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# Stubborn Rectangle

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

Note by Fazla Rabbi
2 years, 10 months ago

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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 · 2 years, 10 months ago

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. · 2 years, 10 months ago

Wow nice approach. Thanks for one more method. · 2 years, 10 months ago

121 · 2 years, 7 months ago

96.04 · 2 years, 9 months ago

96.04 · 2 years, 9 months ago

96.04 · 2 years, 9 months ago

96.04 · 2 years, 9 months ago

96.04 · 2 years, 9 months ago

96.04 · 2 years, 9 months ago

SOLVE THIS PROBLEM · 2 years, 9 months ago

SOLVE THIS PROBLEM · 2 years, 9 months ago

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 · 2 years, 9 months ago

true · 2 years, 9 months ago

(x+y+z)power2 =100 · 2 years, 9 months ago

true · 2 years, 9 months ago

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 · 2 years, 9 months ago

my answer is 96.04 · 2 years, 9 months ago

correct ans : 96.04 · 2 years, 9 months ago

96.04 · 2 years, 9 months ago

96.04 · 2 years, 9 months ago

96.04 · 2 years, 9 months ago

96.04 · 2 years, 10 months ago

96.04 is the ans · 2 years, 10 months ago

answer is 96.04 as here BO=5cm and area boc=68/4=12=1/25*x this means x=24/5=4.8 so value=(5+4.8)square=(9.8)square=96.04 · 2 years, 10 months ago

96.04 · 2 years, 10 months ago

(9.8)^2 means 96.04 is the answer. · 2 years, 10 months ago

96.04 · 2 years, 10 months ago

Ans is 96.04 · 2 years, 10 months ago

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 · 2 years, 10 months ago

(24/5 + 5)^2 · 2 years, 10 months ago

2401/25 · 2 years, 9 months ago

96 should b d answer · 2 years, 10 months ago

121 · 2 years, 10 months ago

The answer is 289 · 2 years, 10 months ago