Can you find area of the Rectangle? | (Justify your answer) |

[@SkinnerRobot]

Wow! You presented a brilliant solution. Thank you, PreMath.

[@beiranvand4066]

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[@PreMath]

Glad to hear that! You are very welcome! Thanks for the feedback ❤️

[@PrithwirajSen-nj6qq]

Join AC diagonal. Now triangle ADC =triangle ABC >🔺 ADC - 🔺 EDC = 🔺 ABC - triangle FBC > 🔺 ACE = 🔺 ACF =S/2 As ACE=S/2=1/2* base 5, height x 🔺 EDC =S/2*2=1*2*(2*5)*height x So y =5+2*5=15 Likewise x = 7+7*2=21 Area of the rectangle =15*21=315 sq units

[@PrithwirajSen-nj6qq]

The big rectangle is an enlargement of 7*5 rectangle . To divide the big rectangle in three equal areas, we have to get trice of length and breadth of 7*5 rectangle It means the length and breadth of the big rectangle will be 7*3=21 and 5*3=15 Area of rectangle =21*15=315 sq units.

[@marioalb9726]

S= xy/3 = ½x(y-5) = ½y(x-7) ⅔xy = xy -5x ⅔y = y-5 ⅓y = 5 --> y = 15 cm ⅔xy = xy -7y ⅔x = x-7 ⅓y = 7 --> x = 21 cm A = xy = 15*21 = 315cm² (Solved √)

[@PreMath]

Excellent! Thanks for sharing ❤️

[@onyemaechiepelle4302]

This is your teaching I enjoyed most; thanks.

[@PreMath]

You are very welcome! Thanks for the feedback ❤️

[@andrewrose3830]

The central shape can be split into two triangles with the rectangle diagonal. This diagonal splits rectangle area in half, so must split the central shape in half two. Each triangle has area S/2. Using "area = (base x height)/2", this means FB must be twice 7 (FBC has twice area of FCA) so AB is 21. Similarly, ED is twice 5, so AD is 15. This area is 21*15 or 315.

[@ИванПоташов-о8ю]

Draw AC. Then Area(ABC)=1,5S, Area(AFC)=0,5S. AF:FB=Area(AFC): Area (FBC)=1:2. So, FB=2AF=14 and AB=21. AD=3AE= 15 (the same proof) Area(ABCD)=15*21=315

[@PreMath]

Excellent! Thanks for sharing ❤️

[@jamestalbott4499]

Thank you! I liked to the exposure to multiple equations when solving.

[@PreMath]

Excellent! Glad to hear that! Thanks for the feedback ❤️

[@yalchingedikgedik8007]

That’s very nice and enjoyable Thanks Sir You are very good With glades ❤❤❤❤❤

[@PreMath]

Thanks for the kind words!🙏

[@allanflippin2453]

I found a very simple approach. I don't see one similar in the comments. My apologies if I missed one. 1) Draw diagonal AC. This divides the rectangle into two equal-sized triangles. 2) The total rectangle area is 3S, so each triangle has area 1.5S 3) Consider triangles ACF and BCF. Both have the same height (BC length), but the area of ACF is half of BCF. That means AF is half the length of BF. Thus BF = 14. 4) Apply the same idea to ACE and DCE. AE is half the length of ED, so ED = 10. 5) AB = 7 + 14 = 21. AD = 5 + 10 = 15. Total area = 21 * 15 = 315.

[@PreMath]

Excellent! Thanks for sharing your solution.

[@uwelinzbauer3973]

Hallo professor! I started like this: I added an auxiliary line connecting points A and C, dividing quadrilateral AECF into two triangles, making up the equation: S=(1/2)*5*x+(1/2)*7*y Remaining proceeding similar, three equations with three unknowns, leading to identical results. Thanks for sharing this interesting geometric puzzle 👍 Happy weekend to you and the friends of the channel 😀

[@soli9mana-soli4953]

Being XY = 3S On 1) X*(Y-5) = 2S we have 5X = S On 2) Y*(X-7) = 2S we have 7Y = S tracing a perpendicular from F to DC and a perpendicular from E to BC we split the main rectangle in 4 small rectangles whose areas are: 1) left high = 5*7 = 35 2) right high = 5X - 35 = S - 35 3) left low = 7Y - 35 = S - 35 4) right low = (X-7)*(Y-5)= XY - 7Y - 5X +35 = 3S - S - S + 35 = 35 + S then multiplying crossed rectangles: 35*(S+35) = (S-35)*(S-35) S = 105 area = 3S = 3*105 = 315

[@beiranvand4066]

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[@imetroangola17]

