Let AD=a & AB=b We know ((a-x).(b-7))/2=7 Expanding and simplifying ab-7a-xb+7x=14. We can see triangles DAB and PFB are similar, so the ratio of the sides are the same, so x/a=(b-7)/b -> xb=ab-7a. Substituting back into the last formula we get xb-xb+7x=14 -> 7x=14 -> x=2
Let FB=y and AD=h. The area of triangle FEB is yh/2 = 7 + xy/2, simplifying to yh=14+xy. The area of triangle ADB is ((7+y)*h)/2=7*(x+h)/2 + xy/2, simplifying to yh+7h=7x+7h+xy, or yh=7x+xy. Setting these two equations equal to each other, we get: 14+xy=7x+xy, or 14=7x, or x=2.
Yes, I used this approach as well and found it to be very easy. That is after I got over the idea of setting up three variables to solve for! Fortunately, solving for x is easy, although h and y are indeterminate.
@@allanflippin2453 True: h must be greater than 2 and y must be greater than 0. Otherwise, they can vary in the relationship: y=14/(h-2). A solution that sort of fits the scale would be h=5.5 and y=4; but there is no definitive information to determine that.
Solution by special case: The problem statement implies that the solution is the same regardless of the length of BF, as long as a valid figure is produced. Let BF = 7 and, if EP is treated as the base of ΔBEP, BF is the height. Area = 7 = (1/2)(EP)(BF) = (1/2)(EP)(7), so EP = 2. We note that ΔDEP and ΔBFP are similar but also have equal corresponding sides, DE = AF = 7 by opposite sides of a rectangle being equal, and BF = 7, so the triangles are congruent as well, and corresponding sides EP and PF are equal, so x = PF = EP = 2. On a multiple choice test, select x = 2 as your answer. On a test where you are not required to show your work, write x = 2 as your answer. If you are required to show your work and have solved the general case, the special case serves as a check that you have solved the problem correctly. If you have not solved the general case, solving the special case may lead you to a solution for the general case as follows: Let EP have length b (base of ΔBEP) and let BF have length h (height of ΔBEP, BF = h instead of 7). So Area ΔBEP = (1/2)bh = 7 and bh =14. By ratio of sides of similar triangles ΔDEP and ΔBFP, BF/DE = PF/EP, h/7 = x/b, x = bh/7 = 14/7 = 2. (Others posted basically the same solution earlier than I did.)
Let the triangle BEP have base EP=b and height FB=h. Note that AB = h+7 and AD = x+b. Since the area of BEP is 7=bh/2, we have b=14/h. Therefore AD = x + 14/h. We can show that triangles ABD and FBP are similar. Therefore AB/AD = FB/FP or (h+7)/(x+14/h) = h/x. Cross multiplying yields x(h+7) = (x+14/h)h, or xh+7x = xh+14. Thus 7x=14 or x=2.
*Solução:* Sejam FB = a e EP = b. Assim, [BPE] = ab/2= 7→ ab = 14. Os triângulos DAB e BPF são semelhantes, logo: AD/FP = AB/FB, como AD=EF, então: (x+b)/x = (7+ a)/a ax + ab = 7x + ax 7x = ab, como ab=14, então: 7x = 14 *x = 2.*
L = the length of the rectangle The lower-right white triangle’s two legs = (L - 7 cm) & x The upper-left triangle is similar to the lower-right white triangle. The two legs of the upper-left white triangle = 7 cm & (7 cm)(x)/(L - 7 cm) The base and height of the yellow triangle = the short leg of the upper-left triangle and the long leg of the lower-right triangle [(7 cm)(x)/(L - 7 cm)] (L - 7 cm)/2 = 7 cm^2 (7 cm)(x) = 14 cm^2 x = 2 cm
Let us draw a perpendicular starting from point P and intersecting AD at a point G. So DG equal to EP. Let FB = y. From the area of the triangle EPB we can say that EP = 14/y, thus DG = 14/y. Since the triangles DGP and PFB are similar. DG/7 = x/y. Substituting DG =14/y we get 2/y = x/y, so x= 2
@ 2:12 is the very one reason I wear a Bowtie, because I like to project an aura of expertise and knowledge. Don't matter if the Bowtie is skewed a bit, it still symbolizes my deep reflection and intellect.😎
Here is something else: We name a = PE and b = FB. The area of the yellow triangle PBE is (1/2).a.b and is also equal to 7, so a.b = 14 and b = 14/a. Triangles DCB and DEP are similar, so DE/EP = DC/CB, so 7/a = (7+b)/(a+x), or 7/a = (7+(14/a))/(a+x), or 7.(a+x) = a.(7+(14/a)), Then we have 7.a +7.x =7.a +14, which gives that 7.x = 14, and so x = 2.
