Can you find area of the Parallelogram ABCD? | (Circles inscribed in a parallelogram) | #math #maths

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Can you find area of the Parallelogram ABCD? | (Circles inscribed in a parallelogram) | #math #maths

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[MrPaulc]

Thank you.
I did it by making the height 6 due to the large circle. I then calculated the base piecemeal by using 30, 60, 90 triangles with at least one already-known side length. You could see that 30, 60, 90 was appropriate due to the relative sizes of the circles (3:1) and that the tangents were colinear from A in both directions.

[الثورة-صق]

لقد إستفذت كثيرا من فيديوهاتك،وأنجزت كثيرا من تمارينك،ولقد أعجبت بطريقتك السلسة في الشرح والتبيان،وتبسيط الأمور،فشكرا لك على مجهوداتك القيمة،وجازاك الله خيرا

[aserilomavatu]

I'm grateful that you showed your skills... I'm learning a lot... Thank you.

[hcgreier]

This is way too complicated. The circles have radiuses R = 3 and r = 1. Distance of midpoints of circles = 3 + 1 = 4, hence horizontal distance of midpoints is √(4² - 2²) = √12 = 2√3. Distance x from little circle's lower touching point to the lower right vertex of parallelogram: → similar triangles x/1 = 2√3/2 → x = √3. Due to symmetry one can put the same small circle in the top left corner as well, then you see that the left vertex distance is the same as x. So parallelogram area = base × height = (2√3 + 2√3)·(2·3) = 6·4·√3 = 24√3 ≈ 41.5692...

[jimlocke]

At 5:50, drop a perpendicular from P to OF and label the intersection G. PGFE is a rectangle, so GF = PE = 1, and OG = OF - GF = 3 - 1 = 2. Consider right ΔPGO. Hypotenuse OP = 4 and short side OG = 2, ratio 2 to 1. It is, therefore, a special 30°-60°-90° right triangle. ΔAOF is similar, so we can apply the ratios for a special 30°-60°-90° right triangle, knowing that the short side OF has length 3. So AF = 3√3 and OA = 6. We can skip ahead to approximately 10:00.

[ybodoN]

Draw a parallel to DA passing through P and cutting OF at Q. ⊿OQP ~ ⊿OFA ~ ⊿OFD.
Then we have OQ = 2 and OP = 4 ⇒ QP = 2√3 ⇒ FA = 3√3 ⇒ DF = 3√3 / 3. So DA = 4√3.
The height of the parallelogram (rhombus) is equal to the diameter of the green circle.
Therefore, the area of the parallelogram is 6 · 4√3 = 24√3 ≈ 41.57 cm²

[davidellis]

I solved this very quickly. With points labeled as in the video, draw a perpendicular from P to OF, and let its foot be G. OP=3+1=4 and OG=3-1=2, which makes OPG a 30-60-90 triangle, since the hypotenuse is twice the shorter leg. It follows that angle BAD is 60 degrees, so the parallelogram ABCD is two equilateral triangles back to back. The altitude is twice the radius of the green circle, that is 6, so each side is 6*2/sqrt(3), or 4*sqrt(3). Then the area of ABCD is 24*sqrt(3).

[Irishfan]

I thought this was much simpler than finding the area of two triangles. The area is the base times the height. The height is clearly the diameter of the circle. The base is figured out by determining one side of the triangle made-up of the base and the intersecting diagonals drawn from the corners. Once you figure the hypotanuse of that triangle and the diameter of the circle, you have the base and height of the parallelogram and multiplied together, you have the area.

[andydaniels]

Per Steiner’s formula for the external symmetry point, we find that AP = (1*(-4)-(3*0))/(1-3) = 2. From the tangent-secant theorem, we then obtain AE = √3. Draw a perpendicular to AC through P and call its intersection with AD point G. Using similar triangles we can find that GP = 2/√3 and so the area of △APG is 2/√3. This triangle is similar to △AOD. The latter’s area is (6/2)^2 * 2/√3 = 18/√3 and four of these make up the paralellogram, for a total area of 4*18/√3 = 24√3.

