Find area of the blue shaded Trapezoid | Pitot Theorem | Tangential Quadrilateral

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Find area of the blue shaded Trapezoid | Pitot Theorem | Tangential Quadrilateral | Trapezium

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Жыл бұрын

Pitot Theorem proof! Learn how to find the area of blue shaded Trapezoid whose bases are 37 and 74. Important Geometry skills are also explained: Pythagorean Theorem; area of the Trapezoid formula; Two-tangent theorem; Pitot Theorem; Tangential Quadrilateral. Step-by-step tutorial by PreMath.com
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Find area of the blue shaded Trapezoid | Pitot Theorem | Tangential Quadrilateral | Trapezium
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Пікірлер: 56

@Shanu_malik_ Жыл бұрын

Thanks sir ❤

@PreMath Жыл бұрын

You are very welcome! So nice of you. Thank you! Cheers! 😀

@dakshsaxena6978 Жыл бұрын

@@PreMathsir I have a doubt please give me your e mail I can mail you my doubt

@misterenter-iz7rz Жыл бұрын

Let h=2r be the height of the trapezoid, where r is the radius, we see that 37-r+74-r=111-2r=the length of hypotenuse, its square, by pythagorean theorem, is also equal to (2r)^2+(74-37)^2=4r^2+37^2, then 111^2-444r=37^2, so h=2r=(111^2-37^2)/222=8x37^2/222=4x37^2/111, therefore the answer is ((37+74)x4x37^2/111)/2=2x37^2=2738.🙂

@Copernicusfreud Жыл бұрын

That is how I solved it, too.

@williamwingo4740 Жыл бұрын

Here's a more pedestrian method: Let x = the radius of the circle. Draw radii from the center to the contact points of the circle on sides AB, BC, and CD. Call these points E, F, and G respectively; and let H be the center of the circle. This gives us two sets of congruent triangles: HEB & HFB and HFC & HGC. It's then easy to see that FC = 37 - x and FB = 74 - x; so BC = 37 + 74 - 2x = 111 - 2x. Now setting up the same triangle you did (EBC on your diagram but your E isn't the same as mine) and invoking Pythagoras, we get: (2x)^2 + 37^2 = (111 - 2x)^2; multiplying out (I confess I used a calculator for this step): 4x^2 + 1369 = 12321 - 444x + 4x^2; subtracting the 4x^2 term out on both sides, simplifying, and rearranging: 444x = 12321 - 1369 = 10952; dividing both sides by 444: x = 10952/444 = 24.666666... = 24-2/3. And finally, the area of the trapezoid is (1/2)(2x)(37 + 74) = (24.666666...)(111) = 2738. Cheers. 🤠

@murdock5537 Жыл бұрын

Nice and awesome, many thanks, Sir! M = center of the circle → ∆ BCM → sin⁡(CMB) = 1 → tan⁡(φ) = (37 - r)/r = r/(74 - r) → r^2 = (37 - r)(74 - r) → r = 74/3 → (37)2r = (1/3)(74)^2 = A (1/2)37(2r) = (1/6)(74)^2 = B → A + B = (1/2) (74)^2 = 2730 sq. units (blue shaded area) btw: (CM)^2 = (37 - r)^2 + r^2 ↔ r = 74/3 → CM = 37√5/3 → cos⁡(φ) = r/CM = 2√5/5 → φ ≈ 26,565°; and: 5476/111 = 148/3

@PreMath Жыл бұрын

Thanks dear

@theoyanto Жыл бұрын

Just loved the BIG REVEAL

@PreMath Жыл бұрын

Thank you! Cheers! 😀

@alster724 Жыл бұрын

I learned something new in this video, the Pitot's Theorem. After seeing the height, the problem became easier

@PreMath Жыл бұрын

Super!

@ghhdcdvv5069 Жыл бұрын

تمرين جيد. رسم واضح مرتب. شرح واضح مرتب. شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم جميعا . تحياتنا لكم من غزة فلسطين .

@PreMath Жыл бұрын

Thanks dear

@ybodoN Жыл бұрын

For the bonus question, let's reuse the diagram from the proof of Pitot's theorem. In the special case of a right trapezoid, the radius of the inscribed circle is a (= c). Area = 2a (a + b + c + d)/2 = a² + *a²* + ab + ad _or_ (a + b) (a + d) = a² + *bd* + ab + ad. ⇒ a² = bd ? Let's mirror CBO using CB as axis of symmetry (O = center of the inscribed circle). We obtain a cyclic quadrilateral where, according to the intersecting chords theorem, a² = bd ! Addendum: this solution assumes that ∠BOC = 90°. So, let's prove it: Let ∠EOB = ∠FOB = α and ∠FOC = ∠GOC = β. Then 2α + 2β = 180° ⇒ α + β = 90° ⇒ ∠BOC = 90°.