*Hebrews 13:6* So we can confidently say "The Lord is my helper; I will not fear; what can man do to me?" *John 14:1* Let not your hearts be troubled. Believe in God; Believe also in me. *Proverbs 28:26* Whoever trusts his own mind is a fool, but he who walks in wisdom will be delivered. *Ephesians 2:8-9* 8 For by grace you have been saved through faith, and this is not from you; it is the gift of God; 9 it is not from works, so no one may boast. *Romans 10:9* If you declare with your mouth, "Jesus is Lord", and believe in your heart that God raised him from the dead, you will be saved. *Romans 3:23-24* [23] for all have sinned and fall short of the glory of God, [24] and all are justified freely by his grace through the redemption that came by Christ Jesus. *John 14:6* Jesus answered, 'I am the way and the truth and the life. No one comes to the Father except through me.' *Acts 4:12* "Salvation is found in no one else, for there is no other name given under heaven by which we must be saved."
MY RESOLUTION PROPOSAL : 01) First we must acknowledge that Triangle (DEP) is similar to Triangle (BFP). 02) By the Law of Proportion we have : 03) (EP / 7) = (X / BF) ; (EP * BF) = 7X 04) As we can easily see : (EP * BF) = 14, because Triangle (BEP) = 7; where Base = EP and height = BF. So : (EP * BF) = 14 sq cm 05) So, 14 = 7X 06) X = 14 / 7 07) X = 2 cm Therefore, MY BEST ANSWER IS : X must be equal to: 2 Centimeters.
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½bh = ½7h = A₂+7
A₁+A₂ = A₂+7
A₁ = 7cm² = ½bh= ½.x.7
x = 2 cm ( Solved √)
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Let AD=a & AB=b
We know ((a-x).(b-7))/2=7
Expanding and simplifying
ab-7a-xb+7x=14.
We can see triangles DAB and PFB are similar, so the ratio of the sides are the same, so x/a=(b-7)/b -> xb=ab-7a.
Substituting back into the last formula we get xb-xb+7x=14 -> 7x=14 -> x=2
Good question sir
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Let FB=y and AD=h. The area of triangle FEB is yh/2 = 7 + xy/2, simplifying to yh=14+xy. The area of triangle ADB is ((7+y)*h)/2=7*(x+h)/2 + xy/2, simplifying to yh+7h=7x+7h+xy, or yh=7x+xy. Setting these two equations equal to each other, we get: 14+xy=7x+xy, or 14=7x, or x=2.
Yes, I used this approach as well and found it to be very easy. That is after I got over the idea of setting up three variables to solve for! Fortunately, solving for x is easy, although h and y are indeterminate.
@@allanflippin2453 True: h must be greater than 2 and y must be greater than 0. Otherwise, they can vary in the relationship: y=14/(h-2). A solution that sort of fits the scale would be h=5.5 and y=4; but there is no definitive information to determine that.
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Thats Sensational, Super excellent
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Muy ingenioso!