[engralsaffar]

Much easier way was to identify that the angle FOA is one of 6 identical angles in the green circle which is equal to 360/6=60 degrees therefore angle FAO =30, and in that special triangle the opposite = 3, therefore the adjacent =3*sqrt(3)

Similarly the angle FOD=30, with the adjacent =3, therefore the opposite = sqrt(3)

Therefore AD=b=4*sqrt(3)
Height=h=2*R=6

Area of parallelogram =h*b=6*4*sqrt(3)=24*sqrt(3)

[Abby-hisf]

Again a great basic review, and beautiful lesson. I enjoyed each minute
After finding OA =6 and learning ABD is equlaterial triangle. We got the two diagonal measure
the diagonal AC =12, and BD = 4√3, Rombos A=pq/2, ==== A= (12)( 4√3, )/2= 24√3

[zipponvr]

it can be solved easier
when you get DA = 12/sqrt(3) = 4*sqrt(3), just multiply it on parallelogram height, which is green diametr = 3*2 = 6, so 4*sqrt(3) * 6 = 24*sqrt(3)
no need sin60

[wackojacko]

This problem is really cool because there are so many things that are obvious and not unexpected...but only if significant concepts of math are adhered to. 🙂

[zdrastvutye]

i have calculated this only using the perpendicular formula without
the thales theorem or trigonometry of the angle bisector:
10 print "premath-can you find area of parallelogram abcd":dim x(3, 2), y(3, 2)
20
30 180
40 xu1=xc:yu1=yc:xu2=xd:yu2=yd
50 gosub 60:goto 80
60
70
80 dg=lo/r1:dg=dg-1:return
90 xpu=xd+sw:gosub 40
100 40:if dg1*dg>0 then 100
110 xpu=(xp1+xp2)/2:gosub 40:if dg1*dg>0 then xp1=xpu else xp2=xpu
120 if abs(dg)>1E-10 then 110 else return
130
140 gosub 90:
150
160
170 df=dfu1-dfu2: return
180 gosub 130
190 130:if df1*df>0 then 190
200 xd=(xd1+xd2)/2:gosub 130:if df1*df>0 then xd1=xd else xd2=xd
210 if abs(df)>1E-10 then 200 else ages=(xa-xd)*h/2:print "die flaeche=";ages
220 print xd:masx=1E3/xa:masy=9E2/h:if masx<masy then mass=masx else mass=masy
230 x(0, 0)=xd:y(0, 0)=yd:x(0, 1)=xa:y(0, 1)=0:x(0, 2)=(xd+xb)/2:y(0, 2)=h/2
240 x(1, 0)=xa:y(1, 0)=0:x(1, 1)=xb:y(1, 1)=h:x(1, 2)=x(0, 2):y(1, 2)=y(0, 2)
250 x(2, 0)=xb:y(2, 0)=h:x(2, 1)=0:y(2, 1)=h:x(2, 2)=x(0, 2):y(2, 2)=y(0, 2)
260 x(3, 0)=0:y(3, 0)=h:x(3, 1)=xd:y(3, 1)=yd:x(3, 2)=x(0, 2):y(3, 2)=y(0, 2):goto 280
270 xbu=x*mass:ybu=y*mass:return
280 x=xm1:y=ym1:gosub 270:circle xbu, ybu, r1*mass:x=xm2:y=ym2:gosub 270:circle xbu, ybu, r2*mass
290 for a=0 to 3:gcola:x=x(a, 0):y=y(a, 0):gosub 270:xba=xbu:yba=ybu:for b=1 to 3:ib=b:if ib=3 then ib=0
300 x=x(a, ib):y=y(a, ib):gosub 270:xbn=xbu:ybn=ybu:goto 320
310 line xba, yba, xbn, ybn:xba=xbn:yba=ybn:return
320 gosub 310:next b:next a
premath-can you find area of parallelogram abcd
die flaeche=20.7846097
3.46410161
>
run in bbc basic sdl and hit ctrl tab to copy from the results window

