Question

The area of a triangle is $57sq.cm$. If the two sides are $8cm$ and $19cm$ respectively, then what is the third side approximately?

Answer 1

The area of triangle $=57sq.cm$

Let the third side is $a$.

Then, $S=\frac{a+8+19}{2}=\frac{27+a}{2}$

Therefore, area of the triangle,

$=\sqrt{S(S-a)(S-b)(S-C)}$

$\sqrt{\left(\frac{27+a}{2}\right)(\frac{27+a}{2}-a)(\frac{27+a}{2}-8)(\frac{27+a}{2}-19)}=57$

$\sqrt{\left(\frac{27+a}{2}\right)\left(\frac{27-a}{2}\right)\left(\frac{11+a}{2}\right)\left(\frac{a-11}{2}\right)}=57$

$\sqrt{\left(\frac{{27}^2-a^2}{4}\right)\left(\frac{a^2-{11}^2}{4}\right)}=57$

$\left(\frac{729-a^2}{4}\right)\left(\frac{a^2-121}{4}\right)=3249$

$(729-a^2)(a^2-121)=51984$

$729a^2-88209-a^4-121a^2=51984$

$a^4-850a^2+140193=0$

Put $x=a^2$

$x^2-850x+140193=0$

Therefore,

$x=\frac{850\pm \sqrt{{850}^2-4\times 1\times 140193}}{2}$

$x=\frac{850\pm \sqrt{722500-560772}}{2}$

$x=\frac{850\pm \sqrt{161728}}{2}$

$x=\frac{850\pm 402}{2}$

Consider the positive sign,

$x=\frac{850+402}{2}$

$x=626$

Therefore,

$a=25cm$

Consider the negative sign,

$x=\frac{850-402}{2}$

$x=224$

Therefore,

$a=15cm$

So, the value of $a$ is $15cm$ or $25cm$.