@johnfilak4751 Жыл бұрын

What a nice simple proof professor. Thank you so much : )

@KAvi_YA666 Жыл бұрын

Thanks for video.Good luck sir!!!!!!!!

@PreMath Жыл бұрын

Thank you too

@allanflippin2453 Жыл бұрын

Yes, the area A = a*b (37*74 in this case). This can be proven by going through the same procedure you showed, but retaining "a" and "b" in place of 37 and 74. The pythagorean equation then becomes ((b-a)^2 + 4r^2) = (a+b-2r)^2. Solving for r, r = a*b/(a+b). Then A = ((a+b)/2) * 2r. With r = a*b/(a+b), A becomes a*b.

@ybodoN Жыл бұрын

In (b − a)² + 4r² = (a + b − 2r)², the right side of the equation was a bit surprising. But since BOC is a right triangle (with O being the center of the inscribed circle): (b − a)² + 4r² = (b − r)² + r² + (a − r)² + r² = a² + b² − 2br − 2ar + 4r² = (a + b − 2r)². Otherwise, I don't quite see how to discover that, amazingly, BC = (a + b − 2r)²🤔

@allanflippin2453 Жыл бұрын

@@ybodoN BC = a+b-2r is based on the Pitot theorem that he quoted in the video. The sums of the opposite sides of an inscribed quadrilateral are equal: a+b = h+x. h=2r, so x=a+b-2r

@ybodoN Жыл бұрын

@@allanflippin2453 Thank you very much! It's now absolutely obvious💡

@JSSTyger Жыл бұрын

I just formed the most complicated and WRONG quartic equation for the radius R and quite amazingly got 37.4.

@ahmedrafea8542 Жыл бұрын

Very nice problem. Excellent explanation, as usual. Thanks

@user-ve7un8hy7m Жыл бұрын

nice method..thank you professor..♪

@PreMath Жыл бұрын

You are welcome!

@jaapcorstanje Жыл бұрын

Make trapeze smaller DC=1 AB=2..makes it much easier to derive the answer.. in the end multiply with 37.

@WernHerr Жыл бұрын

My solution without Pitots's theorem: Let E, F and G be the tangent points AB, BC and DC; x the radius of the circle. Then: GC=CF=37-x and BE=BF=74-x and thus CB=CF+BF=111-2x. (2x)²+37²=(111-2x)² and x=74/3. Trapezoidal formula: A=(74+37)/2 * 2*74/3 = 2738

@honestadministrator Жыл бұрын

Designating side -length BC by a one gets CD = a - r + 2 a - r = 3 a - 2 r Hereby a^2 + (2 r) ^2 = ( 3 a - 2 r) ^2 a^2 = (3 a) ( 3 a - 4 r) 2 r = (3 a - a /3) /2 = 4 a /3 Hereby area of Trapezoid ABCD = (4 a /3 ) (a + 2 a) /2 = 2 a^2

@raya.pawley3563 Жыл бұрын

Thank you!

@marioalb9726 Жыл бұрын

inclined side c : c = (37-R)+(74-R) c = 111-2R c = 111- h Pythagorean theorem: (b-a)²+h²=c² (74-37)²+h²=(111-h)² 37² + h² = 111² -222h+h² 222h = 111²-37² h = 49,33 cm Area = base x height Area = 49,33 ( 74+37)/2 Area = 2738 cm² ( Solved √ )

@ybodoN Жыл бұрын

In the special case of a right trapezoid, reusing the diagram from the proof of Pitot's theorem: AH = AE = DH = DG = r (radius of the inscribed circle whose center we label with the letter O). Let ∠EOB = ∠FOB = α and ∠FOC = ∠GOC = β. Then 2α + 2β = 180° ⇒ α + β = 90° ⇒ ∠BOC = 90°. Therefore, ∠OBF = ∠OBE = β and ∠OCF = ∠OCG = β ⇒ △OBC ~ △OBF ~ △OCF ⇒ △OEB ~ △OGC. So, if AB = a and DC = b then (a − r) / r = r / (b − r) ⇒ ab = r (a + b). Since r = ½ h ⇒ ab = ½ h (a + b), answering the bonus question.

