Question

A square PQRS is enclosed in another square ABCD. Find the ratio of the area of PQRS to the area of ABCD.

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• A. 1/2
• B. 1/4
• C. 1/3
• D. 2/3
• E. 1/\(\sqrt{2}\)

Hint:

When a square is inscribed in another square such that its vertices touch the sides of the outer square, if the corner triangles are isosceles (as indicated by the 45-degree angle), the vertices of the inner square are at the midpoints of the sides of the outer square. The area of such an inscribed square is always half the area of the outer square.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to find the ratio of the area of an inner square to an outer square. The problem provides a diagram showing the orientation of the inner square. The key is to establish a relationship between the side lengths of the two squares.

Step 2: Key Formula or Approach:
- Area of a square = \( (\text{side})^2 \).
- Pythagorean theorem in a right-angled triangle: \( a^2 + b^2 = c^2 \).
Let the side length of the outer square ABCD be \(s\). The area of ABCD is \(s^2\).
The vertices of the inner square PQRS lie on the sides of the outer square ABCD. This creates four congruent right-angled triangles in the corners (e.g., \(\triangle PBQ\)).

Step 3: Detailed Explanation:
Let's analyze the corner triangle \(\triangle PBQ\). It is a right-angled triangle with the right angle at B. Let \(PB = x\) and \(BQ = y\). The side of the outer square is \(AB = s\). Since Q is on BC, we have \(BC = s\). Then \(QC = s-y\). Similarly, \(AP = s-x\).
Due to the symmetry of a square inscribed in another square, the four corner triangles (\(\triangle PBQ, \triangle QCR, \triangle RDS, \triangle SAP\)) are congruent.
Therefore, \(PB = QC = RD = SA = x\) and \(BQ = CR = DS = AP = y\).
This gives us \(s-y = x\), or \(x+y=s\).
The diagram has an angle marked as 45 degrees, which appears to be \(\angle BPQ\). In \(\triangle PBQ\), we have \(\angle PBQ = 90^{\circ}\). If \(\angle BPQ = 45^{\circ}\), then \(\triangle PBQ\) must be an isosceles right-angled triangle, meaning \(PB = BQ\).
So, \(x = y\).
Since \(x+y=s\) and \(x=y\), we have \(x+x=s \implies 2x=s \implies x = s/2\).
This means the vertices of the inner square are at the midpoints of the sides of the outer square.
Now we find the side length of the inner square, PQ, using the Pythagorean theorem in \(\triangle PBQ\):
\[ PQ^2 = PB^2 + BQ^2 \] \[ PQ^2 = (s/2)^2 + (s/2)^2 = \frac{s^2}{4} + \frac{s^2}{4} = \frac{2s^2}{4} = \frac{s^2}{2} \] The area of the inner square PQRS is equal to \(PQ^2\).
\[ \text{Area(PQRS)} = \frac{s^2}{2} \] The area of the outer square ABCD is \(s^2\).
The ratio is:
\[ \frac{\text{Area(PQRS)}}{\text{Area(ABCD)}} = \frac{s^2/2}{s^2} = \frac{1}{2} \]

Step 4: Final Answer:
The ratio of the area of square PQRS to the area of square ABCD is 1/2.