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I recently came across a puzzling question:

Two rectangles ABCD and DBEF are as shown in the figure. The area of DBEF is:
Figure (hand-made): Figure

I know that through Pythagoras, we get DB = 5 cm.

Now I thought that through Heron's formula, I can make an equation:

$$\frac{CG * DB}{2} = \sqrt{s(s - CD)(s - CB)(s - DB)}$$

And get CG, which is equal to EB and EB * FE(5) would give the answer. But I got the answer in decimals, which is far from from the expected answer.

Can anyone please help me.


Funny! I tried again and solved it !! Thanks to both the answers. Upvoted.

Thanks a lot.

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Let intersection on $BD$ be $G$. Then $BCG=BCE$ and $CDG=CDF$ (equal as in congruent), since they are right angle triangles with same height and base. Therefore area of $DBEF$ is 2 times of triangle $BCD$, but that is the same as $ABCD$. So area of $ABCD=DBEF=3\cdot 4=12$.

We know that $BCG=BCE$, since they are right angle triangles with equal base and height.
Area of $BECG=BEC+BCG=2BCG$.

Similarly, we have $CDG=CDF$; they are also right angle triangles with equal base and height. Now area of $DFCG=DCG+DCF=2DCG$.

But triangles $BCD=ABD$, so we see that area of $$ABCD=BCD+ABD=BCD+BCD=2BCD.$$

This shows that area of $ABCD=2BCD=BEDF$.

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Indeed this seems a new way of solving. Nice! But can you please explain your last line after 'therefore'. I am having trouble in understanding how area of DBEF comes twice of BCD. Thanks a lot. – Gaurang Tandon Jan 2 '14 at 16:09
@GaurangTandon I have added an explanation. The point is to be able to show (independently) that area $ABCD=BCD+BCD=2BCD$ and $BEDF=BCD+(BEC+CDF)=BCD+BCD=2BCD$. – Yong Hao Ng Jan 2 '14 at 16:22
Very very thank you for showing a new approach :D Upvoted. – Gaurang Tandon Jan 2 '14 at 16:28
@GaurangTandon You're welcome. =D – Yong Hao Ng Jan 2 '14 at 16:34

Letting $DF=x, FC=y$, we have $CE=5-y.$

Considering two triangles $DFC, CEB$, we have $$4=\sqrt{x^2+y^2},\ \ \ \ 3=\sqrt{x^2+(5-y)^2}.$$

So, you can get $x,y$. The answer is $5x.$

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Thank You. See question edit. – Gaurang Tandon Jan 2 '14 at 15:58
You are welcome! and Good job! – mathlove Jan 2 '14 at 15:59

Strategy sounds reasonable. You get to solve $$ 2.5 CG = \sqrt{6 \cdot 3 \cdot 2 \cdot 1} $$ so $CG = 2.4$ and the area of the second one is $2.4 \times 5 = 12$... It is very intuititve that the areas of both rectangles are the same...

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Thank You. See question edit. – Gaurang Tandon Jan 2 '14 at 15:59

This is a simple trigonometric problem. Referring to the figure below:- enter image description here

In ⊿DAB, sin θ = 3 / 5

In ⊿DGC, sin θ = y / 4

∴ y = 3 * 4 / 5

Area of rectangle DBEF = y * 5 = … = 3 * 4 = 12

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Really nice! Never thought we could do like this! Thanks :) – Gaurang Tandon Feb 1 '14 at 15:04
you are welcome. – Mick Feb 1 '14 at 18:40
@GaurangTandon Maybe you are interested to know that even if ABCD and DBEF are not rectangles but parallelograms, they still equal in area. The proof of it requires the use of the geometric theorem - [the areas of 2 triangles are the same if their altitudes are the same and their bases are the same]. – Mick Feb 2 '14 at 18:13
IIRC, both rectangles and parallelograms have the area $b*h$ (and $l * b$), but never had thought of proving it. So, I will definitely try this stuff today :) – Gaurang Tandon Feb 3 '14 at 11:00
@GaurangTandon In case you want to know, I have the mentioned problem answered in #664374. – Mick Feb 6 '14 at 5:00

area of rectangle $DBEF = 2 \times \text{area of triangle }BCD$ (they are on the same base and between the same parallels)

area of rectangle $ABCD = 2\times\text{ area of triangle }BCD$ (a diagonal divides a parallel into $2$ congruent triangles)

$$\Rightarrow\text{ area of }DBEF = \text{area of }ABCD = 3\times 4 = 12.$$

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