[jan-willemreens]

... Good day, I cut ABCD over AC in 2 triangles, and used obtuse triangle ADC to do my calculations ... the perpendicular height (outside the triangle) is 2 * 3 = 6 cm and finding I AO I = 3/COS(60) = 6 cm ... then I AD I = I AO I/COS(30) = 6 * 2 /SQRT(3) = 4 * SQRT(3) cm ... finally A( ABCD ) = HEIGHT * BASE I AD I = 6 * 4 * SQRT(3) = 24 * SQRT(3) cm .... and now taking some more time to watch your strategy presentation (lol) ... thank you again for another geometric exercise ... best regards, Jan-W

[santiagoarosam]

G es la proyección ortogonal de P sobre OF y T es el punto de tangencia de ambos círculos.
Radio del círculo pequeño =r =1 → Radio del círculo grande =r√9 =3 → Tenemos tres triángulos semejantes de ángulos 60º/90º/30º: OGP, de lados (3-1=2)/(2√3)/(3+1=4) ; PEA, (1)/(√3)/(2) ; DFO, (√3)/(3)/(2√3) → DA =√3 +2√3 +√3 =4√3 → Área ABCD =DA*(2OF) =(4√3)(2*3) =24√3
Gracias y un saludo cordial.

[phungpham]

1/ We have R=OF=3 and r=PE=1
Moreover the parallelogram is also a diamond(CD=CB) so the the height of the diamond=6 the base=DA
2/ Consider the 2 similar trisngles AOF and APE
AP/AO =PE/OF=1/3—-> AO/3=AP/1=(AO-AP)/(3-1)=4/2=2
So AO=6, AP=2
——-> the triangle AOF is a 30-90-60 triangle —— > FA=3sqrt3
Because sq OF= DF.FA—>DF=sqrt3
Area=6.(sqrt3 +3sqrt3)=24 sqrt3 sq units

[humbertorodriguezperez]

Once you have the side of the triangle ABD and its height (6 cm) you can calculate its area, and the total area is twice the area of the triangle ABD.
a' = 6 * (12/√3)/2 = 3 * 12/√3 = 36/√3
a = 2*a' = 2 * 36/√3 = 72/√3
72/√3 = (72√3)/3 = 24√3

[arnavkange]

Acoording to aa test of similarity of triangles in AEP~AFO the ratio should be AE/AF=EP/FO=AP/AO

[quigonkenny]

By observation, ABCD is a rhombus, which means diagonals AC and BD are perpendicular bisectors at O. By Circle Theorem, OF and PE are perpendicular to AD. By observation of complementary angles, ∆OFA and ∆PEA are similar, as is ∆DOA. Let R be the radius of the green circle and r be the radius of the yellow:

Circle O:
A = πr²
9π = πR²
R² = 9
R = 3

Circle P:
A = πr²
π = πr²
r² = 1
r = 1

By observation, OA is equal to R + 2r + x (unknown distance between circumference of yellow circle and A), and PA is equal to r + x. As ∆OFA and ∆PEA are similar:

OA/OF = PA/PE
(3+2+x)/3 = (1+x)/1
5 + x = 3(1+x) = 3 + 3x
2x = 2
x = 1 ∴ OA = 6, PA = 2

Triangle ∆OFA:
a² + b² = c²
3² + FA² = 6²
FA² = 36 - 9 = 27
FA = √27 = 3√3

As ∆OFA and ∆DOA are similar:

DA/OA = OA/FA
DA/6 = 6/3√3
DA = 36/3√3 = 12/√3 = 4√3

Rhombus ABCD:
A = bh = DA(2R) = (4√3)6 = 24√3 cm²