# What is the area of the ABCD rectangle? [closed]

$$(1) \quad \dfrac{256}{17}$$

$$(2) \quad \dfrac{39}{4}$$

$$(3) \quad \dfrac{483}{8}$$

$$(4) \quad \dfrac{52}{4}$$

$$(5) \quad \dfrac{492}{18}$$

• I don't think there is enough information to know. The triangle could be drawn at any angle, giving different bounding box rectangles. – Jaap Scherphuis Oct 22 at 9:57
• The triangle is not unique. However, the area of rectangle is bounded between $12$ and $16$. So only the first or fourth answer is possible. – achille hui Oct 22 at 11:25
• Interesting problem, but if you want your question to remain open, please see how to ask a good question and edit your work accordingly. It's not just 'include your work' on there. – Toby Mak Oct 22 at 11:28

A unique triangle cannot be constructed so no area can be associated with it.

Shown here are two cases of rectangles (red and blue borders). The circles have radii $$(5,4)$$ for side lengths and a side of tangent length $$3$$ units between them.

EDIT1:

Area is not invariant. To calculate when hypotenuse makes an angle $$u$$ to vertical,

$$(3 \cos u + 4 \sin u)(3 \sin u + +4 \cos u)=(12+12.5 \sin2u)$$

It has maximum value 24.5 when hypotenuse makes $$45^{\circ}$$ to vertical, and minimum 12 at time of vertical/horizontal positions of the shorter sides.

• True. The next question would be: is the area an invariant? – cvanaret Oct 22 at 11:12
• Please ask that in another question. – Narasimham Oct 22 at 11:14
• It's not my post, I'm just reacting. – cvanaret Oct 22 at 11:16
• The rigid yellow triangle triangle has constant area 6 sq. units..but let me formulate – Narasimham Oct 22 at 11:18
• No. when hypotenuse makes $45^0$ to vertical,the area is maximum at 24.5; and minimum 12 at time of vertical/horizontal positions of the shorter sides. Generally the area is $(12 +12.5 \sin 2u)$. – Narasimham Oct 22 at 11:27

Let $$AE=a$$, $$BE=b$$, $$AF=c$$, $$DF=d$$.

Now

\begin{align} a^2+c^2 &= 3^2 \tag{1} \\ d^2+(a+b)^2 &= 4^2 \tag{2} \\ b^2+(c+d)^2 &= 5^2 \tag{3} \\ a^2+c^2+d^2+(a+b)^2 &= b^2+(c+d)^2 \tag{3^2+4^2=5^2} \\ a(a+b) &= cd \tag{4} \\ d^2+\left( \frac{cd}{a} \right)^2 &= 4^2 \\ d^2(a^2+c^2) &= 4^2 a^2 \\ 3^2d^2 &= 4^2a^2 \\ d &= \frac{4a}{3} \\ c &= \sqrt{3^2-a^2} \\ a+b &= \frac{4c}{3} \\ (a+b)(c+d) &= \frac{4}{3} \sqrt{3^2-a^2} \left( \frac{4a}{3}+\sqrt{3^2-a^2} \right) \\ \end{align}

Unless $$a+b=c+d$$, that is a square, you cannot have a proper choice.

For the square case,

$$(a+b)(c+d)=\frac{256}{17}$$

I'm leaving the missing steps for above result as an exercise.

• If you are going to post an answer, it would be better to post an incomplete solution since the OP hasn't shown any work yet. – Toby Mak Oct 22 at 11:18
• This is a problematic problem, it's normal for most people unable to solve it. That's an exceptional case. Also, I leave the special case as an exercise. – Ng Chung Tak Oct 22 at 11:21
• It seems likely that the missing information is that the triangle is inscribed in a square, since this gives one of the choices. – David K Oct 22 at 11:46
• It took me a while to realize you meant "if" when you wrote "unless". – David K Oct 22 at 11:47