# Math Help - Find the area in contact with water

1. ## Find the area in contact with water

Diagram IV show the cross-section of the trough after it is filled with water. The cross-section has centre O and diameter 1.2m. From diagram IV, determine the area that is in contact with the water.

2. Hello, Punch!

Some information is missing.

The diagram shows the cross-section of the trough partially filled with water.
The cross-section has centre O and diameter 1.2 m.
Determine the area of the side that is in contact with the water.

Code:
  :     0.6     O     0.6     :
A o ------------*-------------o B
*          *  |  *          *
*     |0.3  * 0.6
*   *       E|        *   *
C o-----------o-----------o D
*         |0.3      *
*      |     *
* o *
F

The cross-section is a semicircle with radius:
. . $OA = OB = OC = OD = OF = 0.6$

The trough has water to a height of: . $EF = 0.3 \quad\Rightarrow\quad OE = 0.3$

Right triangle $OED$ has: . $OE = 0.3,\;OD = 0.6$

. . $\cos(\angle DOE) \:=\:\dfrac{0.3}{0.6} \:=\:0.5 \quad\Rightarrow\quad \angle DOE = 60^o = \dfrac{\pi}{3}$

Hence: . $\angle COD \:=\:\dfrac{2\pi}{3}$

Length of arc: . $s \:=\:r\theta$
We have: . $\text{arc(CFD)} \;=\;(0.6)\left(\dfrac{2\pi}{3}\right) \:=\:0.4\pi$

$\text{Therefore: }\;\text{Area of side} \;=\;(0.4\pi) \times \underbrace{\text{(length of trough)}}_{\text{not given}}$

3. Hi Soroban

I found a way to solve this question!!

I shall use your code to explain
Code:
 :     0.6     O     0.6     :
A o ------------*-------------o B
*          *  |  *          *
*     |0.3  * 0.6
*   *       E|        *   *
C o-----------o-----------o D
*         |0.3      *
*      |     *
* o *
F
Find the area of triangle AOC,ODB and OCD.

The sum of area for the three triangles will give the total area not in contact with water.

Next, find the area of the whole semi circle.

Substract the area of the three triangles from the area of the whole semi circle.

U r done!!!!

4. Originally Posted by Punch
Hi Soroban

I found a way to solve this question!! <=== NO.

I shall use your code to explain
Code:
 :     0.6     O     0.6     :
A o ------------*-------------o B
*          *  |  *          *
*     |0.3  * 0.6
*   *       E|        *   *
C o-----------o-----------o D
*         |0.3      *
*      |     *
* o *
F
Find the area of triangle AOC,ODB and OCD. <=== AOC and ODB are sectors and not triangles

The sum of area for the three triangles will give the total area not in contact with water. <=== nearly (see above) but of the cross-section

Next, find the area of the whole semi circle.

Substract the area of the three triangles from the area of the whole semi circle.

U r done!!!!
According to your consideration you haven't understood your own question. Read Soroban's post again to see where you are mistaken.

5. I do not see what is confusing about this question or what information is missing so please tell me what I did wrong:

Using Sorobans code:
Code:
 :     0.6     O     0.6     :
A o ------------*-------------o B
*          *  |  *          *
*     |0.3  * 0.6
*   *       E|        *   *
C o-----------o-----------o D
*         |0.3      *
*      |     *
* o *
F
$\angle COD = \frac {2 \pi}{3} = 120^{\circ}$

$\text{area of sector COD} = \pi \cdot 0.6^2 \cdot \frac{120}{360} \approx 0.37699m^2$

$\text{area of triangle COD} = \frac{1}{2} \cdot 0.6 \cdot 0.6 \cdot \sin 120^{\circ} \approx 0.15588m^2$

$\text{Area in contact with water} = 0.37699m^2 - 0.15588m^2 = 0.2211m^2$

Did I do anything wrong?

6. Originally Posted by Educated
I do not see what is confusing about this question or what information is missing so please tell me what I did wrong:

Using Sorobans code:
Code:
 :     0.6     O     0.6     :
A o ------------*-------------o B
*          *  |  *          *
*     |0.3  * 0.6
*   *       E|        *   *
C o-----------o-----------o D
*         |0.3      *
*      |     *
* o *
F
$\angle COD = \frac {2 \pi}{3} = 120^{\circ}$

$\text{area of sector COD} = \pi \cdot 0.6^2 \cdot \frac{120}{360} \approx 0.37699m^2$

$\text{area of triangle COD} = \frac{1}{2} \cdot 0.6 \cdot 0.6 \cdot \sin 120^{\circ} \approx 0.15588m^2$

$\text{Area in contact with water} = 0.37699m^2 - 0.15588m^2 = 0.2211m^2$

Did I do anything wrong?
1. You calculated the area of the waterfilled cross-section.

2. According to the text of the question you should calculate the area of the trough (not it's cross-section) which is in contact with the water. (see attachment)

7. Ah, I see. Sorry about that I misunderstood the question, and interpreted it completely wrong.

Too many exam questions that ask for the area of the cross-sections.

EDIT: Maybe it's a spherical trough?

8. Originally Posted by earboth
1. You calculated the area of the waterfilled cross-section.

2. According to the text of the question you should calculate the area of the trough (not it's cross-section) which is in contact with the water. (see attachment)
It was probably a mistake on my part in posting an "edited" question. The question is... determine the area of the side that is in contact with the water.

Educated should be right