<span>A=12h(a+b)
Solve for h.
</span>All these answers are incorrect
<span>A. h=2Aa+b
B.h=a+b2A
C.h=A2(a+b)
D.h=2(a+b)A
The answer is
h = A / 12(a+b)

</span>
Step-by-step explanation:
what is the question?????
Check the picture below.
let's recall that in a Kite, the diagonals meet at 90° angles, therefore, we know the height of each of those 4 triangles, is 2.5 and 6, now, since the pair of triangles above are 45-45-90 triangles, we can use the 45-45-90 rule, as you see there, so, if the height is 2.5, then the base is also 2.5.
so, we really have 2 pair of triangles whose base is 2.5 and height of 2.5, and another pair of triangles whose base is 2.5 and height is 6, let's add their areas.
![\bf \stackrel{\textit{area of 2 triangles above}}{2\left[\cfrac{1}{2}(2.5)(2.5) \right]}~~+~~\stackrel{\textit{area of 2 triangles below}}{2\left[ \cfrac{1}{2}(2.5)(6) \right]}\implies 6.25+15\implies 21.25](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Barea%20of%202%20triangles%20above%7D%7D%7B2%5Cleft%5B%5Ccfrac%7B1%7D%7B2%7D%282.5%29%282.5%29%20%5Cright%5D%7D~~%2B~~%5Cstackrel%7B%5Ctextit%7Barea%20of%202%20triangles%20below%7D%7D%7B2%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%282.5%29%286%29%20%5Cright%5D%7D%5Cimplies%206.25%2B15%5Cimplies%2021.25)