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I appreciate Myself I got a very worthy teacher ❤
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[EPB] = ½EP*FB=7
∆DEP is Similar to ∆PFB.
EP/ED=x/FB
x= EP*FB/ED
x = 14/7=2
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Solution by special case: The problem statement implies that the solution is the same regardless of the length of BF, as long as a valid figure is produced. Let BF = 7 and, if EP is treated as the base of ΔBEP, BF is the height. Area = 7 = (1/2)(EP)(BF) = (1/2)(EP)(7), so EP = 2. We note that ΔDEP and ΔBFP are similar but also have equal corresponding sides, DE = AF = 7 by opposite sides of a rectangle being equal, and BF = 7, so the triangles are congruent as well, and corresponding sides EP and PF are equal, so x = PF = EP = 2.
On a multiple choice test, select x = 2 as your answer. On a test where you are not required to show your work, write x = 2 as your answer. If you are required to show your work and have solved the general case, the special case serves as a check that you have solved the problem correctly. If you have not solved the general case, solving the special case may lead you to a solution for the general case as follows: Let EP have length b (base of ΔBEP) and let BF have length h (height of ΔBEP, BF = h instead of 7). So Area ΔBEP = (1/2)bh = 7 and bh =14. By ratio of sides of similar triangles ΔDEP and ΔBFP, BF/DE = PF/EP, h/7 = x/b, x = bh/7 = 14/7 = 2. (Others posted basically the same solution earlier than I did.)
Thanks for sharing your solution! 🙏
Let the triangle BEP have base EP=b and height FB=h. Note that AB = h+7 and AD = x+b. Since the area of BEP is 7=bh/2, we have b=14/h. Therefore AD = x + 14/h. We can show that triangles ABD and FBP are similar. Therefore AB/AD = FB/FP or (h+7)/(x+14/h) = h/x. Cross multiplying yields x(h+7) = (x+14/h)h, or xh+7x = xh+14. Thus 7x=14 or x=2.
*Solução:*
Sejam FB = a e EP = b. Assim,
[BPE] = ab/2= 7→ ab = 14.
Os triângulos DAB e BPF são semelhantes, logo:
AD/FP = AB/FB, como AD=EF, então:
(x+b)/x = (7+ a)/a
ax + ab = 7x + ax
7x = ab, como ab=14, então:
7x = 14
*x = 2.*
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L = the length of the rectangle
The lower-right white triangle’s two legs = (L - 7 cm) & x
The upper-left triangle is similar to the lower-right white triangle. The two legs of the upper-left white triangle = 7 cm & (7 cm)(x)/(L - 7 cm)
The base and height of the yellow triangle = the short leg of the upper-left triangle and the long leg of the lower-right triangle
[(7 cm)(x)/(L - 7 cm)] (L - 7 cm)/2 = 7 cm^2
(7 cm)(x) = 14 cm^2
x = 2 cm
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@@PreMath Thank YOU 😊
Let us draw a perpendicular starting from point P and intersecting AD at a point G. So DG equal to EP. Let FB = y. From the area of the triangle EPB we can say that EP = 14/y, thus DG = 14/y. Since the triangles DGP and PFB are similar. DG/7 = x/y. Substituting DG =14/y we get 2/y = x/y, so x= 2
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FB=b,AD=h...h/(7+b)=x/b...bh=(7+b)x=7x+bx...=>bh-bx=7x...Area(triangolo)=b(h-x)/2=7...bh-bx=14=>x=14/7=2
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∆DEP подобен ∆BFP. DE=7, FB=y, S(EPB)=EP*FB/2=7, EP=14/y. DE/FB=EP/FP, 7/y=(14/y)/x. 7/y=14/(x*y). x=2.
