ABCD is a rectengle.\(AE \perp BD\) and \(CF \perp BD\).Let AE=x unit,ED=BF=y unit and OE=OF=z unit.AB=\(8\) unit;BC=\(6\) unit.What is the value of \(\left \lfloor (x+y+z)^2 \right \rfloor\)

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In triangle ADB, by using Pythagoras theorem, we can say that BD=10.Area of triangle ABD=24=1/2(10)(AE), thus, AE=x=4.8.And as O is the mid-point of BD, DO=z+y=5. Therefore, (x+y+z)^2=(4.8+5)^2=96.04

The diagonals of a rectangle bisect each other. Furthermore, by Pythagorean's theorem, $AC = 10$, so $AO = DO = 5$. Using these same results, we get

$y + z = 5$

$x^2 + z^2 = 25$

$x^2 + y^2 = 36$

Subtracting the last two equations we get $y^2 - z^2 = 11$, so $(y - z)(y+z) = 11$. Substituting the first equation, we get $y-z = \frac{11}{5}$. Doing a linear combination with $y+z = 5$ we get $2y = \frac{36}{5}$ so $y = \frac{18}{5}$ and $z = \frac{7}{5}$. Using one of the equations to find $x$ we get that $(x+y+z)^2 = (\frac{49}{5})^2 = \frac{2401}{25}$, and the greatest integer less than or equal to that is $96$.

Motivations in this problem: Nothing really too fancy, just finding what we're given using pythagorean's theorem (which should be obvious given the right angles and side lengths of $6$ and $8$) and then solving a system of equations.

I thnk answer is 96.04
In triangle ABC = AB^2 + BC^2 = AC^2
64 + 36 = 100
AC = 10
AC = BD = 10
then 2(y+z) = 10
y+z = 5
z = 5-y (1)
AD = BC= 6
then AD^2 = DE2 + AE2
36 = X2 + Y2 (2)
In Triangle AOE and COF
OE = OF
ANGLE AEO = ANGLE CFO
ANGLE AOE = ANGLE COF
TRIANGLE AOE CONGRUENT TO TRIANGLE COF
THEN AO = OC
so AC= 10 then AO = 5
Also In Triangle AOE AO2 = AE2+OE2
25 = x2 + z2 (3)
25 = x2 + (y-5)2
25 = x2 + y2 + 25 - 10y
0 = x2+y2 - 10y(4)
subtracting 4 from 2 we get
10y = 36
y = 3.6
from (1)
we get z = 1.4
putting value of z in 3rd equation we get
25 = x2 + 1.96
x2 = 23.04
x = 4.8
x+y+z = 1.4+3.6+4.8 = 9.8
(x+y+z)2 = (9.8)2 = 96 .04

Well this problem looks simple
First we see by Pythagoras theorem in triangle $ADB$ that $BD=10$
So We conclude that $y+z=10/2=5$
Now area of triangle $ADB$ is 24
So $24=\frac{1}{2}×BD×AE$ i.e.$24=\frac{1}{2}×10×AE$
i.e. $AE=\frac{48}{10}$
So we get what we desire

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## Comments

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96.04

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96.04

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96.04

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96.04

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2401/25

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96.04

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96.04

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SOLVE THIS PROBLEM

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SOLVE THIS PROBLEM

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2z+2y=10 & (z+y=5)

_(1) ((2z+y)^2)+(x^2)=64 4z^2+y^2+4yz+x^2=64 ((5-z)^2 -z^2=11 25+z^2-10z=11 z=7/5 y=(18/5) x=(24/5) z+y+x=(7/5)+(18/5)+(24/5)) z+x+y=10(z+x+y)^2=100 unit

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true

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(x+y+z)power2 =100

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true

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In triangle ADB, by using Pythagoras theorem, we can say that BD=10.Area of triangle ABD=24=1/2(10)(AE), thus, AE=x=4.8.And as O is the mid-point of BD, DO=z+y=5. Therefore, (x+y+z)^2=(4.8+5)^2=96.04

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my answer is 96.04

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correct ans : 96.04

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96.04

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96.04

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96.04

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96.04

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96.04 is the ans

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answer is 96.04 as here BO=5cm and area boc=6

8/4=12=1/25*x this means x=24/5=4.8 so value=(5+4.8)square=(9.8)square=96.04Log in to reply

96.04

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96 should b d answer

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(9.8)^2 means 96.04 is the answer.

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121

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96.04

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The diagonals of a rectangle bisect each other. Furthermore, by Pythagorean's theorem, $AC = 10$, so $AO = DO = 5$. Using these same results, we get

$y + z = 5$

$x^2 + z^2 = 25$

$x^2 + y^2 = 36$

Subtracting the last two equations we get $y^2 - z^2 = 11$, so $(y - z)(y+z) = 11$. Substituting the first equation, we get $y-z = \frac{11}{5}$. Doing a linear combination with $y+z = 5$ we get $2y = \frac{36}{5}$ so $y = \frac{18}{5}$ and $z = \frac{7}{5}$. Using one of the equations to find $x$ we get that $(x+y+z)^2 = (\frac{49}{5})^2 = \frac{2401}{25}$, and the greatest integer less than or equal to that is $96$.

Motivations in this problem: Nothing really too fancy, just finding what we're given using pythagorean's theorem (which should be obvious given the right angles and side lengths of $6$ and $8$) and then solving a system of equations.

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Wow nice approach. Thanks for one more method.

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Ans is 96.04

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I thnk answer is 96.04 In triangle ABC = AB^2 + BC^2 = AC^2 64 + 36 = 100 AC = 10 AC = BD = 10 then 2(y+z) = 10 y+z = 5 z = 5-y (1) AD = BC= 6 then AD^2 = DE2 + AE2 36 = X2 + Y2 (2) In Triangle AOE and COF OE = OF ANGLE AEO = ANGLE CFO ANGLE AOE = ANGLE COF TRIANGLE AOE CONGRUENT TO TRIANGLE COF THEN AO = OC so AC= 10 then AO = 5 Also In Triangle AOE AO2 = AE2+OE2 25 = x2 + z2 (3) 25 = x2 + (y-5)2 25 = x2 + y2 + 25 - 10y 0 = x2+y2 - 10y(4) subtracting 4 from 2 we get 10y = 36 y = 3.6 from (1) we get z = 1.4 putting value of z in 3rd equation we get 25 = x2 + 1.96 x2 = 23.04 x = 4.8 x+y+z = 1.4+3.6+4.8 = 9.8 (x+y+z)2 = (9.8)2 = 96 .04

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(24/5 + 5)^2

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The answer is 289

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ans is 84

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Well this problem looks simple First we see by Pythagoras theorem in triangle $ADB$ that $BD=10$ So We conclude that $y+z=10/2=5$ Now area of triangle $ADB$ is 24 So $24=\frac{1}{2}×BD×AE$ i.e.$24=\frac{1}{2}×10×AE$ i.e. $AE=\frac{48}{10}$ So we get what we desire

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