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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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 – Dinesh Chavan · 3 years ago

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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. – Michael Tong · 3 years ago

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– Dinesh Chavan · 3 years ago

Wow nice approach. Thanks for one more method.Log in to reply

121 – Naila Rahma Jamilah · 2 years, 10 months ago

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96.04 – Gamal Adli · 3 years ago

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96.04 – Khaled Saeed · 3 years ago

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96.04 – Ashutosh Singh · 3 years ago

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96.04 – Ashutosh Singh · 3 years ago

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96.04 – Anupam Gahoi · 3 years ago

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96.04 – Srinjoy Mukherjee · 3 years ago

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SOLVE THIS PROBLEM – Brajesh Raj · 3 years ago

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SOLVE THIS PROBLEM – Brajesh Raj · 3 years ago

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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 – Abdo Hamdy · 3 years ago

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– Abdo Hamdy · 3 years ago

trueLog in to reply

(x+y+z)power2 =100 – Abdo Hamdy · 3 years ago

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– Abdo Hamdy · 3 years ago

trueLog 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 – Subramaniam Sivasailam · 3 years ago

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my answer is 96.04 – Andy Leonardo · 3 years ago

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correct ans : 96.04 – Yash Bhansali · 3 years ago

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96.04 – Liên Lee · 3 years ago

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96.04 – Rakesh Kumar · 3 years ago

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96.04 – Magdy Essafty · 3 years ago

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96.04 – Archana Panigrahi · 3 years ago

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96.04 is the ans – Rohan Kumar · 3 years ago

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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=96.04 – Vishudh Nagpal · 3 years agoLog in to reply

96.04 – Azizul Islam · 3 years ago

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(9.8)^2 means 96.04 is the answer. – Rubayet Tusher · 3 years ago

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96.04 – Andrew Tiu · 3 years ago

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Ans is 96.04 – Achint Gupta · 3 years ago

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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 – Azhar Ahmad · 3 years ago

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(24/5 + 5)^2 – Pulkit Kogta · 3 years ago

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2401/25 – Hassan Bohra · 3 years ago

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96 should b d answer – Ratnesh Kumar · 3 years ago

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121 – Joyce Yeo · 3 years ago

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The answer is 289 – Muh. Amin Widyatama · 3 years ago

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ans is 84 – Suvadip Sana · 3 years ago

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