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IPMAT 2025 Exam Solved Papers

Question : 10 of 90

Marks: +1, -0

Solution:

The radius of circle is 13 cm. Thus,

\mathrm{BM}=\mathrm{BN}=13 \mathrm{~cm}

and

\mathrm{MA}=\mathrm{DT}=\mathrm{CN}=\mathrm{OT}=5 \mathrm{~cm}

Let the length of CP be x cm. Since

\mathrm{DT}=5 \mathrm{~cm}

, thus

\mathrm{PT}=(13-\mathrm{x}) \mathrm{cm}

In triangle OTP,

O T^{2}+T P^{2}=O P^{2}

\begin{aligned} & \Rightarrow 5^{2}+(13-x)^{2}=13^{2} \\ & \Rightarrow 25+169-26 x+x^{2}=169 \\ & \Rightarrow x^{2}-26 x+25=0 \end{aligned}

\Rightarrow x=1

x=25

. But x cannot be 25 cm, because then the length of the side will be negative. Thus;

=1

Thus,

\mathrm{PD}=13-\mathrm{x}+5=17 \mathrm{~cm}

, and MT = 18 cm

Thus, area of triangle

\mathrm{PMD}=\frac{1}{2} \times 17 \times 18=153 cm^{2}

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