In the figure, BC is a diameter of the circle, where $BC=\sqrt{901}, BD=1$, and $DA=16$. If $EC=x$, what is the value of x? [asy][asy]size(2inch); pair O,A,B,C,D,E; B=(0,0); O=(2,0); C=(4,0); D=(.333,1.333); A=(.75,2.67); E=(1.8,2); draw(Arc(O,2,0,360)); draw(B--C--A--B); label("$A$",A,N); label("$B$",B,W); label("$C$",C,E); label("$D$",D,W); label("$E$",E,N); label("Figure not drawn to scale",(2,-2.5),S); [/asy][/asy]
Problem
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Tags: Pythagorean Theorem, geometry
hoangmath
04.11.2012 11:22
Binomial-theorem wrote: In the figure, BC is a diameter of the circle, where $BC = \sqrt{901}, BD = 1$, and $DA = 16$. If $EC = x$, what is the value of x? (look at part b problem #3 for a diagram) By using Pythagorean theorem, we get $CD = 30, AC = 17 = AB$. Thus, $CE = BD = 1$.
Farenhajt
04.11.2012 12:07
Wrong. $AC=\sqrt{30^2+16^2}=34$, hence by Power of a point, $34(34-x)=16\cdot 17\iff 34-x=8\iff x=26$
10000th User
04.11.2012 17:19
To avoid confusion here is a diagram, reproduced with Asy: [asy][asy]size(2inch); pair O,A,B,C,D,E; B=(0,0); O=(2,0); C=(4,0); D=(.333,1.333); A=(.75,2.67); E=(1.8,2); draw(Arc(O,2,0,360)); draw(B--C--A--B); label("$A$",A,N); label("$B$",B,W); label("$C$",C,E); label("$D$",D,W); label("$E$",E,N); label("Figure not drawn to scale",(2,-2.5),S); [/asy][/asy]
NamelyOrange
12.04.2024 14:26
Let $AE = y$, so that by Power of a Point $y(x+y)=16(16+1) = 272$.
Next, draw $BE$. By the Inscribed Arc Theorem, $CEB$ is a right angle, so by Pythag $BE^2 = (16+1)^2-y^2 = (\sqrt{901})^2-x^2$ $\implies x^2-y^2=612$, or $(x+y)(x-y)=612$.
Dividing the two equations gives $\frac{x-y}{y}=\frac{9}{4}$, or $y = \frac{4}{13}x$. Plugging this in to the second equation gives $x^2-\left(\frac{4}{13}x\right)^2=612$ $\implies \frac{153}{169}x^2=612$ $\implies x = \boxed{26}$.
I'm an algebra main, and that really shows here lol
miyukina
12.04.2024 16:05
In the figure, BC is a diameter of the circle, where $BC=\sqrt{901}, BD=1$, and $DA=16$. If $EC=x$, what is the value of x? If BC = √901 and BD = 1, then DC is just √(901 – 1) = 30, and similarly AC = √(16^2 + 30^2) = 34 (1 + 16)^2 = EB^2 + (34 – x)^2 (√901)^2 = EB^2 + x^2 901 – 289 = –34^2 + 2 × 34 × x 17 × (53 – 17) = 17 × (4x – 68) 4x = 68 + 53 – 17 = 104 ===> x = 26