Find the lengths of the sides of a triangle

[anemone]

Find the lengths of the sides of a triangle with 14, 22 and 28 as the lengths of its altitude.

 

[kaliprasad]

Find the lengths of the sides of a triangle with 14, 22 and 28 as the lengths of its altitude.

Spoiler

let the lengths opposite to altitudes of lengths 14,22,28 be x,y,z
then 2 times area
$14x = 22 y = 28 z$
or $x:: y:: z = \frac{1}{14}::\frac{1}{22}:: \frac{1}{28} = 44::28:: 22$(note all kept even to avoid fraction in computation )
so let x= 44t, y = 28t, z = 22t
area = $\dfrac{14x}{2}= 7x = 7 * 44 t = 308t$
using heros formula ( we get $s= \frac{44+28+22}{2}=47$)
$area = \sqrt{47*(47-44)*(47-28)*(47-22)}t^2 = \sqrt{47* 3 * 19* 25}t^2$
or $area = 5\sqrt{47*57}= 5\sqrt{2679}t^2$
so $308t = 5 \sqrt{2679}t^2$
or $t=\frac{308}{5\sqrt{2679}}$
so the sides are
$44t,28t,22t$ where t is as above

 

[anemone]

Spoiler

let the lengths opposite to altitudes of lengths 14,22,28 be x,y,z
then 2 times area
$14x = 22 y = 28 z$
or $x:: y:: z = \frac{1}{14}::\frac{1}{22}:: \frac{1}{28} = 44::28:: 22$(note all kept even to avoid fraction in computation )
so let x= 44t, y = 28t, z = 22t
area = $\dfrac{14x}{2}= 7x = 7 * 44 t = 308t$
using heros formula ( we get $s= \frac{44+28+22}{2}=47$)
$area = \sqrt{47*(47-44)*(47-28)*(47-22)}t^2 = \sqrt{47* 3 * 19* 25}t^2$
or $area = 5\sqrt{47*57}= 5\sqrt{2679}t^2$
so $308t = 5 \sqrt{2679}t^2$
or $t=\frac{308}{5\sqrt{2679}}$
so the sides are
$44t,28t,22t$ where t is as above

Well done, kaliprasad!

 

[loquetedigo]

Spoiler

a = 52.3657142876
b = 33.3236363648
c = 26.1828571438

From Wikipedia:
"Next, denoting the altitudes from sides a, b, and c respectively as ha, hb, and hc, and denoting the semi-sum of the reciprocals of the altitudes as H = (h_a^{-1} + h_b^{-1} + h_c^{-1})/2 we have[11]
A^{-1} = 4 \sqrt{H(H-h_a^{-1})(H-h_b^{-1})(H-h_c^{-1})}."