@zdrastvutye Жыл бұрын

i have applied the law of symmetry = angle bisection and interpolation: 10 lu=74:lo=37:xg11=0:yg1=0:xg1=lu:ng=lu^2+lo^2:sw=.01:r=sw 20 dim x(3),y(3):goto 60 30 ls2=sqr((lu-lo)^2+4*r^2):xg2=(lu-ls2+lo)/2:yg2=r 40 xm=r:ym=r:dlu1=(xm-xg1)*(yg2-yg1):dlu2=(xg2-xg1)*(ym-yg1) 50 dg=(dlu1-dlu2)/ng:return 60 gosub 30 70 dg1=dg:r1=r:r=r+sw:if r>20*lu then stop 80 r2=r:gosub 30:if dg1*dg>0 then 70 90 r=(r1+r2)/2:gosub 30:if dg1*dg>0 then r1=r else r2=r 100 if abs(dg)>1E-10 then 90 110 print r 120 mass=10:goto 140 130 xb=x*mass:yb=y*mass:return 140 x(0)=0:y(0)=0:x(1)=lu:y(1)=0:x(2)=lo:y(2)=2*r:x(3)=0:y(3)=2*r 150 x=x(0):y=y(0):gosub 130:xba=xb:yba=yb:for a=1 to 3 160 x=x(a):y=y(a):gosub 130:xbn=xb:ybn=yb:line xba,yba,xbn,ybn 170 xba=xbn:yba=ybn:next a:x=xm:y=ym:gosub 130:circle xb,yb,r*mass this routine calculates r as radius, have fun using bbc basic sdl

@misterenter-iz7rz Жыл бұрын

Generally, the answer is ab, where a,b are lower and upper widths of trapezoidal.

@Copernicusfreud Жыл бұрын

Yay,I solved it.

@PreMath Жыл бұрын

Bravo

@aksiiska9470 Жыл бұрын

i suggest the bisectors of angles in every point

@batavuskoga Жыл бұрын

I solved it in another way. In my opinion my solution is much easier 37²+(2r)²=(111-2r)² - in the video : 37²+h²=x² hypothenus=111-2r --> 37-r+74-r=111-2r, according to a theorem in geometry 37²+4r²=111²-2*2r*111+4r² 1369=12321-444r 444r=12321-1369=10952 r=74/3 - height trapozoid = 2r = 148/3 area trapezoid=(small base=big base)*height/2 area=(37+74)*(148/3)/2 area trapezoid=2738

@harikatragadda Жыл бұрын

If O is the center, and F, G are the top and bottom contact points of the circle, H being the right most contact point, then Kite FOHC is Similar to Kite GOHA since ∠GOH is Supplimentary to ∠FOH. Each of these Kites have two congruent right triangles. Hence, the right triangle FOC is Similar to the right triangle AOG. R/(74-R) = (37-R)/R R= 37*74/111 Trapezoid Area= ½*(2R)*(37+74) = R*111 = 37*74 = 2738

@user-sw9lb2zs6e 2 ай бұрын

Where do you get (111-2r)^2?

@bigm383 Жыл бұрын

Thanks, Professor!🥂👍

@PreMath Жыл бұрын

My pleasure! So nice of you. Thank you! Cheers! 😀

@bigm383 Жыл бұрын

@@PreMath 😀❤️

@halitiskender1324 Жыл бұрын

Hocam seni seviyorum ❤

@PreMath Жыл бұрын

Thank you! Cheers! 😀

@abdelrahmantarek9704 Жыл бұрын

What software you are using please

@batavuskoga Жыл бұрын

Is this always the case that the area of a tangential trapezoid is equal to the product of its bases ? Or is this just a coincidence in this exercise ?

@wackojacko3962 Жыл бұрын

PITOT THEOREM: 4 equal pairs of tangent segments of an Inscribed circle of Right Trapezoid and sum of two side lengths can be derived from the 4 tangent segments....soooooooo cool!🙂

@aksiiska9470 Жыл бұрын

8:46a strange theorem x=|||-h

@giuseppemalaguti435 Жыл бұрын

Per i triangoli simili r:(2r-37)=2r:(74-37)...r=111/4

@RadhaSingh-fn4nz Жыл бұрын

2738 ans.

@asishdas-wy8uv Жыл бұрын

The ans will be yes.

@shantanusrivastava9898 Жыл бұрын

Sir, you have missed to prove that AB is parallel to CD. Without this , you cannot use the Pythagoras theorem .

@PreMath Жыл бұрын

They are making 90 deg angle => They are parallel. Cheers

@williamwingo4740 Жыл бұрын

Here's a more pedestrian method: Let x = the radius of the circle. Draw radii from the center to the contact points of the circle on sides AB, BC, and CD. Call these points E, F, and G respectively; and let H be the center of the circle. This gives us two sets of congruent triangles: HEB & HFB and HFC & HGC. It's then easy to see that FC = 37 - x and FB = 74 - x; so BC = 37 + 74 - 2x = 111 - 2x. Now setting up the same triangle you did (your EBC, but your point E is different from mine) and invoking Pythagoras, we get: (2x)^2 + 37^2 = (111 - 2x)^2; multiplying out: 4x^2 + 1369 = 12321 - 444x + 4x^2; subtracting out the 4x^2 term on both sides, simplifying, and rearranging: 444x = 12321 - 1369 = 10952; dividing both sides by 444: x = 10952/444 = 24.666666... = 24-2/3. And finally, the area of the trapezoid is (1/2)(2x)(37 + 74) = (24.666666...)(111) = 2738. Cheers.🤠