-1

enter image description here

I am not Able to proceed with this question .

Abhinav Kumar Singh
  • 265
  • 1
    A great way to approach these kinds of questions is to list what you know. What information does the diagram give you? Next time, please include this information in your question itself –  Mar 09 '19 at 04:29

2 Answers2

1

how can PQ be independent of a? I am getting .6 aenter image description here

maveric
  • 2,168
0

I get that the length of "?" is 3a/5.

This is probably the longest possible way, so here goes.

Let the two points on the left and right ends of the line labled "?" be P and Q.

Put the origin at the base of the circle near "a".

Let $\angle DAP = \angle ADQ =g$, and let $\sin g = s, \cos g = c, \tan g = t $.

The equation of $AP$ is $y = t(a/2-x) $ and the equation of $BQ$ is $y = t(x+a/2) $.

These intersect when $x = 0$ at $(0, at/2)$. This is the center of the circle. Call this $R$.

The distance $AP = a$, and

$\begin{array}\\ AR^2 &=(a/2)^2+(ta/2)^2\\ &=(a/2)^2(1+t^2)\\ &=(a/2)^2(1+(s/c)^2)\\ &=(a/2)^2(1/c)^2\\ \text{so}\\ AR &=a/(2c) \end{array} $

Therefore the radius of the circle is $a-a/(2c) =a(1-1/(2c)) $.

But this has to be $at/2$, so $at/2 =a(1-1/(2c)) $ or $t =2-1/c $ or $s/c = 2-1/c$ or $s = 2c-1$.

Squaring, $4c^2-4c+1 =s^2 =1-c^2 $, so $5c^2 = 4c$ so $c = 0$ or $c = 4/5$.

From this, $s = \sqrt{1-16/25} =3/5$, and, indeed, $s = 2c-1$.

Also, $t = s/c = 3/4$.

If point $P$ is at $(-u, v)$, then $(u+a/2)^2+v^2 = a^2 $ and $v/(u+a/2) =t$ or $v = t(u+a/2) $ so

$\begin{array}\\ a^2 &=(u+a/2)^2+t(u+a/2)^2 \\ &=(u+a/2)^2(1+t^2)\\ &=(u+a/2)^2(1/c^2)\\ \text{so}\\ ac &=u+a/2\\ \text{or}\\ u &=a(c-1/2)\\ &=3a/10\\ \end{array} $

Finally, the length of $?$ is $2u = 6a/10 =3a/5 $.

marty cohen
  • 107,799