A workman is told to make a triangular enclosure of sides $50$, $41$ and $21$ m respectively. Having made the first side one meter too long, what length must he make the other two sides in order to enclose the prescribed area with the prescribed length of fencing?
Actual lengths of sides: $\;$ $a = 50 \; m$, $\;$ $b = 41 \; m$, $\;$ $c = 21 \; m$
Actual perimeter $= a + b + c = 50 + 41 + 21 = 112 \; m$
Actual semi-perimeter $= s = \dfrac{a + b + c}{2} = 56 \; m$ $\;\;\; \cdots \; (1a)$
Actual area $= \Delta = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$
i.e. $\;$ $\Delta = \sqrt{56 \left(56 - 50\right) \left(56 - 41\right) \left(56 - 21\right)} = 420 \; m^2$ $\;\;\; \cdots \; (1b)$
Incorrect length $= a_1 = 51 \; m$
Let the remaining lengths be $b_1$ and $c_1$
New semi-perimeter $= s_1 = \dfrac{a_1 + b_1 + c_1}{2} = \dfrac{51 + b_1 + c_1}{2}$ $\;\;\; \cdots \; (2a)$
Given: $\;$ $s_1 = s$
$\therefore \;$ We have from equations $(1a)$ and $(2a)$
$\dfrac{51 + b_1 + c_1}{2} = 56$ $\implies$ $b_1 + c_1 = 61$ $\;\;\; \cdots \; (3)$
New Area $= \Delta_1 = \sqrt{s_1 \left(s_1 - a_1\right) \left(s_1 - b_1\right) \left(s_1 - c_1\right)}$
i.e. $\;$ $\Delta_1 = \sqrt{s \left(s - a_1\right) \left(s - b_1\right) \left(s - c_1\right)}$ $\;\;\;$ $\left[\because \; s_1 = s\right]$
Substituting the values of $s$ and $a_1$ we have,
$\Delta_1 = \sqrt{56 \left(56 - 51\right) \left(56 - b_1\right) \left(56 - c_1\right)}$
i.e. $\;$ $\Delta_1 = \sqrt{56 \times 5 \left(56 - b_1\right) \left(56 - c_1\right)}$ $\;\;\; \cdots \; (2b)$
Also given: $\;$ $\Delta_1 = \Delta$
$\therefore \;$ We have from equations $(1b)$ and $(2b)$
$\sqrt{56 \times 5 \left(56 - b_1\right) \left(56 - c_1\right)} = 420$
i.e. $\;$ $\sqrt{70 \left(56 - b_1\right) \left(56 - c_1\right)} = 210$
i.e. $\;$ $70 \left(56 - b_1\right) \left(56 - c_1\right) = \left(210\right)^2$
i.e. $\;$ $\left(56 - b_1\right) \left(56 - c_1\right) = 630$
i.e. $\;$ $3136 - 56 \left(b_1 + c_1\right) + b_1 c_1 = 630$
Substituting the value of $\left(b_1 + c_1\right)$ from equation $(3)$ we have,
$b_1 c_1 = 630 - 3136 + \left(56 \times 61\right) = 910$ $\;\;\; \cdots \; (4)$
Now, $\;$ $\left(b_1 - c_1\right)^2 = \left(b_1 + c_1\right)^2 - 4 b_1 c_1$ $\;\;\; \cdots \; (5)$
In view of equations $(3)$ and $(4)$, equation $(5)$ becomes
$\left(b_1 - c_1\right)^2 = \left(61\right)^2 - \left(4 \times 910\right) = 81$
$\implies$ $b_1 - c_1 = 9$ $\;\;\; \cdots \; (6)$
Adding equations $(5)$ and $(6)$ we have,
$2b_1 = 70$ $\implies$ $b_1 = 35$
Substituting the value of $b_1$ in equation $(3)$ gives, $c_1 = 61 - 35 = 26$
$\therefore \;$ The new lengths are $35 \; m$ and $26 \; m$