Can you find area of the Blue shaded Rectangle? | (Two Methods) | #math #maths | #geometry

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Can you find area of the Blue shaded Rectangle? | (Two Methods) | #math #maths | #geometry

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[predator]

This is awsome 👍, thank you teacher 🙏.

[montynorth]

Note that point E is not at a fixed position on the circumference, the only requirement is that AE =18.
This means that both the radius & dimensions of the rectangle are not fixed.
If Point E coincides with point F, then we have a semicircle of radius 9, and a rectangle of height 9 & width 18.
Hence 9 x 18 = 162.

[jimlocke]

The tricky part here is that the lengths of the sides of rectangle ABCD are variable but their product is a constant. We can not solve for the value of r because there is a range of values for r which are valid.

[TimothyCizadlo]

Given that the area of ABDC = AD*(AO+OB) ∴ ABDC=(r)*(r+OB). ∴ (AO+AB)^2+(BE)^2=AE^2 & (OE)^2=(OB)^2+(BE)^2 ∴ 2*(r^2+(r+OB))=324 ∴ ABCD = 162.

[rey-dqnx]

Let l= length
r= width
Area=lr
EB²=r²-(l-r)²
EB²=18²-l²
r²-l²+2lr-r²=324-l²
2lr=324
lr=162 sq units

[ivorauh]

Just solved it via Pythagoras: 18^2= (r+OB)^2 + BE^2 and r^2= OB^2 + BE^2 and A= r*(r+OB) => A=r^2+r*OB and 18^2=r^2 +2*r*OB +OB^2 + BE^2; substitute OB^2+BE^2 by r^2 results in 18°2=2(r^2+rOB) and finally substituting A=r^2+r*OB results in 18^2=2*A => A=9*18=162. Makes sense?

[hongningsuen]

Method using Thales theorem, Pythagoras theorem and similar triangles:
1. Let radius be R.
2. Triangle AEF is right-angled triangle by Thales theorem.
Hence EF^2 = (2R)^2 - 18^2 = 4R^2 - 324 by Pythagoras theorem. (equation 1)
3. Triangles AEF and EBF are similar (AAA)
Hence EF/BF = AF/EF
(EF)^2 = (AF)(BF) = (2R)(BF) (equation 2)
4. From equations (1) and (2)
4R^2 - 324 = (2R)(BF)
4R^2 - 2(R)(BF) = 324
2R^2 - (R)(BF) = 162
4. When DC extends to DC' such that DC' = AF, area of rectangle ADC'F = 2R^2 and
Area of rectangle BCC'F = (R)(BF)
Area of ABCD = area ADC'F - area BCC'F = 2R^2 - (R)(BF) = 162

[michaelkouzmin]

3rd method:
1) Let r=AO, x=OB, h=OE; y=EF;
2) triangle ABE: 18^2=h^2+(r+x)^2 => h^2 = 18^2-(r+x)^2; (1)
triangle OBE : r^2=h^2+x^2; => h^2 = r^2-x^2; (2)
3) compare (1) and (2) => r^2-x^2 = 18^2-(r+x)^2;
r^2-x^2 = 324 - (r^2+2rx+x^2);
r^2-x^2 = 324 - r^2-2rx-x^2;
2r^2+2rx=324;
2*(r^2+rx)=324;
r^2+rx=162
4) Let us observe that Ablue = r^2+rx= 162sq units.

[yatesfletcher]

Let R be the radius OA and (x, y) be the coordinates of B. So B is the intersection of the semi-circle with the circle of radius 18 centered at A. Thus (x, y) must satisfy
1) ( x+R)^2 + y^2 = 18^2 and
2) x^2 + y^2 = R^2
Subtracting 2) from 1) eliminates the x^2 and y^2 terms and simplifies to
(X+R)R = 18^2/2 = 162.

But (x+R)R is the blue area!

[zdrastvutye]

