₦airaland Forum

Show your maths skills

Please show your workings not just answers

Anyone?

Shapes are getting tougher.

Martinez39:
Shapes are getting tougher.

Lol, then we are getting stronger,

Don't worry by tomorrow I will be post one with difficulty low and another high.
But let's kill this first sir

15.1948232 units².

Let me cross-check, I will post my workings soon.

NB: I don't know basic geometry too well like ifada123 so I used coordinate geometry.

1) Place the diagram on a rectangle coordinate plane such that A is at the origin and AB lies on the positive x-axis. Finally let θ be the angle ∠EAF

2) Using θ, we have the coordinates of all points :
A(0, 0), E(4Cosθ, 4Sinθ), D(7Cosθ, 7Sinθ), C(7Cosθ + 3, 7Sinθ) & B(7Cosθ + 3, 0). The coordinates for F isn't given.

3) Using the aforementioned coordinates, the slope of the line through EC is
m = Sinθ/(1 + Cosθ)
The line through EF is perpendicular to the line through EC and hence has a slope
m' = -(1 + Cosθ)/Sinθ
Using the slope-point form, the equation of the line through EF is
(y - 4Sinθ) = [-(1 + Cosθ)/Sinθ](x - 4Cosθ) ---- (1)
The x-intercept of this line would give you the x-coordinate of F. Making y = 0 in (1), we have x = 4.
Therefore F is the point (4, 0)

4) Using the coordinates of point B and F, we have
(7Cosθ + 3) - 4 = 5.
Therefore Cosθ = 6/7. We have Sinθ = (√13)/7
Using this, we can evaluate the coordinates of points B, C, F and E. Doing this, we have
B(9, 0), C(9, √13), F(4, 0) & E( 24/7, (4√13)/7)

5) The area of the required region is the sum of the areas of EFC and CFB. Using these coordinates together with the distance formula,
EFC = ½(|EF|)(|EC|) = (2√22932)/49
CFB = ½(|CB|)(|BF|) = (5√13)/2
EFC + CFB = ( (5√13)/2 ) + ( (2√22932)/49 ) ≈ 15.1948232 units².

The area is 15.1948232 units².

@ifada123

I am getting busy recently and may not participate in the challenges as frequently. Still keep the shapes coming. Your work is appreciated.

24.5 sq. units.

ifada123:
Show your maths skills

are u a teacher

Martinez39:
NB: I don't know basic geometry too well like ifada123 so I used coordinate geometry.

1) Place the diagram on a rectangle coordinate plane such that A is at the origin and AB lies on the positive x-axis. Finally let θ be the angle ∠EAF

2) Using θ, we have the coordinates of all points :
A(0, 0), E(4Cosθ, 4Sinθ), D(7Cosθ, 7Sinθ), C(7Cosθ + 3, 7Sinθ) & B(7Cosθ + 3, 0). The coordinates for F isn't given.

3) Using the aforementioned coordinates, the slope of the line through EC is
m = Sinθ/(1 + Cosθ)
The line through EF is perpendicular to the line through EC and hence has a slope
m' = -(1 + Cosθ)/Sinθ
Using the slope-point form, the equation of the line through EF is
(y - 4Sinθ) = [-(1 + Cosθ)/Sinθ](x - 4Cosθ) ---- (1)
The x-intercept of this line would give you the x-coordinate of F. Making y = 0 in (1), we have x = 4.
Therefore F is the point (4, 0)

4) Using the coordinates of point B and F, we have
(7Cosθ + 3) - 4 = 5.
Therefore Cosθ = 6/7. We have Sinθ = (√13)/7
Using this, we can evaluate the coordinates of points B, C, F and E. Doing this, we have
B(9, 0), C(9, √13), F(4, 0) & E( 24/7, (4√13)/7)

5) The area of the required region is the sum of the areas of EFC and CFB. Using these coordinates together with the distance formula,
EFC = ½(|EF|)(|EC|) = (2√22932)/49
CFB = ½(|CB|)(|BF|) = (5√13)/2
EFC + CFB = ( (5√13)/2 ) + ( (2√22932)/49 ) ≈ 15.1948232 units².

The area is 15.1948232 units².

Try again sir

shegzy2009:
24.5 sq. units.

Try again sir

Sorry I reply late
Let's keep the workings coming

ifada123:
Try again sir

Nice to have you back. Show us your working. This question don all of us strong thing.