*Apontamento:* xy - 5x=2S, xy - 7y=2S e xy+ 7y=4S. Ora, xy = 3S. Então: 3S - 5x=2S, 3S - 7y=2S e 3S+ 7y=4S. Assim, S= 5x e S=7y → *x=7y/5.* Logo, [ABCD] = xy= 7y²/5 e, por outro lado, [ABCD] = 3S = 21y. Assim, 7y²/5 = 21y → y = 15. Portanto, [ABCD] = 21×15 = *_315 U.Q_*

[@beiranvand4066]

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[@PreMath]

Excellent! Thanks for sharing ❤️

[@marcgriselhubert3915]

With ED = a and FB = b, we have: rectangle area = 3.S = (7 +b).(5 +a); 2.triangle EDC area = 2.S = a.(7 +b); 2.triangle CBF area = 2.S = b.(5 +a). We replace (7+ b) by (3.S)/(5+ a) in the second equation and obtain (a.(3.S))/ (5+ a) = 2.S, we simplify by S and obtain (3.a)/(5+ a) = 2 giving a = 10 In the same way we replace (5+ b) by (3.S)/(7+ b) in the third equation and obtain (3.b)/(7+ b) = 2, giving b = 14 The side lengths of the rectangle are then AD = 5 + a = 15 and AB = 7 + b = 21, and the rectangle area is 15.21 = 315.

[@beiranvand4066]

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[@PreMath]

Excellent! Thanks for sharing ❤️

[@cyruschang1904]

Rectangle area = 3S = WD = (7 + x)(5 + y) y/(5 + y) = 2/3 3y = 10 + 2y y = 10 x/(7 + x) = 2/3 3x = 14 + 2x x = 14 Rectangle area = (7 + x)(5 + y) = 21(15) = 315

[@PreMath]

Excellent! Thanks for sharing ❤️

[@cyruschang1904]

@@PreMath Thank YOU

[@Ibrahimfamilyvlog2097l]

Thanks sir good ❤❤

[@PreMath]

You are very welcome! Thanks for the feedback ❤️

[@JoanRosSendra]

Otra forma: Trazamos la diagonal AC y la mediatriz desde el vértice C de los triángulos CFD y CED La figura queda dividida en 6 triángulos y cada uno de ellos tiene la misma área: S/2 Por tanto, los que tienen la altura "y" han de tener la misma base (7), así que x=7*3=21 Del mismo modo, los triángulos que tienen la altura "x" han de tener la misma base (5), así que y=5*3=15 Área solicitada = 21*15 = 315 u²

[@santiagoarosam430]

bh/2= S+{5b/2)=S+(7h/2) ---> b=7h/5---> 35n²/3=35n ---> n=3---> b*h=(3*7)*(3*5) =21*15=315. Gracias y saludos

[@beiranvand4066]

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[@PreMath]

Excellent! Thanks for sharing ❤️

[@adept7474]

S(ABC) = 1,5S, S(FBC) = S. AF : BF = AE : DE = 1:2. (7 × 3) × (5 × 3) = 315.

[@PrithwirajSen-nj6qq]

Sir 3*🔺 CBE(S) =3[1/2*x*(y -5])=.xy (area of rectangle) > y =15 Now3 🔺 CBF =3[1/2*y*(x-7)]=xy >x =21 Area of the rectangle =21*15=315 sq units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@Birol731]

My way of solution ▶ Let's divide the quadrilateral AECF into two triangles : ΔAEC + ΔACF [AF]= 7 [FB]= y [AE]= 5 [ED]= x A(AECF)= S ⇒ A(ΔAEC)= 5*(7+y)/2 A(ΔACF)= 7*(5+x)/2 ⇒ S= 5*(7+y)/2 + 7*(5+x)/2 S= (70+5y+7x)/2 b) the triangle ΔEDC: A(ΔAEC)= x*(7+y)/2 A(ΔAEC)= (7x+xy)/2 c) the triangle ΔFCB: A(ΔFCB)= y*(5+x)/2 A(ΔAEC)= (5y+xy)/2 d) A(ΔAEC) = A(ΔAEC) = S (7x+xy)/2 = (5y+xy)/2 7x+xy= 5y+ xy 5y= 7x y= 7x/5 e) A(AECF)= A(ΔAEC)= S (70+5y+7x)/2 = (7x+xy)/2 70+5y= xy y= 7x/5 ⇒ 70+ 5*(7x/5)= x*(7x/5) 70+7x= 7x²/5 both sides multiplied by 5 350+35x= 7x² ⇒ 50+5x= x² x²-5x-50=0 Δ= 25-4*1*(-50) Δ= 225 √Δ= 15 x₁= (5+15)/2 x₁= 10 x₂= (5-15)/2 x₂= -5 ❌ x₂ < 0 ⇒ x= 10 y= 7*10/5 y= 14 S= (7x+xy)/2 S= (7*10+10*14)/2 S= 105 square units Arectangle= 3S Arectangle= 3*105 Arectangle= 315 square units ✅

