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.
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.
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:
. . $\displaystyle OA = OB = OC = OD = OF = 0.6$
The trough has water to a height of: .$\displaystyle EF = 0.3 \quad\Rightarrow\quad OE = 0.3$
Right triangle $\displaystyle OED$ has: .$\displaystyle OE = 0.3,\;OD = 0.6$
. . $\displaystyle \cos(\angle DOE) \:=\:\dfrac{0.3}{0.6} \:=\:0.5 \quad\Rightarrow\quad \angle DOE = 60^o = \dfrac{\pi}{3}$
Hence: .$\displaystyle \angle COD \:=\:\dfrac{2\pi}{3}$
Length of arc: .$\displaystyle s \:=\:r\theta$
We have: .$\displaystyle \text{arc(CFD)} \;=\;(0.6)\left(\dfrac{2\pi}{3}\right) \:=\:0.4\pi$
$\displaystyle \text{Therefore: }\;\text{Area of side} \;=\;(0.4\pi) \times \underbrace{\text{(length of trough)}}_{\text{not given}} $
Hi Soroban
I found a way to solve this question!!
I shall use your code to explain
Find the area of triangle AOC,ODB and OCD.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 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!!!!
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:
$\displaystyle \angle COD = \frac {2 \pi}{3} = 120^{\circ}$Code:: 0.6 O 0.6 : A o ------------*-------------o B * * | * * * |0.3 * 0.6 * * E| * * C o-----------o-----------o D * |0.3 * * | * * o * F
$\displaystyle \text{area of sector COD} = \pi \cdot 0.6^2 \cdot \frac{120}{360} \approx 0.37699m^2$
$\displaystyle \text{area of triangle COD} = \frac{1}{2} \cdot 0.6 \cdot 0.6 \cdot \sin 120^{\circ} \approx 0.15588m^2$
$\displaystyle \text{Area in contact with water} = 0.37699m^2 - 0.15588m^2 = 0.2211m^2$
Did I do anything wrong?