Question

Maggie puts together two isosceles triangles so that they share a base, creating a kite. Each leg of the upper triangle measures 41 inches and each leg of the lower one measures 50 inches.
If the length of the base of both triangles measures 80 inches, what is the length of the kite’s shorter diagonal?

30 inches
39 inches
inches
inches

Answer 1

Answer: Shortest diagonal = 39 inches

Step-by-step explanation:

We consider ΔAOD,  from the figure mentioned below:

To get the diagonal we use the pythagorus theorem,

$AO^2+OD^2=AD^2\\\\AO^2+40^2=41^2\\\\AO^2+1600=1681\\\\AO^2=1681-1600\\\\AO^2=81\\\\AO=\sqrt{81}\\\\AO=9$

Similarly,

We consider ΔCOD,

By Pythagorus theorem,

$CD^2=CO^2+DO^2\\\\50^2=CO^2+40^2\\\\2500=CO^2+1600\\\\2500-1600=CO^2\\\\900=CO^2\\\\\sqrt{900}=CO\\30=CO$

So, by adding both two i.e.

$\text{Shortest diagonal of kite }=AC=AO+CO=9+30=39\text{ inches }$

Answer 2

(See attached)
AO^2 = side a^2 - BO^2
AO^2 = 41^2 -40^2
AO^2 = 1,681 -1600
AO^2 = 81
AO = 9

DO^2 = side b^2 -BO^2
DO^2 = 50^2 -40^2
DO^2 = 2,500 -1,600
DO^2 = 900
DO = 30

Diagonal AD = AO + DO = 9 + 30 = 39 inches

Source:
http://www.1728.org/quadkite.htm