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

[@luigipirandello5919]

Beautiful solution. Thank you.

[@PreMath]

Glad to hear that! You are very welcome! Thanks ❤️

[@MrPaulc222]

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-ص7ق]

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

[@PreMath]

Thanks ❤️

[@jimlocke9320]

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.

[@PreMath]

Thanks ❤️

[@aserilomavatu5143]

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

[@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

[@rabotaakk-nw9nm]

👍❤️ The Best Solution!

[@PreMath]

Thanks ❤️

[@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.

[@PreMath]

Thanks ❤️

[@allanflippin2453]

Yes, I did it that way as well.

[@andydaniels6363]

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.

[@hcgreier6037]

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...

[@davidellis1929]

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).

[@Ibrahimfamilyvlog2097l]

Very nice sharing❤❤❤

[@PreMath]

Glad to hear that! Thanks ❤️

[@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)

[@PreMath]

Thanks ❤️

[@humbertorodriguezperez1214]

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

[@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 a1=9*pi:a2=pi:r1=sqr(a1/pi):r2=sqr(a2/pi):sw=r1/(r1+r2)/10 30 xc=0:h=2*r1:yc=h:yd=0:xd=sw:ym1=r1:goto 180 40 xu1=xc:yu1=yc:xu2=xd:yu2=yd 50 gosub 60:goto 80 60 dx=xu2-xu1:dy=yu2-yu1:zx=dx*(xpu-xu1):zy=dy*(ypu-yu1):k=(zx+zy)/(dx^2+dy^2) 70 dxk=dx*k:dyk=dy*k:xlo=xu1+dxk:ylo=yu1+dyk:lo=sqr((xpu-xlo)^2+(ypu-ylo)^2) 80 dg=lo/r1:dg=dg-1:return 90 xpu=xd+sw:gosub 40 100 dg1=dg:xp1=xpu:xpu=xpu+sw:xp2=xpu:gosub 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 xu1=xc:yu1=yc:xu2=xd:yu2=yd:ypu=ym1 140 gosub 90: xm1=xpu:xb=2*(xm1-xd):xb=xb+xd:xa=xb+xd:ya=0 150 ym2=r2:xm2=sqr((r1+r2)^2-(r1-r2)^2)+xm1:xpu=xc:ypu=yc:xu1=xd:yu1=yd:xu2=xm1:yu2=ym1 160 dfu1=(xa-xm1)/ym1:dfu2=(xa-xm2)/ym2 170 df=dfu1-dfu2: return 180 gosub 130 190 df1=df:xd1=xd:xd=xd+sw:xd2=xd:gosub 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 run in bbc basic sdl and hit ctrl tab to copy from the results window

[@zipponvr7043]

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

[@PreMath]

Thanks ❤️

[@Abby-hi4sf]

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

[@PreMath]

Excellent! Thanks ❤️

[@jan-willemreens9010]

... 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

[@PreMath]

Excellent! You are very welcome! Thanks ❤️

[@wackojacko3962]

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. 🙂

[@PreMath]

Very true! Thanks ❤️

[@phungpham1725]

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

[@PreMath]

Thanks ❤️

[@santiagoarosam430]

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.

[@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

[@PreMath]

Thanks ❤️

[@arnavkange1487]

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

[@PreMath]

Thanks ❤️

[@mahinnazu5455]

Nice math solution. Thank u Sir

[@PreMath]

Glad to hear that! You are very welcome! Thanks ❤️

[@laxmikantbondre338]

Please show how to prove that diagonals of rhombus pass through center of circles.

[@GetMeThere1]

It's interesting that the area of triangle ABD must equal the area of triangle ABC (since 2 times either one must give the area of the full rhombus).

[@HeywoodUmanoff]

How do we know that O, the center of the large circle, lies on the diagonals of the rhombus?

[@jasongoldman3850]

How did you determine the parallelogram is a rhombus? I'm confused.

[@LuisdeBritoCamacho]

Let's do it!! The height of the Parallelogram is 6; as the Radius of the Green Circle is 3. 1) The line that passes the points COPA, divides the Parallelogram in two equal triangles. 2) The angle [OPF'] being F' the point between the line OF and the intersection of a line passing by P and parallel to AD. 3) Now we have a Triangle Rectangle with lengths: OF' = 2 ; OP = 4 ; F'P = sqrt(12) 4) The angle [OPF'] = 30º and the angle [BAD] = 60º. Now, we want to find the length AD 5) Drop the point C vertically, in order to intercept the extending line AD and find C' 6) Now we have a triangle [ACC'] and a triangle [CDC']. 7) C'D = 6 / tan(60º) ~ 3,464 lu 8) C'A = 6 / tan(30º) ~ 10,392 lu 9) DA = C'A - C'D = 10,392 - 3,464 ~ 6,928 lu 10) 6 * 6,928 ~ 41,5692 su My Answer is: The Area of the Parallelogram is approx. 41,5692 square units.

[@misterenter-iz7rz]

The right-angled triangle is 2:4:2sqrt(3)=1:2;sqrt(3), thus angle ratio is 30:60:90), so AP=2, OA=6, OD=3/sqrt(3)×2=2sqrt(3), therefore the area is 2×6×2sqrt(3)=24sqrt(3).😊

[@PreMath]

Thanks ❤️

[@bhaskarps]

Area = AB x BC x Sin60 Area= (12/Sqrt(3)) x (12/Sqrt(3)) x Sqrt(3)/2

[@rajendraameta7993]

Formula of Area of equilateral triangle can also be used.

