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2 Problems

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1:  The line $$y = (3x + 7)/4$$ intersects the circle $$x^2 + y^2 = 25$$ at A and B. Find the length of chord $$\overline{AB}$$.

and

2:  Points A(0,0), B(9,6) and C(6,12) are vertices of triangle ABC. Point D is on segment AB such that 2(AD) = DB, point E is on segment BC such that 2(BE) = EC and point F is on segment CA such that 2(CF) = FA. What is the ratio of the area of triangle DEF to the area of triangle ABC? Express your answer as a common fraction.

Thanks!

Jun 14, 2018

#3
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2:  Points A(0,0), B(9,6) and C(6,12) are vertices of triangle ABC.
Point D is on segment AB such that 2(AD) = DB,
point E is on segment BC such that 2(BE) = EC and
point F is on segment CA such that 2(CF) = FA.
What is the ratio of the area of triangle DEF to the area of triangle ABC?

$$\begin{array}{|rcll|} \hline D &=& \dbinom{9}{6}\cdot \dfrac13 \\ &=& \dbinom{3}{2} \\\\ E &=& \dbinom{9}{6} + \left( \dbinom{6}{12} - \dbinom{9}{6} \right)\cdot \dfrac13 \\ &=& \dbinom{9}{6} + \dbinom{-3}{6}\cdot \dfrac13 \\ &=& \dbinom{9}{6} + \dbinom{-1}{2} \\ &=& \dbinom{8}{8} \\\\ F &=& \dbinom{6}{12}\cdot \dfrac23 \\ &=& \dbinom{12}{24}\cdot \dfrac13 \\ &=& \dbinom{4}{8} \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \text{Area}_\text{ABC} &=& \frac12 \left( 0\cdot 6 - 9\cdot 0 + 9 \cdot 12 - 6 \cdot 6 + 6\cdot 0 - 0 \cdot 12 \right) \\ &=& \frac12 \left( 9 \cdot 12 - 6 \cdot 6 \right) \\ &=& \frac12 \left( 72 \right) \\ &=& 36 \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \text{Area}_\text{DEF} &=& \frac12 \left( 3\cdot 8 - 8 \cdot 2 + 8 \cdot 8 - 4 \cdot 8 + 4\cdot 2 - 3 \cdot 8 \right) \\ &=& \frac12 \left( 24-16+64-32+8-24 \right) \\ &=& \frac12 \left( 24 \right) \\ &=& 12 \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \dfrac { \text{Area}_\text{DEF} } {\text{Area}_\text{ABC}} &=& \dfrac{12}{36} = \dfrac13 \\ \hline \end{array}$$  Jun 15, 2018
edited by heureka  Jun 15, 2018
edited by heureka  Jun 15, 2018
edited by heureka  Jun 15, 2018
#4
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1:

The line

$$y = (3x + 7)/4$$

y = (3x + 7)/4

intersects the circle

$$x^2 + y^2 = 25$$

x^2 + y^2 = 25

at A and B.

Find the length of chord .

$$\begin{array}{|rcll|} \hline x^2+y^2 &=& 25 \quad & | \quad y = \dfrac{3x+7}{4} \\ x^2+ \left(\dfrac{3x+7}{4}\right)^2 &=& 25 \\\\ x^2+ \dfrac{(3x+7)^2}{16} &=& 25 \quad & | \quad \cdot 16 \\\\ 16x^2+ (3x+7)^2 &=& 25\cdot 16 \\ 16x^2+ (3x+7)^2 &=& 400 \\ 16x^2+ 9x^2+42x + 49 &=& 400 \\ 25x^2+42x + 49-400 &=& 0 \\ 25x^2+42x - 351 &=& 0 \\ x &=& \dfrac{-42\pm \sqrt{42^2-4\cdot 25\cdot(-351) } } {2\cdot 25} \\\\ x &=& \dfrac{-42\pm \sqrt{1764+100\cdot(351) } } {50} \\\\ x &=& \dfrac{-42\pm \sqrt{36864} } {50} \\\\ x &=& \dfrac{-42\pm 192} {50} \\\\ x_a & =& \dfrac{-42 + 192} {50} \\ \mathbf{x_a} &\mathbf{=}& \mathbf{3} \\ y_a &=& \dfrac{3\cdot 3+7}{4} \\ \mathbf{y_a} & \mathbf{=}& \mathbf{ 4 } \\\\ \mathbf{x_b} & \mathbf{=}& \mathbf{\dfrac{-42 - 192} {50} } \\ x_b &=& -4.68 \\ y_b &=& \dfrac{3\cdot (-4.68)+7}{4} \\ \mathbf{y_b} & \mathbf{=}& \mathbf{-1.76} \\ \hline \end{array}$$ The length of chord AB:

$$\begin{array}{|rcll|} \hline && \sqrt{(x_a-x_b)^2+(y_a-y_b)^2} \\ &=& \sqrt{[3-(-4.68)]^2+[4-(-1.76)]^2} \\ &=& \sqrt{(3+4.68)^2+(4+1.76)^2} \\ &=& \sqrt{7.68^2+5.76^2} \\ &=& \sqrt{92.16} \\ &=& 9.6 \\ \hline \end{array}$$

The length of chord AB is 9.6 Jun 15, 2018
#5
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Thanks so much Heureka!  Still not sure what happened on the other one haha!

AnonymousConfusedGuy  Jun 15, 2018