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DFB=EFB
DPF+PFB=EPB+PFB
DPF=7cm^2
x*7/2=7 x=2(cm)
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b/x=( b+7)/h
bh/2-bx/2=7
bh-bx=14
bh=14+bx
bh=(b+7)x
14+bx = bx+7x
14=7x
x=2
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EP=2*7/FB=14/FB---> EP/ED=FP/FB---> 14/FB*7=FP/FB---> FP=14/7=2=X.
Gracias y un saludo cordial.
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Let's find x:
.
..
...
....
.....
The right triangles BFP and ABD are obviously similar, so we can conclude:
FP/AD = BF/AB
FP/EF = BF/(AF + BF)
FP/(EP + FP) = BF/(AF + BF)
x/(EP + x) = BF/(AF + BF)
x*(AF + BF) = BF*(EP + x)
AF*x + BF*x = BF*EP + BF*x
AF*x = BF*EP
⇒ x = BF*EP/AF = 2*(1/2)*BF*EP/AF = 2*(1/2)*EP*h(EP)/AF = 2*A(BEP)/AF = 2*7cm²/(7cm) = 2cm
Best regards from Germany
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@ 2:12 is the very one reason I wear a Bowtie, because I like to project an aura of expertise and knowledge. Don't matter if the Bowtie is skewed a bit, it still symbolizes my deep reflection and intellect.😎
😀
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شكرا لكم على المجهودات
يمكن استعمال
BC=L
BF=T
T(L-x)=2×7
tan(DBA)=x/T = L/(7+T)
......
x/T = 2/T
x=2
Here is something else: We name a = PE and b = FB. The area of the yellow triangle PBE is (1/2).a.b and is also equal to 7, so a.b = 14 and b = 14/a.
Triangles DCB and DEP are similar, so DE/EP = DC/CB, so 7/a = (7+b)/(a+x), or 7/a = (7+(14/a))/(a+x), or 7.(a+x) = a.(7+(14/a)),
Then we have 7.a +7.x =7.a +14, which gives that 7.x = 14, and so x = 2.
1/ Consider the yellow triangle: FBxEP = 14 (1)
2/ Because of similarity: FB/7= x/EP --> FB= 7x/EP (2)
3/ From (1) and (2) we have: 7x= 14
--> x= 2 units😅😅😅
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Got it
Using x/(FB) = (EP)/7
And 1/2 • (EP•FB) = 7
2
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1st Comment
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*Hebrews 13:6*
So we can confidently say "The Lord is my helper; I will not fear; what can man do to me?"
*John 14:1*
Let not your hearts be troubled. Believe in God; Believe also in me.
*Proverbs 28:26*
Whoever trusts his own mind is a fool, but he who walks in wisdom will be delivered.
*Ephesians 2:8-9*
8 For by grace you have been saved through faith, and this is not from you; it is the gift of God; 9 it is not from works, so no one may boast.
*Romans 10:9*
If you declare with your mouth, "Jesus is Lord", and believe in your heart that God raised him from the dead, you will be saved.
*Romans 3:23-24*
[23] for all have sinned and fall short of the glory of God, [24] and all are justified freely by his grace through the redemption that came by Christ Jesus.
*John 14:6*
Jesus answered, 'I am the way and the truth and the life. No one comes to the Father except through me.'
*Acts 4:12*
"Salvation is found in no one else, for there is no other name given under heaven by which we must be saved."
Jesus Christ is Lord. Turn to Christ before its too late
Wait, didn't the world end a couple of days back?
MY RESOLUTION PROPOSAL :
01) First we must acknowledge that Triangle (DEP) is similar to Triangle (BFP).
02) By the Law of Proportion we have :
03) (EP / 7) = (X / BF) ; (EP * BF) = 7X
04) As we can easily see : (EP * BF) = 14, because Triangle (BEP) = 7; where Base = EP and height = BF. So : (EP * BF) = 14 sq cm
05) So, 14 = 7X
06) X = 14 / 7
07) X = 2 cm
Therefore,
MY BEST ANSWER IS :
X must be equal to: 2 Centimeters.
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