Can you solve for X and Y? | (Rectangle) |

Learn how to solve for X and Y in the rectangle. Important Geometry and Algebra skills are also explained: Pythagorean theorem; area of the Triangle formula. Step-by-step tutorial by PreMath.com
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Can you solve for X and Y? | (Rectangle) | #math #maths | #geometry
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[@vandine47]

Beautiful

[@PreMath]

Thanks for watching! ❤️🙏

[@SaurabhYadav-hr9nk]

Sir please keep​@@PreMath

[@AzouzNacir]

Let H be the projection of point E on CD. We put BC=a and HC=b, and from this a*b=60 and a*(21-b)=192. Solving these two equations, we get a=12 and b=5, and from this x=√(12²+16²)=20 and y=√(12²+5²)=13.

[@rohitgupta6845]

Same approach bro🎉

[@marioalb9726]

Triangle CDE: A = ½b.h = (96+30)= 126cm² h = 2A/b= 2*126/21= 12cm b₁=b(A₁/A)=16cm ; b₂=b(A₂/A)=5cm Pytagorean theorem: x²= b₁²+h² --> x = 20 cm y²= b₂²+h² --> y = 13 cm

[@MrPaulc222]

That triangle configuration means that the white triangle occupies half the rectangle. Rectangle area = 252 cm^2. Therefore, h = 252/21 = 12. 192/12 = 16 and 60/12 = 5. 16^2 + 12^2 = 256 + 144 = 400 = x^2 Therefore, x = 20. 5^2 + 12^2 = 25 + 144 = 169 = y^2 Therefore, y = 13. On reflection, 12, 16, 20 and 5, 12, 13 are both Pythagorean triples or multiples thereof.

[@marioalb9726]

A₁=½b₁h= 96cm² ; A₂=½b₂.h=30cm² A₁+A₂= ½h(b₁+b₂)= 96+30= 126 cm² h= 2(A₁+A₂)/(b₁+b₂)= 2*126/21= 12cm b₁=2A₁/h= 16cm ; b₂= 21-b₁= 5cm x²= b₁²+h² --> x = 20 cm y²= b₂²+h² --> y = 13 cm

[@鈞齊]

Notice that ABCD is rectangle, so ∆AED and ∆BEC have the same height(AD=BC) That means the following formula holds Area(∆AED):Area(∆BEC)=AE:BE => AE:BE=96:30 => AE:BE=16:5 and we have AE+BE=AB=CD=21 => AE=16, BE=5 5×BC=2×30 => BC=12=AD x=√(16²+12²)=20, y=√(5²+12²)=13

[@mohammedkhettab9965]

❤❤❤❤❤

[@alexniklas8777]

AE/BE= 96/30= 16/5; AE= 16, BE= 5, AD= BC= 2×30(96)/5(16)= 12. DE= √(16^2+12^2)= √400= 20; CE= √(5^2+12^2)= 13.

[@Kanad-q3b]

x= 20 y= 13 easy

[@jamestalbott4499]

Thank you!

[@alster724]

Pretty straightforward

[@santiagoarosam430]

AD=(96+30)*2/21=12---> EB=2*30/12=5---> EA=21-5=16 ---> EC=√(5²+12²)=13=y ; ED=4*5=20=x. Gracias y un saludo cordial

[@AsifAbdullah-jo1eb]

Since two shaded triangles have same height, then a/(21-a)=96/30=16/5 Comparing both sides of equation, a=16 21-a=5 Heiight = 12cm By Pythagorean theorem x=20cm y=13cm

[@marcgriselhubert3915]

Be h = AD = BC; a = AE and b = EB. The area of the green triangle is (1/2).a.h = 96, so a.h = 198. The area of the yellow triangle is (1/2).b.h = 30, so b.h = 60 We then have a/b = 198/60 = 16/5. There is a real k as a = 16.k and b = 5.k. Now knowing hat a + b = DC = 21, we have 16.k + 5.k = 21 and so k = 1 We then have a = 16 and b = 5. We now calculate h: b.h = 60, so h = 60/b = 60/5 = 12. In the green triangle x^2 = a^2 + h^2 = 16^2 + 12^2 = 400, so x = 20. In the yellow triangle y^2 = b^2 + h^2 = 5^2 + 12^2 = 169, so y = 13.

[@AmirgabYT2185]

(20; 13)

[@michaelstahl1515]

Great video again . Thanks . I choosed another solution and AD = a , AB = b . Than EB = 21 - b and so on . I got without the white triangle x = 20 LE and y = 13 LE.

[@PreMath]

Nice work! Thanks for the feedback ❤️🙏

[@cyruschang1904]

Rectangle area = 2(96 + 30) cm^2 = 252 cm^2 = (21 cm)(width) width = (252/21) cm = 12 cm x^2 = (12 cm)^2 + [(21 cm)(96)/(96 + 30)]^2 x = √(12^2 + 16^2) cm = 20 cm y^2 = (12 cm)^2 + [(21 cm)(30)/(96 + 30)]^2 y = √(12^2 + 5^2) cm = 13 cm

[@giuseppemalaguti435]

20,13

[@Waldlaeufer70]

BE = 21 / 126 * 30 = 5 AE = 21 - 5 = 16 AD = BC = 2 * 30 / 5 = 12 x² = 16² + 12² = 400 x = 20 y² = 5² + 12² ? 25 + 144 = 169 y = 13

[@himo3485]

(96+30)*2=252 252/21=12 12*AE/2=96 AE=16 12*EB/2=30 EB=5 x=√[12^2+16^2]=20 y=√[5^2+12^2]=13

[@SkinnerRobot]

Beautifully concise writeup of the solution. Thanks.

