Question

A circle of radius 13 cm touches the adjacent sides AB and BC of a square ABCD at M and N, respectively. If AB = 18 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 ________

Hint:

In geometry problems involving shapes like squares and circles, establishing a coordinate system is often the most straightforward approach. Place one vertex at the origin to simplify coordinates and equations.

Answer 1

Correct Answer: 153

Step 1: Understanding the Concept:
This problem can be solved using coordinate geometry. We can place the square on a 2D Cartesian plane, determine the equation of the circle, find the coordinates of the vertices of the triangle, and then calculate its area.
Step 2: Key Formula or Approach:
1. Equation of a circle with center (h, k) and radius r: \((x-h)^2 + (y-k)^2 = r^2\).
2. Area of a triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\): \( \text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \).
Alternatively, if one side of the triangle is parallel to an axis, the area is \( \frac{1}{2} \times \text{base} \times \text{height} \).
Step 3: Detailed Explanation:
Let's set up a coordinate system with the vertex B at the origin (0, 0).
Since ABCD is a square with side length 18 cm, the coordinates of the vertices are B(0, 0), A(0, 18), C(18, 0), and D(18, 18).
The sides are AB (on the line x=0), BC (on the line y=0), CD (on the line x=18), and DA (on the line y=18).
The circle touches sides AB (x=0) and BC (y=0) and has a radius \(r=13\) cm. Therefore, the center of the circle is (r, r) = (13, 13).
The equation of the circle is: \[ (x-13)^2 + (y-13)^2 = 13^2 = 169 \] The circle touches AB at M. For point M on AB, x=0. \[ (0-13)^2 + (y-13)^2 = 169 \implies 169 + (y-13)^2 = 169 \implies y=13 \] So, the coordinates of M are (0, 13).
The circle intersects side CD at P. For point P on CD, x=18. \[ (18-13)^2 + (y-13)^2 = 169 \implies 5^2 + (y-13)^2 = 169 \] \[ 25 + (y-13)^2 = 169 \implies (y-13)^2 = 144 \implies y-13 = \pm 12 \] So, \(y = 13+12 = 25\) or \(y = 13-12 = 1\). Since P lies on the side CD, its y-coordinate must be between 0 and 18. Thus, y=1.
The coordinates of P are (18, 1).
We need to find the area of triangle PMD. The vertices are P(18, 1), M(0, 13), and D(18, 18).
Let's consider the side PD as the base of the triangle. The points P and D have the same x-coordinate (18), so the base PD is a vertical line segment.
Length of the base PD = \( |18 - 1| = 17 \) cm.
The height of the triangle with respect to the base PD is the perpendicular distance from vertex M(0, 13) to the line x=18.
Height = \( |18 - 0| = 18 \) cm.
Area of triangle PMD = \( \frac{1}{2} \times \text{base} \times \text{height} \). \[ \text{Area} = \frac{1}{2} \times 17 \times 18 = 17 \times 9 = 153 \] Step 4: Final Answer:
The area of triangle PMD is 153 sq. cm.