[@PreMath]

Excellent! Thanks for sharing ❤️

[@phungpham1725]

1/ Label a and b as the width and lenght of the rectangle respectively. Focus on the area of the triangle EDC: (a-5).b/2=ab/3-> (a-5)/2=a/3-> a= 15 Similarly, (b-7).a/2= ab/3 -> b=21 Area of the rectangle=15x21= 315 sq units😅😅😅

[@RealQinnMalloryu4]

(5)^2 (7)^2={25+49}=74 {90°A+90°B+90°C+90°D}=360°ABCD/74=4.64 2^2.2^3^2^2 1^1.1^3^1^2 3^2 ((ABCD ➖ 3ABCD+2)

[@JoanRosSendra]

Yo tracé una perpendicular desde F al punto G del segmento DC, dividiendo la figura en dos rectángulos. El área de FBCG = 2S Por tanto el área de AFGD = S Si igualamos tenemos que (x-7)*y=2(7*y) xy-7y=14y xy=21y x=21 Igualando las áreas de los triángulos rectángulos: (x-7)y/2=(y-5)x/2 xy-7y=xy-5x y=5x/7 y=5*21/7=15 Área solicitada = 21*15 = 315 u²

[@misterenter-iz7rz]

Creative puzzle🎉. a(5+b)=b(7+a)=5(7+a)+7(5+b), 5a+ab=7b+ab=70+5a+7b, a(5¹×⁹1⁰/7)=70+5a, 』),

[@beiranvand4066]

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[@PreMath]

Excellent! Glad to hear that! Thanks for the feedback ❤️

[@zdrastvutye]

it is once again a nested calculation: 10 print "premath-can you find area of the rectangle":dim x(5),y(5) 20 l1=5:l2=7:sw=sqr(l1^2+l2^2)/57:n=l1*l2:goto 130 30 da1=l2*(l1+l3):da2=l4*(l1+l3)/2:a1=l3*(l2+l4)/2:a2=da1+da2-a1 40 a3=l4*(l1+l3)/2:dg=(a2-a1)/n:return 50 l4=sw:gosub 30 60 l41=l2:dg1=dg:l4=l4+sw:if l4>20*l1 then 110 70 l42=l4:gosub 30:if dg1*dg>0 then 60 80 l4=(l41+l42)/2:gosub 30:if dg1*dg>0 then l41=l4 else l42=l4 90 if abs(dg)>1E-10 then 80 110 return 120 gosub 50:df=(a3-a2)/n:return 130 l3=sw 140 gosub 120:if l4>20*l1 then else 160 150 l3=l3+sw:goto 140 160 l31=l3:df1=df:l3=l3+sw:l32=l3:gosub 120:if df1*df>0 then 160 170 l3=(l31+l32)/2:gosub 120:if df1*df>0 then l31=l3 else l32=l3 180 if abs(df)>1E-9 then 170 else print a1,"%",a2,"%",a3:goto 200 190 xbu=x*mass:ybu=y*mass:return 200 masx=1200/(l2+l4):masy=850/(l1+l3):if masx run in bbc basic sdl and hit ctrl tab to copy from the results window.

[@himo3485]

AD=BC=x AB=DC=y ED=x-5 DC=y FB=y-7 BC=x (x-5)y/2=(y-7)x/2 xy-5y=xy-7x 5y=7x y=7x/5 AECF=5*7x/5*1/2 + 7*x*1/2 = 7x S = 7x S*3 = 7x*3 = 21x = x*7x/5 = 7x²/5 7x²/5 - 21x = 0 7x² -105x = 0 7x(x - 15) = 0 x>0 , x=15 Rectangle area = 21*15 = 315

[@beiranvand4066]

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[@PreMath]

Excellent! Thanks for sharing ❤️

[@pranavamali05]

👋

[@PreMath]

Excellent!😀 Thanks for the feedback ❤️

[@WernHerr]

My solution:1/3 (ABCD) = (ECD), so x*(y-5)/2=x*y/3, 3xy-15x=2xy , xy=15x and y=15, x=21

[@marcgriselhubert3915]

OK, you obtain y = 15, but you need something else to obtain x. With only one equation you cannot find two unknown values.

[@WernHerr]

@@marcgriselhubert3915 You are absolutely right. I also used: 1/3 (ABCD) = (FBC) with y*(x-7)/2=x*y/3 and x=21.

