Question - Area Calculation of Overlapping Squares
Solution:
Let the side of the larger square be \( x \) cm.
The area of the larger square is \( x^2 \) cm\(^2\).
Given the side of the smaller square is 10 cm, the area of the smaller square is \( 10^2 = 100 \) cm\(^2\).
The difference in area between the shaded part B and the unshaded part A is given as 24 cm\(^2\).
We know that the unshaded part A is a smaller square with an area of 100 cm\(^2\) minus the part overlapping with the larger square.
Let the side of the overlapped square be \( y \) cm.
The area of the shaded part B includes two rectangles and the overlapped square:
\( 2 \times x \times y + y^2 \).
The difference given is:
\( 2xy + y^2 - 100 = 24 \).
As the side of the larger square is \( x \) and the side of the overlapped square is \( y \), then \( x = 10 + y \).
Replace \( x \) with \( 10 + y \) in the difference equation:
\( 2(10 + y)y + y^2 - 100 = 24 \)
\( 20y + 2y^2 + y^2 - 100 = 24 \)
\( 3y^2 + 20y - 124 = 0 \)
Solving this quadratic equation by the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = 20 \), and \( c = -124 \):
\( y = \frac{-20 \pm \sqrt{400 + 1488}}{6} \)
\( y = \frac{-20 \pm \sqrt{1888}}{6} \)
\( y = \frac{-20 \pm 2\sqrt{472}}{6} \)
\( y = \frac{-10 \pm \sqrt{472}}{3} \)
Since \( y \) must be positive and it is the smaller dimension, we take the positive root:
\( y = \frac{-10 + \sqrt{472}}{3} \)
Now we find \( x \):
\( x = 10 + y \)
\( x = 10 + \frac{-10 + \sqrt{472}}{3} \)
The area of the unshaded part A is the side of the smaller square squared:
Area of A is \( 10^2 = 100 \) cm\(^2\).
The area of the shaded part B is the side of the larger square squared minus the area of A.
Using the values of \( x \) and \( y \) found above, we can calculate the exact area of B.
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com