As shown in the figure, the vertex of square mnpq is on the edge of triangle ABC. When SC = A and high ad = h satisfy what conditions, square Mn

As shown in the figure, the vertex of square mnpq is on the edge of triangle ABC. When SC = A and high ad = h satisfy what conditions, square Mn

a=h

As shown in the figure, in the triangle ABC, BC = 16, height ad = 8, N and P of its inscribed rectangle mnpq are on BC. M and Q are on AB and AC respectively. Find the area of this rectangle

There's no way to answer. The area of this rectangle is a variable. I can work out the variable formula. Please wait a moment
The answer is Mn * (16-2mn)
When Mn is 4, the largest area is 32

As shown in the figure, in RT △ ABC, M is the midpoint of hypotenuse AB, Mn ⊥ AB, n is on BC, if AB = 10cm, AC = 6cm, then the perimeter of △ BMN is, and the area of △ BMN is

Connect an
⊙ is the midpoint of the hypotenuse AB, Mn ⊥ ab
We can set an = BN = X
In RT △ ABC, BC = √ 10 & # 178; - 6 & # 178; = 8
∴CN=8-X
In RT △ ACN
x²-(8-x)²=6²
The solution is x = 25 / 4
MN=√BN²-BM²=15/4
Then the circumference of △ BMN = 25 / 4 + 15 / 4 + 5 = 15
The area of △ BMN = 1 / 2 * 15 / 4 * 5 = 75 / 8

M is the midpoint of the right side AC of the RT triangle ABC, Mn is perpendicular to the hypotenuse AB, and the perpendicular foot is n
Prove: BC's Square = BN's Square - An's Square

In the RT triangle BCM, BC ^ 2 = BM ^ 2-cm ^ 2
In the RT triangle BNM, BM ^ 2 = BN ^ 2 + Mn ^ 2
In the RT triangle anm, am ^ 2 = Mn ^ 2 + an ^ 2
M is the midpoint of AC, am = cm,
Therefore, BC ^ 2 = BN ^ 2 + Mn ^ 2 - (Mn ^ 2 + an ^ 2) = BN ^ 2-An ^ 2