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IPMAT Indore 2025

Geometry
>
Circles

Medium

A circle of radius 1313 cm touches the adjacent sides AB and BC of a square ABCD at M and N, respectively. If AB = 1818 cm and the circle intersects the other two sides CD and DA at P and Q, respectively, then the area, in sq. cm, of triangle PMD is

Entered answer:

✅ Correct Answer: 153
We need to find the area of triangle PMD in a square ABCD where a circle of radius 1313 cm touches sides AB and BC at points M and N respectively, and intersects sides CD and DA at points P and Q.
Let's place the square in a coordinate system with A at origin (0,0)(0,0), B at (18,0)(18,0), C at (18,18)(18,18), and D at (0,18)(0,18).
The center of the circle must be at a distance of 1313 cm from both sides AB and BC.
Since the circle touches side AB (x-axis) and side BC (line x=18x = 18), the center is at: \newline (18−13,13)=(5,13)(18-13, 13) = (5, 13)
Point M is where the circle touches side AB. \newline Since AB is along the x-axis, M is directly below the center: M=(5,0)M = (5, 0)
To find point P on side CD, we need to find where the circle intersects the line y=18y = 18.
The equation of the circle is: \newline (x−5)2+(y−13)2=132(x-5)^2 + (y-13)^2 = 13^2
Substituting y=18y = 18: \newline (x−5)2+(18−13)2=132(x-5)^2 + (18-13)^2 = 13^2 \newline (x−5)2+25=169(x-5)^2 + 25 = 169 \newline (x−5)2=144(x-5)^2 = 144 \newline x−5=±12x-5 = \pm 12
So x=17x = 17 or x=−7x = -7 \newline Since P must be on CD, P=(17,18)P = (17, 18)
Now we have the coordinates: \newline M=(5,0)M = (5, 0) \newline P=(17,18)P = (17, 18) \newline D=(0,18)D = (0, 18)
Using the formula for area of a triangle with coordinates:
Area = 12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣\dfrac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|
Area = 12∣5(18−18)+17(18−0)+0(0−18)∣\dfrac{1}{2}|5(18-18) + 17(18-0) + 0(0-18)|
= 12∣0+17(18)+0∣\dfrac{1}{2}|0 + 17(18) + 0|
= 12(306)\dfrac{1}{2}(306)
= 153153 sq. cm
Therefore, the area of triangle PMD is 153153 sq. cm.

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