Martinez39:
Nice to have you back. Show us your working. This question don all of us strong thing.

Ok sir
I will prepare solution soon,
If get home.

ifada123:
Try again sir

Chai....This is serious.
Pls show us your solution.

ifada123:
Ok sir
I will prepare solution soon,
If get home.

We are waiting.

Martinez39:
NB: I don't know basic geometry too well like ifada123 so I used coordinate geometry.

1) Place the diagram on a rectangle coordinate plane such that A is at the origin and AB lies on the positive x-axis. Finally let θ be the angle ∠EAF

2) Using θ, we have the coordinates of all points :
A(0, 0), E(4Cosθ, 4Sinθ), D(7Cosθ, 7Sinθ), C(7Cosθ + 3, 7Sinθ) & B(7Cosθ + 3, 0). The coordinates for F isn't given.

3) Using the aforementioned coordinates, the slope of the line through EC is
m = Sinθ/(1 + Cosθ)
The line through EF is perpendicular to the line through EC and hence has a slope
m' = -(1 + Cosθ)/Sinθ
Using the slope-point form, the equation of the line through EF is
(y - 4Sinθ) = [-(1 + Cosθ)/Sinθ](x - 4Cosθ) ---- (1)
The x-intercept of this line would give you the x-coordinate of F. Making y = 0 in (1), we have x = 4.
Therefore F is the point (4, 0)

4) Using the coordinates of point B and F, we have
(7Cosθ + 3) - 4 = 5.
Therefore Cosθ = 6/7. We have Sinθ = (√13)/7
Using this, we can evaluate the coordinates of points B, C, F and E. Doing this, we have
B(9, 0), C(9, √13), F(4, 0) & E( 24/7, (4√13)/7)

5) The area of the required region is the sum of the areas of EFC and CFB. Using these coordinates together with the distance formula,
EFC = ½(|EF|)(|EC|) = (2√22932)/49
CFB = ½(|CB|)(|BF|) = (5√13)/2
EFC + CFB = ( (5√13)/2 ) + ( (2√22932)/49 ) ≈ 15.1948232 units².

The area is 15.1948232 units².

You maybe right
I later found out my solution is inconsistent with geometry theorem.

I think the question requires more details,
I just got the Shape, maybe there are more write up to it.
Sorry for the delay,
Further light gotten will be communicated
To you guys

ifada123:
You maybe right
I later found out my solution is inconsistent with geometry theorem.

Yeah. I was surprised when you said that I was incorrect because I exhaustively checked my workings.

Tumbs up,
Look forward to the next shape,
I am posting now

Martinez39:
NB: I don't know basic geometry too well like ifada123 so I used coordinate geometry.

1) Place the diagram on a rectangle coordinate plane such that A is at the origin and AB lies on the positive x-axis. Finally let θ be the angle ∠EAF

2) Using θ, we have the coordinates of all points :
A(0, 0), E(4Cosθ, 4Sinθ), D(7Cosθ, 7Sinθ), C(7Cosθ + 3, 7Sinθ) & B(7Cosθ + 3, 0). The coordinates for F isn't given.

3) Using the aforementioned coordinates, the slope of the line through EC is
m = Sinθ/(1 + Cosθ)
The line through EF is perpendicular to the line through EC and hence has a slope
m' = -(1 + Cosθ)/Sinθ
Using the slope-point form, the equation of the line through EF is
(y - 4Sinθ) = [-(1 + Cosθ)/Sinθ](x - 4Cosθ) ---- (1)
The x-intercept of this line would give you the x-coordinate of F. Making y = 0 in (1), we have x = 4.
Therefore F is the point (4, 0)

4) Using the coordinates of point B and F, we have
(7Cosθ + 3) - 4 = 5.
Therefore Cosθ = 6/7. We have Sinθ = (√13)/7
Using this, we can evaluate the coordinates of points B, C, F and E. Doing this, we have
B(9, 0), C(9, √13), F(4, 0) & E( 24/7, (4√13)/7)

5) The area of the required region is the sum of the areas of EFC and CFB. Using these coordinates together with the distance formula,
EFC = ½(|EF|)(|EC|) = (2√22932)/49
CFB = ½(|CB|)(|BF|) = (5√13)/2
EFC + CFB = ( (5√13)/2 ) + ( (2√22932)/49 ) ≈ 15.1948232 units².

The area is 15.1948232 units².

U are correct
I will bring a basic shape solution to it soon,