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.
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.
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.
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
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
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
Simpler: Let x=DC and y=BC. The rectangle's area is xy. Since triangle EDC has area xy/3 and its base DC=x, then its height ED must be 2/3 of AD, so ED:AE=2:1, ED=10 and y=15. Similarly, in triangle FBC, FB:AF=2:1, FB=14 and x=21. The desired area is xy=21*15=315.
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 😀
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.
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
Solution: a = AB = DC, b = AD = BC. I look at the lower left triangle. Its area on one side is a*(b-5)/2. And because the area of this triangle is 1/3 of the entire rectangle area, the equation applies: a*(b-5)/2 = 1/3*a*b |*2/a ⟹ b-5 = 2/3*b |+5-2/3*b ⟹ b/3 = 5 |*3 ⟹ b = 15 I do the same with the upper right triangle and get the equation: (a-7)*b/2 = 1/3*a*b |*2/b ⟹ a-7 = 2/3*a |+7-2/3*a ⟹ a/3 = 7 |*3 ⟹ a = 21 ⟹ Area of the entire rectangle = a*b = 21*15 = 315
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
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.
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😅😅😅
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
Wow! You presented a brilliant solution. Thank you, PreMath.
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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.
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.
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 √)
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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.
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Thank you! I liked to the exposure to multiple equations when solving.
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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
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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
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
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*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_*
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My solution:1/3 (ABCD) = (ECD), so x*(y-5)/2=x*y/3, 3xy-15x=2xy , xy=15x and y=15, x=21
OK, you obtain y = 15, but you need something else to obtain x. With only one equation you cannot find two unknown values.
@@marcgriselhubert3915 You are absolutely right. I also used: 1/3 (ABCD) = (FBC) with y*(x-7)/2=x*y/3 and x=21.
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@@WernHerr OK then.
@@WernHerrAre you sure
Area of the three parts are equal
S(ABC) = 1,5S, S(FBC) = S. AF : BF = AE : DE = 1:2. (7 × 3) × (5 × 3) = 315.
Simpler: Let x=DC and y=BC. The rectangle's area is xy. Since triangle EDC has area xy/3 and its base DC=x, then its height ED must be 2/3 of AD, so ED:AE=2:1, ED=10 and y=15. Similarly, in triangle FBC, FB:AF=2:1, FB=14 and x=21. The desired area is xy=21*15=315.
Nice! x(y - 5) = y(x - 7) = 2xy/3 → y = 5x/7 → y(x - 7) = 2xy/3 → x - 7 = 2x/3 → x = 21 → y = 15
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
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@@PreMath Thank YOU
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
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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 😀
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.
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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
(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)
Solution:
a = AB = DC,
b = AD = BC.
I look at the lower left triangle. Its area on one side is
a*(b-5)/2. And because the area of this triangle is 1/3 of the entire rectangle area, the equation applies:
a*(b-5)/2 = 1/3*a*b |*2/a ⟹
b-5 = 2/3*b |+5-2/3*b ⟹
b/3 = 5 |*3 ⟹
b = 15
I do the same with the upper right triangle and get the equation:
(a-7)*b/2 = 1/3*a*b |*2/b ⟹
a-7 = 2/3*a |+7-2/3*a ⟹
a/3 = 7 |*3 ⟹
a = 21 ⟹ Area of the entire rectangle = a*b = 21*15 = 315
I solved it simply by using the third equation: 3S =(x)*(y).
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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
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
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.
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
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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
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Can you find area of the Rectangle? Answer : yes, I can
(Justify your answer) : because I'm clever.
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شكرا لكم على المجهودات
يمكن استعمال
S(ABCD)= 3S
S=y(x-7=x(y-5)
........
S(ABCD)=315
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😅😅😅
3S
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 ✅
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315
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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
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S=105)(15×21=315fullarea
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I have mentioned that it is a trapezium NOT A TRAPEZOID. Trapezoid is something else. Please correct your videos!
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.
Excellent!
Thanks for sharing ❤️
Too long
315 Sq. Units