x)(y)$
$2xy = 480 --------(2)$
now, we know that , 
                  $(x-y)^{2} = x^{2} + y^{2} -2xy$
                    [tex] 14^{2} = x^{2} + y^{2} -480 [/tex]
                    [tex] x^{2} + y^{2} = 196+480 = 676 ----------(3) [/tex]
           $(x+y)^{2} = x^{2} + y^{2} +2xy$
              $(x+y)^{2} = 676 + 480 = 1156$
              x+y = √1156 = 34   ---------------(4)
From (1) & (4)
  2x = 48   therefore x= 24cm
from (1)
   y = x-14 = 24-14 = 10cm
therefore AB =24 cm,BC = 10cm
$Ac = \sqrt{ x^{2} + y^{2} }$
$AC = \sqrt{ 24^{2}+ 10^{2} } = \sqrt{676} = 26 cm$
therefore perimetre = AB+AC +BC = 24+26+10 =60 cm