[@phungpham1725]

1/ h= 12 2/ 1/2 . h.DF= 96-> DF= 16-> DE= x= 20 ( 3-4-5 triples) 3/ FC= 5-> sq y= sq5 + sq12= 25+144=169 -> y= 13 😅😅😅

[@Christopher-e7o]

X,2x+5=8

[@pas6295]

Not D but E

[@sergioaiex3966]

Solution: Green Area = ½ b h 96 = ½ a h ah = 192 ... ¹ Yellow Area = ½ b h 30 = ½ (21 - a) h 60 = 21h - ah ... ² Replacing Equation ¹ in Equation ², we will get: 60 = 21h - 192 21h = 252 h = 252/21 h = 12 Substituting h in Equation ¹ a (12) = 192 a = 16 Applying Pythagorean Theorem to calculate "x": (16)² + (12)² = x² x² = 256 + 144 x² = 400 x = 20 cm ✅ Applying Pythagorean Theorem, once again, now to calculate "y": (21 - 16)² + (12)² = y² 25 + 144 = y² y² = 169 y = 13 cm ✅ Thus, x = 20 cm ✅ y = 13 cm ✅

[@pas6295]

First we name the rectangle as ABCD. , Let AD meet BC at D. We r given AD ss 21 cms. And so also BC as 21 cms. D is the point where the line AE meets BC. From the triangles ABE and ECD we r given the areas as 96 sq Cms and 30 Sq Cms. Let us assume the distance BE as Z Cms. Then EC is 21-Z Cms. With the help of the areas of two triangles we by Applying Pythagoras therem as Angle at B and at C. We cansolve for Z as two triangles give u two equations involvinv Z. Similarly AB and CD we can get once we know the Z. Total Area of Rectangle is sum of 96+30+the Area of AED. AE is X and ED as Y Cms.Area of AED can be calculatdd invollvinv Heron theorem But u have Xand Y as two unknowns. Fro. The area of 96 and 30 sq. Cms having got Z we can get the bredth SB and CD. So Area of Rectangle is 21×Bredth AB. From the rectangle Area -(96+30) give u the area of AED. Once u get Area of AED. Where u have the two u knowns and each X and Y. We can solve for Xand Y. Hence the answer

[@DB-lg5sq]

شكرا لكم على المجهودات يمكن استعمال BC=t EB=60/t AE=192/t 60/t + 192/t =21 t = 12 x=20 y=13

[@unknownidentity2846]

Let's find x and y: . .. ... .... ..... From the given diagram we can conclude: A(ADE) + A(BCE) = (1/2)*AE*h(AE) + (1/2)*BE*h(BE) = (1/2)*AE*AD + (1/2)*BE*AD = (1/2)*(AE + BE)*AD = (1/2)*AB*AD = (1/2)*CD*AD 96cm² + 30cm² = (1/2)*(21cm)*AD 126cm² = (1/2)*(21cm)*AD ⇒ BC = AD = 2*126cm²/(21cm) = 12cm A(ADE) = (1/2)*AE*AD ⇒ AE = 2*A(ADE)/AD = 2*96cm²/12cm = 16cm ⇒ BE = AB − AE = 21cm − 16cm = 5cm Now we can apply the Pythagorean theorem to the right triangles ADE and BCE: x² = DE² = AD² + AE² = (12cm)² + (16cm)² = 144cm² + 256cm² = 400cm² ⇒ x = √(400cm²) = 20cm y² = CE² = BC² + BE² = (12cm)² + (5cm)² = 144cm² + 25cm² = 169cm² ⇒ y = √(169cm²) = 13cm Best regards from Germany

[@SkinnerRobot]

Very nice and thorough writeup.

[@ManojkantSamal]

X=20, y=13......May be Explain later

[@texitaliano64]

Numero i vertici del rettangolo in senso antiorario partendo dal vertice in basso a sinistra ABCD assegno E al vertice del triangolo sul lato CD, chiamo x il lato AE e y il lato BE del triangolo, chiamo a il lato DE e b il lato EC, l'altezza del rettangolo la chiamo h. area triangolo EDA S1=a*h/2=96 area triangolo BCE S2=b*h/2=30 b=21-a S2=(21-a)*h/2=30 area rettangolo ABCD S=21*h area triangolo ABE S3=21*h/2 S=S3+S1+S2 sostituendo abbiamo che 21*h=21*h/2+96+30 42*h=21*h+126*2 42*h-21*h-252=0 21*h-252=0 h=252/21=12 h=12 a*h/2=96 a*12/2=96 a=96/6=16 a=16 b=21-16=5 b=5 con pitagora ricavo x x=sqrt(a^2+h^2) x=sqrt(16^2+12^2) x=sqrt(400)=20 x=20 con pitagora ricavo y y=sqrt(b^2+h^2) y=sqrt(5^2+12^2) y=sqrt(169)=13 y=13

[@fhffhff]

((x-1)/y')'*y'²+5(x/y)'*y²=cosx ((x-1)/y')'(y'/y)²+5(x/y)'=0 x/y=z y=x/z z(x²-x)z''+(-2x²+2x)z'²+(5x²+x-2)zz'+z²=0

[@LuisdeBritoCamacho]

This Problem was quite easy. Area of Triangle [ABCD] = (2 * 96) + (2 * 30) = 192 + 60 = 252 AD = BC = 252 / 21 ; AD = BC = 12 AE = X BE = Y X * 12 = 192 ; X = 16 Y * 12 = 60 ; Y = 5 Using PT we get : Triangle [AED] = (12 ; 16 ; 20) = ((4 * 3) ; (4 * 4) ; (4 * 5)) Triangle [BCE] = (5 ; 12 ; 13) Answer x = 20 and y = 13