IPMAT Indore 2025 · Quantitative Ability · Question 15

Step 1 — put coordinates on the square. Let B=\left( 0,0 \right), C=\left( 18,0 \right), D=\left( 18,18 \right), A=\left( 0,18 \right). Then side AB is the line x=0 and side BC is the line y=0.

Step 2 — locate the centre. The circle touches both x=0 and y=0, so its centre is 13 away from each:

O=\left( 13,\,13 \right), M=\left( 0,13 \right), N=\left( 13,0 \right)

Circle: {\left( x-13 \right)}^{2}+{\left( y-13 \right)}^{2}=169.

Step 3 — find P on side CD (x=18).

{\left( 18-13 \right)}^{2}+{\left( y-13 \right)}^{2}=169 \Rightarrow{\left( y-13 \right)}^{2}=144 \Rightarrow y=25\text{or}1

Only y=1 lies on the square, so P=\left( 18,1 \right).

(Similarly Q=\left( 1,18 \right) on side DA — not needed for this question.)

Step 4 — area of triangle PMD with P\left( 18,1 \right), M\left( 0,13 \right), D\left( 18,18 \right). Use the coordinate formula

\text{Area}=1/2|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)|

=1/2|18\left( 13-18 \right)+0\left( 18-1 \right)+18\left( 1-13 \right)|=1/2|-90-216|=306/2=153

Shortcut: PD is vertical, of length 18-1=17, and M is a horizontal distance 18 from it, so area =1/2\times 17\times 18=153.

Answer: 153