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A backyard is in the shape of a triangle ABC with right angle at B. AB = $7$ m and BC = $15$ m. A circular pit was dug inside it such that it touches the walls AC, BC and AB at P, Q and R respectively such that AP = $x$ m.
Based on the above information, answer the following questions :
(i) Find the length of AR in terms of $x$.
(ii) Write the type of quadrilateral BQOR.
(iii) (a) Find the length PC in terms of $x$ and hence find the value of $x$.
OR
(b) Find $x$ and hence find the radius $r$ of circle.
Based on the above information, answer the following questions :
(i) Find the length of AR in terms of $x$.
(ii) Write the type of quadrilateral BQOR.
(iii) (a) Find the length PC in terms of $x$ and hence find the value of $x$.
OR
(b) Find $x$ and hence find the radius $r$ of circle.
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Sol. (i) AR = $x$ m
(ii) Quad. ORBQ is a square.
(iii) (a) PC = $8 + x$
AC$^2 = (8 + 2x)^2 =49 + 225 =274$
$\Rightarrow 8 + 2x = \sqrt{274}$
$x = \frac{-8+\sqrt{274}}{2}$ or $4.28$ approx.
OR
(iii) (b) AC$^2 = (8 + 2x)^2 =49 + 225 =274$
$8 + 2x = \sqrt{274}$
$x = \frac{-8+\sqrt{274}}{2}$ or $4.28$ approx.
Hence, radius $r= 7 - x = 7 - \left(\frac{-4 + \sqrt{274}}{2}\right)$
$= \left(11 - \frac{\sqrt{274}}{2}\right)$ or $2.72$ approx.
Therefore, radius of the circle is $\left(11 - \frac{\sqrt{274}}{2}\right)$ m or $2.72$ m approx.
(ii) Quad. ORBQ is a square.
(iii) (a) PC = $8 + x$
AC$^2 = (8 + 2x)^2 =49 + 225 =274$
$\Rightarrow 8 + 2x = \sqrt{274}$
$x = \frac{-8+\sqrt{274}}{2}$ or $4.28$ approx.
OR
(iii) (b) AC$^2 = (8 + 2x)^2 =49 + 225 =274$
$8 + 2x = \sqrt{274}$
$x = \frac{-8+\sqrt{274}}{2}$ or $4.28$ approx.
Hence, radius $r= 7 - x = 7 - \left(\frac{-4 + \sqrt{274}}{2}\right)$
$= \left(11 - \frac{\sqrt{274}}{2}\right)$ or $2.72$ approx.
Therefore, radius of the circle is $\left(11 - \frac{\sqrt{274}}{2}\right)$ m or $2.72$ m approx.