Right Triangle Within a Rectangle, Hypotenuse Shares Entire Side of Rectangle and Third Point Intersects Rectangle Opposite Side

Question

I know the Length (L) and Width (W) of a rectangle is 8 and 3.5 meters respectively.

Triangle ABL is a right triangle, therefore angle AB is 90-degrees.

L = C + D = 8

W will fall within this range: 0 < W <= L/2; I'll assume W = 3.5 for this math.

I need to solve for A, B, C and D.

I'm stuck, I don't know why, likely stress and lack of sleep, but I'm stuck. If you can offer help I'd appreciate it. Thanks.

Answer 1

EDIT1:

The semi-circle that contains right triangle cuts the lower side of rectangle, since height is less than radius. Construction below :

By symmetry $(A,B)$ and so $(C,D)$ can be interchanged.

Pythagoras theorem and height of right triangle $$A^2+B^2=L^2\,;\,\dfrac{1}{W^2}=\dfrac{1}{A^2}+\dfrac{1}{B^2} \,;$$

You must commit to memory the second formula if not already done so; (it is often useful in geometrical calculation of right triangles). Solve the quadratic equation getting:

$$ (2 A^2,2 B^2)=L^2\pm \sqrt{L^4-4L^2W^2};\,$$ $$ C^2=A^2-W^2;\, D^2=B^2-W^2;\, $$

Evaluate them for given $(L,W)$ as

$$(A,B,C,D) \approx (6.89144,4.06301,5.93649,2.06351),\,$$ and $$(A,B,C,D) \approx (4.06301,6.89144,2.06351,5.93649).\,$$

Answer 2

Note that

$$A^2+B^2 = L^2,\>\>\>\>\> AB = WL$$

Assume $A>B$. Then,

$$A-B = \sqrt{A^2+B^2 -2AB} = \sqrt{L^2 -2WL} $$ $$A+B = \sqrt{A^2+B^2 + 2AB} = \sqrt{L^2 +2WL} $$

which yield the solutions for $A$ and $B$,

$$A = \frac {1}2 \left(\sqrt{L^2 +2WL}+\sqrt{L^2 -2WL}\right)$$ $$B = \frac {1}2 \left(\sqrt{L^2 +2WL}-\sqrt{L^2 -2WL}\right)$$

Also, note that

$$C+D = L, \>\>\>\>\>CD = W^2$$

So, $C$ and $D$ satisfy the quadratic equation

$$x^2 -Lx+W^2=0$$

which yields,

$$C = \frac12(L+\sqrt{L^2-4W^2}), \>\>\>\>\>\>D = \frac12(L-\sqrt{L^2-4W^2})$$

Answer 3

According to your data you have a right triangle of hypotenusa $8$ and height $5$ so you can calculate the legs $A, B$ from the system $$AB=8\cdot5\space \text{(areas)}\\A^2+B^2=64\space \text{(Pythagoras)}$$

You can verify solving the system it has no real solutions or, easier, plotting the hyperbola $xy=40$ and the circle $x^2+y^2=64$ so you can see there are no common points.

Answer 4

As long as $W \lt \frac L2$ there is a solution. If you draw a semicircle with $L$ the diameter it shows the points where there will be a right angle. Where the semicircle intersects the lower edge of the window is the point you want. $AWC$, $BDW$ and $ABL$ are similar triangles.

I get the equations $$c^2+3.5^2=a^2\\ d^2+3.5^2=b^2\\ a^2+b^2=8^2\\ c+d=8$$ Which Alpha gives a solution $$a≈4.06301, b≈6.89144, c≈2.06351, d≈5.93649$$