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Mathematics · 2024 · 4 marks
CBSE 2024 · Region 1 · Set 1 · Q38
A backyard is in the shape of a triangle ABC with right angle at B. $\displaystyle \mathrm{AB}=7 \mathrm{~m}$ and $\displaystyle \mathrm{BC}=15 \mathrm{~m}$. 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 $\displaystyle \mathrm{AP}=x \mathrm{~m}$.
Based on the above information, answer the following questions :(i)Find the length of AR in terms of $\displaystyle x$.(ii)Write the type of quadrilateral BQOR.(iii)Find the length PC in terms of $\displaystyle x$ and hence find the value of $\displaystyle x$.Find $\displaystyle x$ and hence find the radius $\displaystyle r$ of circle.
A backyard is in the shape of a triangle ABC with right angle at B. $\displaystyle \mathrm{AB}=7 \mathrm{~m}$ and $\displaystyle \mathrm{BC}=15 \mathrm{~m}$. 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 $\displaystyle \mathrm{AP}=x \mathrm{~m}$.
Based on the above information, answer the following questions :
(i)
Find the length of AR in terms of $\displaystyle x$.
(ii)
Write the type of quadrilateral BQOR.
(iii)
Find the length PC in terms of $\displaystyle x$ and hence find the value of $\displaystyle x$.
Find $\displaystyle x$ and hence find the radius $\displaystyle r$ of circle.
Marking-scheme solution
(i)
$\displaystyle \mathrm{AR}=x \mathrm{~m}$
(ii)
Quad. ORBQ is a square.
(iii)
$\displaystyle \mathrm{PC}=8+x$
\[\begin{aligned}
& \mathrm{AC}^{2}=(8+2 x)^{2}=49+225=274 \\
\Rightarrow & 8+2 x=\sqrt{274} \\
\Rightarrow & x=\frac{-8+\sqrt{274}}{2} \text { or } 4.28 \text { approx. }
\end{aligned}
\]
$\displaystyle \mathrm{AC}^{2}=(8+2 x)^{2}=49+225=274$
\[\begin{array}{l}
\Rightarrow 8+2 x=\sqrt{274} \\
\Rightarrow x=\frac{-8+\sqrt{274}}{2} \text { or } 4.28 \text { approx. }
\end{array}
\]
Hence, radius $\displaystyle r=7-\mathrm{x}=7-\left(-4+\frac{\sqrt{274}}{2}\right)$
\[=\left(11-\frac{\sqrt{274}}{2}\right) \text { or } 2.72 \text { approx. }
\]
Therefore, radius of the circle is $\displaystyle \left(11-\frac{\sqrt{274}}{2}\right) \mathrm{m}$ or $\displaystyle 2.72$ m approx.
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