the result of the area does not depend on the radius, see line 30:
10 l1=18:l4=13:if l4<l1/2 then else 30
20 print "ungueltige eingabe":end
30 x(3), y(3):goto 60
40
60 gosub 40
70 40:if dg1*dg>0 then 70
80 xs=(xs1+xs2)/2:gosub 40:if dg1*dg>0 then xs1=xs else xs2=xs
90 if abs(dg)>1E-10 then 80
100 print xs;"%";"die gesuchte
110 if masx<masy then mass=masx else mass=masy
120
120 goto 140
130 xbu=x*mass:ybu=y*mass:return
140 x=r*2:y=0:gosub 130:xba=xbu:yba=ybu:for a=1 to
150 gosub 130:xbn=xbu:ybn=ybu:goto 170
160 line xba, yba, xbn, ybn:xba=xbn:yba=ybn:return
170 gosub 160:next a:gcol 12:xba=0:yba=0:for a=1 to 4:ia=a:if ia=4 then ia=0
180 x=x(ia):y=y(ia):gosub 130:xbn=xbu:ybn=ybu:gosub 160:next a:gcol 11
190 xba=0:yba=0:x=xs:y=ys:gosub 130:xbn=xbu:ybn=ybu:gosub 160
12.4615385%die gesuchte flaeche=162
>
run in bbc basic sdl and hit ctrl tab to copy from the results window

[batavuskoga]

I have another method, not necessarily better, but I found the solution.
That's the most important 😆

Let x=BF en y=BE
18²=y²+(2r-x)² --> 324=y²+4r²-4rx+x² --> 324=x²+y²+4r²-4rx

Draw a line from o to E, that's the radius r

r²=OB²+y² --> r²=(r-x)²+y² --> r²=r²-2rx+x²+y² --> 2rx=x²+y²

324=x²+y²+4r²-4rx
324=2rx+4r²-4rx
324=4r²-2rx

162=2r²-rx (divide by 2)

area rectangle=(2r-x)*r=2r²-rx

so, area rectangle=162 square units

[stevetitcombe]

Area=AB . Radius

Radius=AF/2,

Therefore;

Area= AB . AF/2,

Now, AF= AB + BF

So Area= AB . (AB + BF)/2

Therefore;

Area= (AB^2)/2 + (BF . AB)/2

Therefore;

Twice Area= AB^2 + (BF . AB)

By intersecting chord theory;

EB^2 = AB . BF

So Twice Area= AB^2 + EB^2

By Pythag,
AE^2= EB^2 + AB^2

So AB^2= AE^2 - EB^2

By substitution;

Twice Area=AE^2 - EB^2 + EB^2

Simplifies to;

Twice Area= AE^2

So, Area= (AE^2)/2 = 18^2 / 2

Area= 324 / 2 = 162 units^2

[alscents]

Again, since there are no variables, it means that the blue area is constant, take the special case where the semi-circle is inscribed in the rectangle meaning B, F, and E are all the same points and [AE] is now a diameter of the circle, meaning [AE] = 2 r = 18
This gives r = 9 and the sides of the rectangle as r and 2 r
Area of rectangle = 2 r^2 = 2 * 9^2 = 2 * 81 = 162 square units.

[povijarrro]

|OB|=x, |BE|=y
(r+x)^2+y^2=18^2
x^2+y^2=r^2

2r(r+x)=18^2
A=r(r+x)=9*18=162

[solimana-soli]

two alternative methods
first one: similarity
tracing perpendicular to chord AE we have right triangle OAH (H is the midpoint of AE) similar to AEB, then
9 : AB = R : 18
area = AB*R = 9*18 = 162
second one: logical
extending AE untill it lies on diameter AF we have that area = radius*diameter=18*9=162

[santiagoarosam]

Llamaremos G a la proyección ortogonal de O sobreDC. Respetando las condiciones del trazado de la figura propuesta, desplazando E hasta G, el rectángulo original se transforma en un cuadrado de dimensiones r*r y diagonal de longitud 18 ud cuya superficie es igual a la buscada- ---> Área ABCD =18²/2=162 ud².
Gracias y saludos.

[marcgriselhubert]

Let's use an orthonormal, center O, first axis(OB).
Be OB = c, and R the radius of the circle.
The equation of the circle is x^2 + y^2 = R^2
At point E we have x = c and c^2 + y^2 = R^2,
so E(c, sqrt(R^2 - c^2)
Then VectorAE(c + R, sqrt(R^2 - c^2)
AE^2 = (c + R)^2 + (R^2 - c^2) = 2.R^2 + 2.c.R = 18^2
Meaning that: R.[R + c] = (18^2)/2 = 162
Or that the blue area is 162.

[davesusko]

It is interesting that the radius cancels in the final equation which means the area is independent the radius and any rectangle with the similar configuration (ncluding a square will have the same area.

[JSSTyger]

To me, this appears to be a "minimal information" problem. We cant solve for the radius or the height of the triangle but we know they are dependent on eachother. If I choose h = 0, it will make r = 9. The area of the rectangle would the be 9(18) = 162.

[wackojacko]

Second method is mo bettah!...cuz i take great care to avoid enclosed spaces. Can't think inside a box. 🙂