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A circle, having center at $(2, 3)$ and radius $6$, crosses $y$-axis at the points $P$ and $Q$. The straight line with equation $x= 1$ intersects the radii $CP$ and $CQ$ at points $R$ and $S$ respectively. Find the area of the trapezium $PQSR$.

I am getting stuck to find the length of parallel sides? As area of a trapezium is $A=\frac{a+b}{2} \cdot h$

here $a$ = $SR$ , b = $PQ $

How to find $h$?

Rick
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3 Answers3

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$h = 1$ since the distance between the $y$-axis ($x=0$) and $x=1$ is $1$. The tricky part is finding $a$ and $b$.

Toby Mak
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Hint:

The equation of the circle is $$(x-2)^2 +(y-3)^2 =36$$ To find $P$ and $Q$, set $x=0$ and solve the quadratic in $y$. Now, find the equations of lines $CP$ and $CQ$, and finally set $x=1$ in these equations to find $R$ and $S$. You have everything you need now.

Vishu
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  • @Statistics What do you mean by ‘strange’ values? – Vishu May 25 '20 at 11:33
  • sorry.My calculations were wrong! thanks for your answer.it is indeed helpful –  May 25 '20 at 11:38
  • Setting x =1 ; u will never find R and S . U will find the y value which cut the circle at x = 1.Edit your answer. –  May 25 '20 at 12:00
  • @Statistics Finding the $y$ value is indirectly finding $R$ and $S$ themselves, it should be obvious. – Vishu May 25 '20 at 12:02
  • No. How can u find R and S value after finding (1,y1) & (1,y2)? –  May 25 '20 at 12:03
  • According to me u have to find the eq using y-y1 = m(x-x1) which also will support R and S ,in this way u have to proceed.Did u mean that? –  May 25 '20 at 12:05
  • @Statistics Yes. R and S are precisely $(1,y_1)$ and $(1,y_2)$. No need to do anything else. – Vishu May 25 '20 at 12:05
  • R value can never be (1,y1) the y value of R will be less than y1,same goes to S also –  May 25 '20 at 12:06
  • @ Tavish No, at x = 1 u are getting , y^2-6y-26 = 0 ,upon solving the values u get of y that 'll intersect the circle not CP and CQ. check once more.U will get it –  May 25 '20 at 12:13
  • @Statistics I never told you to put $x=1$ in the equation of the circle. If you read carefully, it says put $x=1$ in the equation of the lines $CP$ and $CQ$. – Vishu May 25 '20 at 12:28
  • i was also saying that. after finding the eqn of CP and CQ u have to put x =1 there,then u will get the required co-ordinates. –  May 25 '20 at 12:42
  • @Statistics So what’s the problem? The answer I wrote was correct and you just misread it. – Vishu May 25 '20 at 13:05
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Using the Pythagorean theorem you can find $PQ$. $$PQ = 2\cdot\sqrt{r^2 - (2h)^2},$$ where $r$ is the radius of the circle, and $h$ is the altitude of the trapezoid $PQSR$. $$RS = \frac{PQ}{2},$$ because the triangles $PCQ$ and $RCS$ are similar and $PC = 2RC.$ So you can plug in the values and find the area of the trapezoid by using $A = h\cdot\frac{PQ + RS}{2}$. $$r = 6, h =\frac{HC}{2} = 1, PH = 4\sqrt 2, PQ = 8\sqrt 2, RS = 4\sqrt 2,$$ $$A = 6\sqrt 2$$ enter image description here

Tigran Harutyunyan
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