[@jamestalbott4499]

Thank you!😀

[@giuseppemalaguti435]

Risulta R=3,r=1...hp=(altezza del parallelogramma)=3*2=6.….BP(base del parallelogramma)=√3+√12+√3=4√3.…Ap(area)=6*4√3=24√3

[@PreMath]

Thanks ❤️

[@prossvay8744]

Area of the parallelogram ABCD=6(4√3)cm^2=41.57cm^2.❤❤❤ thanks

[@PreMath]

You are very welcome! Thanks ❤️

[@robertlynch7520]

Yet another “completely different” way to solve, but which in the end is similar.    𝒂² = (3 + 1)² + (3 - 1)²    𝒂 = √12    𝒂 = 2√3 Then, by proportionality, determining the 𝒃 along base line from center of small circle to right corner is    2 / 2√3 = 1 / 𝒃 … cross multiply    𝒃 = √3 Where '2' was the height (3 - 1) of the little inside triangle. OK, now we know the baseline of (1√3 + 2√3) = 3√3 units. However, this isn't yet a solution. Just the baseline from the righthand lower corner of the rhombus to center of the 9π circle, and down to the base. Because the parallelogram is symmetric around the 9π circle's center (not drawn as a big kite as you did, but rather just around a vertical line!), we now need a couple of more things. Next is to observe that the height 𝒉 = 6, or 2× the radius of the 9π circle. And we'd like to figure out the intersect onto baseline to the upper right corner of the rhombus. Hmmm… Again, a little fiddling and    θ = atan( 3 ÷ 3√3 )    θ = atan( 1 / √3 )    θ = atan( √3/3 )    θ = 30° !! How humiliating. The little △ between centers of the circles had 'rise = 2' and 'hypotenuse = 3 ⊕ 1 = 4', so the 30-60-90 triangle ratios of 1-2-√(3) should have jumped out at me. Oh well. Didn't. ATAN to the rescue. Observing that the angle 2θ = 60° is the lower-right ∠ then allows the virtual △ of height 6 and baseline = 2√(3) to be derives right quick. With that, then the area of the whole thing must be    Area = 2 × ( inside rectangle + virtual triangle )    Area = 2 × ( 6√3 ⊕ ½ 2√3⋅6 )    Area = 2 × ( 6√3 + 6√3 )    Area = 2 × ( 12√3 )    Area = 24√3    Area = 41.57 u² And that would be that. Totally differently. YOUR SOLUTION was way prettier, and way easier. Thank you anyway!!!!! ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅

[@StephenRayWesley]

(4x)^2 (4x) ^2 =16x+16x) =32x^2 (360°-32x^2)={328÷9π}= √364 √6^√2√ 2√^2 3^2√^1 √1√^1 (π+2π-3)

[@PreMath]

Thanks ❤️

[@kiratsinghsaluja9123]

Nice one , uses the concept that circle inscribed in a parallelogram then it must be a Rhombus👍👍

[@PreMath]

Thanks ❤️

[@murdock5537]

φ = 30°; ∆ DAO → AD = DF + EF + AE; AO = AP + PO; sin⁡(AOD) = 1; FO = FT + TO = EP + TO = 1 + 2 = 3 ∆ POT→ PO = 1 + 3; TO = 2 → TP = EF = 2√3 → sin⁡(δ) = TO/PO = 1/2 = sin⁡(φ) = DF/DO = DF/2√3 → DF = √3 → ∆ APE → cos⁡(φ) = √3/2 = AE/2 → AE = √3 → AD = 4√3 → h = 2FO = 6 → area parallelogram = 4h√3 = 24√3

[@PreMath]

Excellent! Thanks ❤️

[@arnavkange1487]

You are awesome

[@PreMath]

Thanks dear❤️

[@trump1678]

🤠🤠🤠🤠🤠

[@PreMath]

Thanks ❤️

[@raya.pawley3563]

Thank you

[@PreMath]

You are very welcome! Thanks ❤️

[@StephenRayWesley]

(360°÷9π)=√40π 2^√20 2^√5^√4 2^1^√2^2 2^√1^√1^2 22 (π+2π-2)

[@PreMath]

Thanks ❤️

[@k.amanda.m]

When you know, the base length of the parallelogram is 12√3 and the height of the parallelogram is 6 (from 2 × r = 2× 3). Then the area of the parallelogram = 12√3 × 6 = 72√3 = 12√3 cm² So that's a quick trick so you don't use the sincostan formula 💫😁

[@PreMath]

Thanks ❤️

[@mihailnaum7642]

72=12=24????

[@k.amanda.m]

@@mihailnaum7642 oww im sorryy.. the base is 12/√3 So, 12/√3 × 6 = 72/√3 = 72√3 × √3/√3 = 12/√3

[@k.amanda.m]

@@PreMath thank youu, teacherrr.. and im so interested with your explain. Thanks for sharing your knowledge ♡

[@manojitmaity7893]

The diagonals of the parallelograms do not intersect each other in 90⁰ angle...please check

[@ybodoN]

Since this parallelogram is a rhombus, they do! 🧐

[@manojitmaity7893]

@@ybodoN But it is mentioned that the given parallelogram is a rhombus.

[@HeywoodUmanoff]

A rhombus is a special type of parallelogram in which all sides are equal and diagonals intersect at right angles.

[@ybodoN]

@@manojitmaity7893 as explained in the video (0:57), since the green circle is fully inscribed in the parallelogram, it is a rhombus. 😉