Answer:
The length of the perpendicular = 20 meters
The length of the base = 48 meters
Step-by-step explanation:
The hypotenuse of the triangle = 52 meters
Let the Length of the perpendicular is = k meters
So, the length of the base = ( k + 28) m
Now, by PYTHAGORAS THEOREM , in a right angled triangle:
![(BASE)^{2} + (PERPENDICULAR)^{2} = (HYPOTENUSE)^{2}](https://tex.z-dn.net/?f=%28BASE%29%5E%7B2%7D%20%20%20%2B%20%28PERPENDICULAR%29%5E%7B2%7D%20%20%3D%20%20%28HYPOTENUSE%29%5E%7B2%7D)
⇒ Here, ![(k)^{2} + (k +28) ^{2} = (52)^{2}](https://tex.z-dn.net/?f=%28k%29%5E%7B2%7D%20%2B%20%28k%20%2B28%29%20%5E%7B2%7D%20%20%3D%20%2852%29%5E%7B2%7D)
Also, by Algebraic Identity:
![(a+b) ^{2} = a^{2} + b ^{2} + 2ab\\ \implies (k+28) ^{2} = k^{2} + (28) ^{2} + 2(28)(k)\\](https://tex.z-dn.net/?f=%28a%2Bb%29%20%5E%7B2%7D%20%20%3D%20a%5E%7B2%7D%20%2B%20b%20%5E%7B2%7D%20%2B%202ab%5C%5C%20%5Cimplies%20%28k%2B28%29%20%5E%7B2%7D%20%20%3D%20k%5E%7B2%7D%20%2B%20%2828%29%20%5E%7B2%7D%20%2B%202%2828%29%28k%29%5C%5C)
So, the equation becomes:
![(k)^{2} +k^{2} + (28) ^{2} + 2(28)(k) = (52)^{2}](https://tex.z-dn.net/?f=%28k%29%5E%7B2%7D%20%2Bk%5E%7B2%7D%20%2B%20%2828%29%20%5E%7B2%7D%20%2B%202%2828%29%28k%29%20%20%3D%20%2852%29%5E%7B2%7D)
or, ![2k^{2} + 784+ 56k = 2704\\\implies k^{2} + 28k - 960 = 0](https://tex.z-dn.net/?f=2k%5E%7B2%7D%20%20%2B%20784%2B%2056k%20%3D%202704%5C%5C%5Cimplies%20k%5E%7B2%7D%20%2B%2028k%20-%20960%20%3D%200)
or,![k^{2} + 48k -20 k - 960 = 0](https://tex.z-dn.net/?f=k%5E%7B2%7D%20%20%2B%2048k%20-20%20k%20-%20960%20%3D%200)
Solving the equation:
⇒ (k+48)(k-20) = 0 , or (k+48) = 0 , or (k-20) = 0
or, either k = -48 , or k = 20
As k is the length of the side, so k ≠ - 48, k = 20
Hence, the length of the perpendicular = k = 20 meters
and the length of the base is k + 28 = 48 meters