Mensuration question

Question

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):

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.

Edit:

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

Thanks a lot.

Answer 1

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 $5x.$

Answer 2

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...

Answer 3

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 $ABCD=DBEF=3\cdot 4=12$.

Edit_1:
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 $DFCG=DCG+DCF=2DCG$.

Area of $BEDF=BECG+DFCG=2BCG+2DCG=2(BCG+DCG)=2BCD$.
But triangles $BCD=ABD$, so we see that area of $$ABCD=BCD+ABD=BCD+BCD=2BCD.$$

This shows that area of $ABCD=2BCD=BEDF$.
End_Edit_1

Answer 4

This is a simple trigonometric problem. Referring to the figure below:-

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