[@beiranvand4066]

kzbin.info/www/bejne/Zn6kZoGKoNx5nKssi=zMDpJAFhS83KjP4b

[@marcgriselhubert3915]

@@WernHerr OK then.

[@srimathisubramaniyam9099]

​@@WernHerrAre you sure Area of the three parts are equal

[@aljawad]

I solved it simply by using the third equation: 3S =(x)*(y).

[@beiranvand4066]

kzbin.info/www/bejne/Zn6kZoGKoNx5nKssi=zMDpJAFhS83KjP4b

[@PreMath]

Thanks for the feedback ❤️

[@КонстантинМирошниченко-щ5о]

5*x=S 7*y=S y=5*3 (т.к. S 3шт.) x=7*3 (т.к. S 3шт.) y=15 x=21 area=15*21=315

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... Let w=AB=CD and h=AD=BC be the width and the height of the rectangle, respectively. Since the triangles BCF and CDE have the same area as the quadrilateral AECF, we can conclude: A(ABCD) = w*h = 3*S ⇒ S = w*h/3 (1) S = w*h/3 (2) S = A(BCF) = (1/2)*BC*BF = (1/2)*BC*(AB − AF) = (1/2)*h*(w − 7) (3) S = A(CDE) = (1/2)*CD*DE = (1/2)*CD*(AD − AE) = (1/2)*w*(h − 5) (4) S = A(AECF) = A(ACE) + A(ACF) = (1/2)*AE*CD + (1/2)*AF*BC = (1/2)*5*w + (1/2)*7*h = 5*w/2 + 7*h/2 The combination of equation (2) and equation (3) results in: (1/2)*h*(w − 7) = (1/2)*w*(h − 5) h*(w − 7) = w*(h − 5) h*w − 7*h = h*w − 5*w −7*h = −5*w ⇒ w = 7*h/5 The combination of this result with equation (1) and equation (4) results in: w*h/3 = S = 5*w/2 + 7*h/2 = 5*(7*h/5)/2 + 7*h/2 = 7*h/2 + 7*h/2 = 7*h ⇒ w = 21 ⇒ h = 5*w/7 = 5*21/7 = 15 Now we are able to calculate the area of the rectangle: A(ABCD) = w*h = 21*15 = 315 Best regards from Germany

[@PreMath]

Excellent! Thanks for sharing ❤️

[@anthonyheaton5798]

3S

[@giuseppemalaguti435]

315

[@PreMath]

Excellent! Thanks for sharing ❤️

[@LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL : 01) AF = 7 lin un 02) FB = X lin un 03) AB = (7 + X) lin un 04) AE = 5 lin un 05) ED = Y lin un 06) AD = (5 + Y) lin un 07) Rectangle Area = (AB * AD) sq un ; RA = (7 + X) * (5 + Y) ; RA = (35 +7Y + 5X + XY) sq un 08) Bottom Triangle [CDE] Area = 2 * S(1) = Y * (7 + X) ; 2 * S(1) = (7Y + XY) sq un 09) Middle Quadrilateral [AECF] Area = 2 * S(2) = (7 * (5 + Y)) + (5 * (7 + X)) ; 2 * S(2) = (35 + 7Y + 35 + 5X) ; 2 * S(2) = (70 + 5X + 7Y) sq un 10) Upper Triangle [BCF] Area = 2 * S(3) = X * (5 + Y) ; 2 * S(3) = (5X + XY) sq un 11) 2 * S(1) = 2 * S(2) = 2 * S(3) 12) 2 * S(1) = 2 * S(2) ; 7Y + XY = 70 + 5X + 7Y ; XY = 70 + 5X 13) 2 * S(1) = 2 * S(3) ; 7Y + XY = 5X + XY ; 7Y = 5X 14) 2 * S(2) = 2 * S(3) ; 70 + 5X +7Y = 5X + XY ; 70 + 7Y = XY 15) Solving this System of Equations I get these Positive Solutions : X = 14 and Y = 10 16) Rectangle Area = (AB * AD) sq un ; RA = (35 + 7(10) + 5(14) + (14)(10)) sq un 17) Rectangle [ABCD] Area = 35 + 70 + 70 + 140 ; Rectangle [ABCD] Area = 280 + 35 ; Rectangle [ABCD] Area = 315 sq un Therefore, OUR FINAL ANSWER : Rectangle Area must be equal 315 Square Units.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@wasimahmad-t6c]

S=105)(15×21=315fullarea

[@beiranvand4066]

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[@PreMath]

Excellent! Thanks for sharing ❤️

[@nuhumaishanu6944]

Too long

[@rajivb9493]

315 Sq. Units