₦airaland Forum

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.

femi4:
x² = 41 -40cosa


x² = 25 - 24cos b

Cos a = (41-x²)/40


Cos b = (25-x²)/24

40 cosa -24 cosb = 16

5 cosa - 3 cosb = 2


Also, (x√2)² = 5²+3² - 30 cos (360 - (a+b))
X² = 31 - 30 cos (a+b)
X² = 31 - 30 cosa cosb - sina sinb

Darivie04:
@ifada123 can you suggest books or sites containing practice problems on shapes?

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

1) label the bottom left vertex, the bottom right vertex, top left vertex and top right vertex as A, B, C and D respectively.

2) Place the square on a coordinate plane with A at the origin, AB on the positive x-axis and AC on the positive y-axis. Hence the coordinates of the vertices are A(0, 0), B(x, 0), C(0, y) and D(x, y). Let the coordinates of the intersection point H in the square be (a, b). Observe that |AB| = |AC| since they are sides of a square. Therefore x = y.

3) Using the distance formula, we have
|HA|² = a² + b² = 5² = 25
|HB|² = (a - x)² + b² = 4² = 16
|HD|² = (a - x)² + (b - y)² = 3² = 9

Expanding, substituting and using the fact that x = y, we
have :
a = (x² + 9)/2x and b = (x² + 7)/2x.
Since a² + b² = 25, substitute and solve for x.
You will get an equation quadratic in form, which is x⁴ - 34x² + 65 = 0. Using the quadratic formula, x² = 17 ± 4√14. The value 17 - √14 yields an area for the square which is too small to contain the line segments hence

x = (17 + 4√14)½ ≈ 5.65390392 units.

Martinez39:
The number (17 + 4√14)½ is exactly what I posted before but I pressed wrongly on my calculator and got an inaccurate decimal approximation. I will post my workings soon.

Martinez39:
Cc. Ifada123

Calculator error.

X = (17 + 4√14)½ ≈ 5.65390392 units.

Martinez39:
I will wait for your solution.

Martinez39:


X = (17 + 4√14)^½ ≈ 4.68685711 units.

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.

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

shegzy2009:
24.5 sq. units.

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.

Martinez39:
Shapes are getting tougher.

Tersorio:
she studied petrochemical engineering bro....I am sure she can handle that.

oyekanwahab:
36 degree, (sum of angle in insolence triangle)