# Please solve q 58 ?

May 20, 2018

Choice 3 is correct

#### Explanation:

Given: $\setminus \frac{\setminus \overline{A B}}{\setminus \overline{B C}} = \setminus \frac{\setminus \overline{C D}}{\setminus \overline{A C}} = \setminus \frac{\setminus \overline{A D}}{\setminus \overline{D E}} = k$

Required: Find ${\left(\setminus \frac{\setminus \overline{A E}}{\setminus \overline{B C}}\right)}^{2}$

Analysis: use Pythagorean Theorem $c = \setminus \sqrt{{a}^{2} + {b}^{2}}$

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Solution: Let, $\setminus \overline{B C} = x$, $\setminus \because \setminus \frac{\setminus \overline{A B}}{\setminus \overline{B C}} = k ,$

$\setminus \overline{A B} = k x$, use Pythagorean Theorem to find the value of $\setminus \overline{A C}$ :

 \overline{AC} = \sqrt{\overline{BC}^2 + \overline{AB}^2 } = \sqrt{ x^2 + k^2x^2} = \sqrt{(x^2)(1+k^2)} = x\sqrt{1+k^2}

$\setminus \overline{A C} = x \setminus \sqrt{1 + {k}^{2}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\setminus \because \setminus \frac{\setminus \overline{C D}}{\setminus \overline{A C}} = k ,$ $\setminus \overline{C D} = \setminus \overline{A C} \cdot k = x k \setminus \sqrt{1 + {k}^{2}}$

Use Pythagorean Theorem to find the value of $\setminus \overline{A D}$ :

 \overline{AD} = \sqrt{\overline{CD}^2 + \overline{AC}^2

$= \setminus \sqrt{{\left(x k \setminus \sqrt{1 + {k}^{2}}\right)}^{2} + {\left(x \setminus \sqrt{1 + {k}^{2}}\right)}^{2}}$

$= \setminus \sqrt{{x}^{2} {k}^{2} \left(1 + {k}^{2}\right) + {x}^{2} \left(1 + {k}^{2}\right)}$

$= \setminus \sqrt{{x}^{2} \left[{k}^{2} \left(1 + {k}^{2}\right) + 1 \left(1 + {k}^{2}\right)\right]}$

$= x \setminus \sqrt{\left({k}^{2} + 1\right) \left(1 + {k}^{2}\right)}$, thus

$\setminus \overline{A D} = x \left(1 + {k}^{2}\right)$

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$\setminus \because \setminus \frac{\setminus \overline{A D}}{\setminus \overline{D E}} = k ,$

$\setminus \overline{D E} = \setminus \frac{\setminus \overline{A D}}{k} = \setminus \frac{x}{k} \cdot \left(1 + {k}^{2}\right)$

Use Pythagorean Theorem to find the value of $\setminus \overline{A E}$ :

 \overline{AE}^2 = \sqrt{\overline{DE}^2 +\overline{AD}^2 =

 = \sqrt{(frac{x}{k} * (1+k^2))^2 + (x(1+k^2))^2

 = \sqrt{(x^2/k^2)(1+k^2)^2 + (x^2)(1+k^2)^2

 = x\sqrt{(1/k^2 + 1)(1+k^2)^2

$= x \setminus \sqrt{\setminus \frac{1 + {k}^{2}}{{k}^{2}} {\left(1 + {k}^{2}\right)}^{2}}$

Thus,
 \overline{AE} = x\sqrt{\frac{(1+k^2)^3}{k^2}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

${\left(\setminus \frac{\setminus \overline{A E}}{\setminus \overline{B C}}\right)}^{2}$

$= {\left(\setminus \frac{x \setminus \sqrt{\setminus \frac{{\left(1 + {k}^{2}\right)}^{3}}{{k}^{2}}}}{x}\right)}^{2}$

$= {\left(\setminus \sqrt{\setminus \frac{{\left(1 + {k}^{2}\right)}^{3}}{{k}^{2}}}\right)}^{2}$

Thus,
${\left(\setminus \frac{\setminus \overline{A E}}{\setminus \overline{B C}}\right)}^{2} = \setminus \frac{{\left(1 + {k}^{2}\right)}^{3}}{{k}^{2}}$

May 21, 2018

I got ${\left({k}^{2} + 1\right)}^{3} / {k}^{2}$ which is choice (3).

#### Explanation:

We're gonna do every problem in Rahul's book!

This one is weird though with a diagram with right angles that aren't. Is it supposed to be 3D? The middle fraction is upside down compared to the others; let's assume that's correct.

Rahul, you deserve a better book.

We'll renotate for sanity:

$b = A B , c = A C , d = A D , e = A E , p = B C , q = C D , r = D E$

We're given

$k = \frac{b}{p} = \frac{q}{c} = \frac{d}{r}$

We want to find ${e}^{2} / {p}^{2} ,$ a hint that we'll never have to write a square root.

$b = p k , \quad \quad q = k c , \quad \quad r = \frac{d}{k}$

${c}^{2} = {b}^{2} + {p}^{2} = {p}^{2} {k}^{2} + {p}^{2} = {p}^{2} \left(1 + {k}^{2}\right)$

${d}^{2} = {c}^{2} + {q}^{2} = {c}^{2} + {\left(k c\right)}^{2} = {c}^{2} \left(1 + {k}^{2}\right) = {p}^{2} {\left(1 + {k}^{2}\right)}^{2}$

${e}^{2} = {d}^{2} + {r}^{2} = {d}^{2} \left(1 + \frac{1}{k} ^ 2\right) = {p}^{2} {\left(1 + {k}^{2}\right)}^{2} \left(1 + \frac{1}{k} ^ 2\right)$

${e}^{2} / {p}^{2} = {\left(1 + {k}^{2}\right)}^{2} \left(1 + \frac{1}{k} ^ 2\right) = {\left({k}^{2} + 1\right)}^{3} / {k}^{2}$

Choice (3)