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 \(\left \lfloor (x+y+z)^2 \right \rfloor\)

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

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.

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

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

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestWell 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

Log in to reply

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.

Log in to reply

Wow nice approach. Thanks for one more method.

Log in to reply

121

Log in to reply

96.04

Log in to reply

96.04

Log in to reply

96.04

Log in to reply

96.04

Log in to reply

96.04

Log in to reply

96.04

Log in to reply

SOLVE THIS PROBLEM

Log in to reply

SOLVE THIS PROBLEM

Log in to reply

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

Log in to reply

true

Log in to reply

(x+y+z)power2 =100

Log in to reply

true

Log in to reply

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

Log in to reply

my answer is 96.04

Log in to reply

correct ans : 96.04

Log in to reply

96.04

Log in to reply

96.04

Log in to reply

96.04

Log in to reply

96.04

Log in to reply

96.04 is the ans

Log in to reply

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

Log in to reply

(9.8)^2 means 96.04 is the answer.

Log in to reply

96.04

Log in to reply

Ans is 96.04

Log in to reply

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

Log in to reply

(24/5 + 5)^2

Log in to reply

2401/25

Log in to reply

96 should b d answer

Log in to reply

121

Log in to reply

The answer is 289

Log in to reply

ans is 84

Log in to reply